Top 45 String Programming Interview Questions in Java (with Tips)

Strings are a fundamental part of Java and a key focus in programming interviews. From basic text manipulation to complex algorithmic challenges, strings appear across all levels of coding assessments. This blog covers everything you need to know about strings in Java, including their definition, typical applications, and a comprehensive set of string programming interview questions in Java. Whether a fresher or a seasoned developer, you’ll find targeted questions for each experience level and preparation strategies to help you succeed in your next Java interview.

What is a String? 

In Java, a string is a sequence of characters in double quotes. It is an immutable object, meaning its value cannot be changed once created. Strings are widely used for text manipulation and are part of the ‘java.lang.String’ class, which provides various methods for operations like concatenation, comparison, and modification.

The key features of strings in Java:

• Immutable: Once a string is created, it cannot be modified.
• Stored in String Pool: It helps in memory optimization.
• Supports Various Methods: It supports various methods, such as ‘length()’, ‘toUpperCase()’, ‘toLowerCase()’, ‘substring()’, and more.

Applications of a String in Java

In Java, a string is an object designed to store and manipulate sequences of characters. It is fundamental to text processing, data management, and communication. Due to its flexibility, the string class is widely utilized across multiple applications and industries. Here is a list of applications where it is used: 

• Text Processing: Strings are fundamental for handling text-based operations, such as searching for words or patterns, replacing specific content, formatting text into uppercase or lowercase, and splitting text into smaller units like words or sentences.
• Data Storage and Retrieval: Strings are commonly used for storing and managing textual data across different formats, including file handling for reading and writing content, database interactions for storing structured data, and text representation within arrays, lists, and maps.
• Network Communication: Strings facilitate encoding and decoding data transmitted over networks. They are essential for HTTP requests and responses, ensuring proper formatting of web communication and data serialization, which converts complex data into a string format for transmission.
• User Interfaces: Interactive applications rely on strings to display text-based content, accept user input, and process commands or responses within software interfaces.
• Security: Strings play an integral role, particularly in password storage and comparison, ensuring secure authentication, and using encryption techniques to protect sensitive data.
• Bioinformatics: In genetic analysis, strings represent and analyze DNA sequences, aiding pattern recognition and computational biology research.
• Search Engines: Search engines utilize strings for indexing, keyword searches, and information retrieval, allowing efficient access to relevant data across the web.
• Compilers and Programming Languages: Strings are necessary for representing source code, parsing commands, and managing identifiers within programming environments.
• Game Development: Strings support text rendering, dialogue handling, user input processing, and storing game-related data, contributing to interactive gameplay experiences.

Java provides built-in tools for efficient string manipulation, such as:

• String: Represents immutable character sequences.
• StringBuilder and StringBuffer: Handle dynamic text modifications.
• Methods for concatenation, comparison, searching, and modification.

String Programming Interview Questions in Java

String manipulation is one of the most common topics in Java interviews, regardless of experience level. Whether you are a fresher brushing up on the basics, an intermediate coder refining problem-solving skills, or an experienced developer preparing for advanced algorithmic challenges, string-based questions are bound to come up. You can strengthen your programming concepts with the help of a core Java course. Here is a curated list of string programming interview questions in Java, categorized to help you prepare effectively based on your experience level.

I. String Programming Interview Questions in Java for Freshers

For freshers, string questions typically focus on the basics—such as string declaration, immutability, standard methods, and simple operations like concatenation or comparison. These foundational string programming interview questions in Java are designed to test your understanding of core string concepts and your ability to use them correctly in code. Here is a curated list of 15 string programming interview questions in Java for freshers: 

 Q1. What is a string in Java? How is it different from other data types?

Sample Answer: A string in Java is an immutable sequence of characters stored as objects. Unlike primitive data types like ‘int’ or ‘char’, a string is a reference type managed internally by Java’s ‘String’ class. Here’s an example of a string in Java:

String greeting = "Hello, World!";
System.out.println(greeting);

Q2. Why are strings immutable in Java?

Sample Answer: Strings are immutable in Java to improve performance, security, and reliability. Since Java strings are stored in a special memory pool called the string Pool, making them immutable helps prevent accidental modifications and optimizes memory use. Here is an example of immutable strings in Java:

public class StringImmutabilityDemo {
    public static void main(String[] args) {
        String original = "Hello";
        String modified = original.concat(" World");

        System.out.println("Original String: " + original); // Prints: Hello
        System.out.println("Modified String: " + modified); // Prints: Hello World
    }
}

Q3. How do you create a string in Java?

Sample Answer: A string can be created in two main ways:

1. Using String Literal: The string is stored in the String Pool for reuse.
2. Using the ‘new’ Keyword: It creates a new String object in the heap.

Here is an example of string creation in Java: 

public class StringCreationDemo {
    public static void main(String[] args) {
        String str1 = "Hello";                // String literal
        String str2 = new String("Hello");   // Using new keyword

        System.out.println(str1);             // Output: Hello
        System.out.println(str2);             // Output: Hello

        System.out.println(str1 == str2);    // false, different objects
        System.out.println(str1.equals(str2)); // true, same content
    }

Q4. What is the difference between ‘==’ and ‘.equals()’ when comparing strings?

Sample Answer: The difference between  ‘==’ and ‘.equals()’is the following: 

• ‘==’ operator: It compares whether two string references point to the same object in memory.
• .equals() method: It compares whether two strings have the same sequence of characters (i.e., same content).

Here’s an example:

public class StringComparisonDemo {
    public static void main(String[] args) {
        String s1 = "Java";
        String s2 = "Java";
        String s3 = new String("Java");

        System.out.println(s1 == s2);       // true: both point to the same string in String Pool
        System.out.println(s1.equals(s3));  // true: content is the same
        System.out.println(s1 == s3);       // false: different objects in memory
    }
}

Q5. How do you concatenate two strings?

Sample Answer: String concatenation can be done using the ‘+’ operator or the ‘.concat()’ method. Here is an example of how to concatenate two strings:

public class StringConcatenationDemo {
    public static void main(String[] args) {
        String first = "Hello";
        String second = " World";

        // Using '+' operator
        String result = first + second;
        System.out.println(result);

        // Using concat() method
        String result2 = first.concat(second);
        System.out.println(result2);
    }
}
Output:
Hello World
Hello World

Q6. How do you find the length of a string?

Sample Answer: To find the length of a string, use the ‘.length()’ method to get the number of characters in a String. Here is an example: 

public class StringLengthDemo {
    public static void main(String[] args) {
        String text = "InterviewPrep";
        System.out.println("Length: " + text.length());
    }
}

Output:
Length: 12

Q7. How do you convert a string to uppercase and lowercase?

Sample Answer: To convert a string to uppercase and lowercase, use the ‘.toUpperCase()’ and ‘.toLowerCase()’ methods. Here is an example of converting a string to uppercase and lowercase:

public class CaseConversionDemo {
    public static void main(String[] args) {
        String word = "Java";
        
        System.out.println(word.toUpperCase()); // Output: JAVA
        System.out.println(word.toLowerCase()); // Output: java
    }
}

String word = "Java";
System.out.println(word.toUpperCase()); // Output: JAVA
System.out.println(word.toLowerCase()); // Output: java

Q8. How do you extract a substring in Java?

Sample Answer: To extract a substring, use the .substring(startIndex, endIndex) method. The startIndex is inclusive, and the endIndex is exclusive. Here is an example of substring extraction in Java:  

public class SubstringDemo {
    public static void main(String[] args) {
        String sentence = "Programming in Java";
        String sub = sentence.substring(0, 11);  // Extracts "Programming"
        System.out.println(sub);
    }
}

Output:
Programming

Q9. How do you check if a string contains a specific word?

Sample Answer: To check if a string contains a specific word, use the ‘.contains()’ method. Here is an example of how to check if a spring contains a specific word:

public class StringContainsDemo {
    public static void main(String[] args) {
        String phrase = "Java is powerful";
        System.out.println(phrase.contains("powerful")); // Output: true
        System.out.println(phrase.contains("Python"));   // Output: false
    }
}

Q10. How do you replace characters in a string?

Sample Answer: To replace characters in a string, use the .replace() method. Here is an example of replaced characters in a string: 

public class StringReplaceDemo {
    public static void main(String[] args) {
        String original = "Java is fun";

        // Replace substring "fun" with "awesome"
        String updated = original.replace("fun", "awesome");
        System.out.println(updated);  // Output: Java is awesome

        // Replace character 'a' with 'o'
        String replacedChars = original.replace('a', 'o');
        System.out.println(replacedChars);  // Output: Jovo is fun
    }
}

Output:
Java is awesome
Jovo is fun

Pro Tip: Read the Java developer job description carefully before applying and preparing for the string programming interview questions in Java. The job description will help you prepare the list of concepts and programming you need to know for the interview.

Q11. How do you split a string in Java?

Sample Answer: To split a string in Java, use the ‘.split(delimiter)’ method, which splits the string around matches of the given regular expression. Here is an example of a split string in Java: 

public class SplitStringDemo {
    public static void main(String[] args) {
        String data = "apple,banana,grape";
        String[] fruits = data.split(",");

        for (String fruit : fruits) {
            System.out.println(fruit);
        }
    }
}

Output:
apple
banana
grape

Q12. How do you convert a string to an integer in Java?

Sample Answer: To convert a string to an integer in Java, use ‘Integer.parseInt()’.Here is an example of a string converted into an integer in Java: 

public class StringToIntegerDemo {
    public static void main(String[] args) {
        String number = "123";
        int num = Integer.parseInt(number);
        System.out.println(num + 10);  // Output: 133
    }
}

Q13.  How do you reverse a string in Java?

Sample Answer:  To reverse a string in Java, use ‘StringBuilder’ or loops. Here is an example of a reverse string in Java using StringBuilder:

public class ReverseStringDemo {
    public static void main(String[] args) {
        String original = "Java";
        String reversed = new StringBuilder(original).reverse().toString();
        System.out.println(reversed);  // Output: avaJ
    }
}

Q14. How do you check if a string is empty or null?

Sample Answer: To check if a string is either null or empty (“”), you should first check if it is not null, then check if it’s empty using .isEmpty(). Here is an example of how to check if a string is empty or null:

public class StringCheckDemo {
    public static void main(String[] args) {
        String text1 = "";
        String text2 = null;

        System.out.println(isNullOrEmpty(text1)); // Output: true
        System.out.println(isNullOrEmpty(text2)); // Output: true
        System.out.println(isNullOrEmpty("Hello")); // Output: false
    }

    static boolean isNullOrEmpty(String str) {
        return str == null || str.isEmpty();
    }
}

Q15. How do you remove white spaces from a string?

Sample Answer: To remove white spaces from a string, use the ‘.trim()’ method. It removes leading and trailing whitespaces, not all whitespaces inside the string. If you want to remove all white spaces (including those inside the string), you should use .replaceAll(“\\s+”, “”). Here is an example of how you can remove white spaces from a string: 

public class RemoveWhitespaceDemo {
    public static void main(String[] args) {
        String spaced = "   Java Programming   ";
        
        // Remove leading and trailing spaces
        System.out.println(spaced.trim());  // Output: "Java Programming"
        
        // Remove all spaces
        System.out.println(spaced.replaceAll("\\s+", ""));  // Output: "JavaProgramming"
    }
}

II. Coding Questions on String in Java for Intermediate Candidates

At the intermediate level, string questions become more logic-driven. Candidates must demonstrate practical coding skills and solve problems like removing duplicates, checking palindromes, or counting character frequency. These questions help evaluate your problem-solving approach and command over Java string manipulation. Here are some coding questions on strings in Java for the intermediate candidates: 

Q16. How do you determine if a string reads the same forwards and backwards, ignoring case and non-alphanumeric characters?

Sample Answer:  To determine if a string reads the same forwards and backwards, ignoring case and non-alphanumeric characters, we need to clean the string by removing non-alphanumeric characters, convert it to lowercase, and compare it with its reverse.

Here is how you can decide  if a string reads the same forwards and backwards, ignoring case and non-alphanumeric characters:

boolean isPalindrome(String text) {
    String clean = text.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();
    return clean.equals(new StringBuilder(clean).reverse().toString());
}

// Example usage:
System.out.println(isPalindrome("A man, a plan, a canal, Panama")); // true
System.out.println(isPalindrome("Racecar")); // true
System.out.println(isPalindrome("Hello")); // false

Q17. How do you check if two strings contain the same characters in a different order, ignoring case and non-alphanumeric characters?

Sample Answer: To check if two strings are anagrams, we need to clean both strings by removing non-alphanumeric characters, convert them to lowercase, sort their characters, and compare the sorted results. Here is an example of how to check if two strings contain the same characters in a different order, ignoring case and non-alphanumeric characters: 

boolean areAnagrams(String str1, String str2) {
    String cleanStr1 = str1.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();
    String cleanStr2 = str2.replaceAll("[^a-zA-Z0-9]", "").toLowerCase();
    if (cleanStr1.length() != cleanStr2.length()) return false;
    
    char[] arr1 = cleanStr1.toCharArray();
    char[] arr2 = cleanStr2.toCharArray();
    java.util.Arrays.sort(arr1);
    java.util.Arrays.sort(arr2);
    
    return java.util.Arrays.equals(arr1, arr2);
}

// Example usage:
System.out.println(areAnagrams("listen", "silent")); // true
System.out.println(areAnagrams("hello", "world")); // false

Q18. How do you reverse the order of words in a string while handling extra spaces?

Sample Answer: To reverse the order of words in a string while handling extra spaces, we need to split the text into words, remove unnecessary spaces, and reconstruct the sentence in reverse order. Here is how you can reverse the order of words in a string while handling extra spaces:

String reverseWords(String text) {
    String[] words = text.trim().split("\\s+");
    StringBuilder reversed = new StringBuilder();

    for (int i = words.length - 1; i >= 0; i--) {
        reversed.append(words[i]).append(" ");
    }

    return reversed.toString().trim();
}

// Example usage:
System.out.println(reverseWords("  Hello   World! ")); // "World! Hello"
System.out.println(reverseWords("Java is powerful")); // "powerful is Java"

 Q19. How do you find the longest common prefix among an array of strings?

Sample Answer: To find the longest common prefix among an array of strings, we compare characters across all strings and shrink the prefix until a match is found. Here is an example of how to find the longest common prefix among an array of strings:

String longestCommonPrefix(String[] strs) {
    if (strs.length == 0) return "";
    String prefix = strs[0];

    for (int i = 1; i < strs.length; i++) {
        while (strs[i].indexOf(prefix) != 0) {
            prefix = prefix.substring(0, prefix.length() - 1);
            if (prefix.isEmpty()) return "";
        }
    }

    return prefix;
}

// Example usage:
String[] words1 = {"flower", "flow", "flight"};
String[] words2 = {"dog", "racecar", "car"};

System.out.println(longestCommonPrefix(words1)); // "fl"
System.out.println(longestCommonPrefix(words2)); // ""

Q20. How do you compress a string by replacing consecutive repeating characters with the character followed by its count?

Sample Answer: To compress a string by replacing consecutive repeating characters with the character followed by its count, we iterate through the string, count occurrences, and construct a compressed version. If the compressed string is not shorter, we return the original. Here is an example of a compressed string by replacing consecutive repeating characters with the character followed by its count:

String compressString(String text) {
    if (text.isEmpty()) return text;

    StringBuilder compressed = new StringBuilder();
    int count = 1;

    for (int i = 0; i < text.length(); i++) {
        if (i + 1 < text.length() && text.charAt(i) == text.charAt(i + 1)) {
            count++;
        } else {
            compressed.append(text.charAt(i));
            if (count > 1) {
                compressed.append(count);
            }
            count = 1;
        }
    }

    return compressed.length() < text.length() ? compressed.toString() : text;
}

// Example usage:
System.out.println(compressString("aaabbcddd")); // "a3b2c1d3"
System.out.println(compressString("abcdef")); // "abcdef"

Pro Tip: Are you preparing for coding questions on strings in Java and drafting your applications? Explore our guide on writing a Java developer cover letter, which includes a sample to draft your unique document based on the job description.

Q21. How do you remove duplicate characters from a string while maintaining the original order?

Sample Answer: To remove duplicate characters from a string while maintaining the original order, we have to use a ‘LinkedHashSet’ to store unique characters while preserving the insertion order. Here is how to remove duplicate characters from a string while maintaining the original order: 

String removeDuplicates(String text) {
    StringBuilder result = new StringBuilder();
    java.util.Set<Character> seen = new java.util.LinkedHashSet<>();

    for (char c : text.toCharArray()) {
        if (seen.add(c)) {
            result.append(c);
        }
    }

    return result.toString();
}

// Example usage:
System.out.println(removeDuplicates("programming")); // "progamin"
System.out.println(removeDuplicates("hello world")); // "helo wrd"

Q22. How do you count the frequency of each character in a string?

Sample Answer: To count the frequency of each character in a string, we iterate through the string and store character counts in a ‘HashMap’. Here is an example of how you can count the frequency of each character in a string:

java.util.Map<Character, Integer> countCharacterFrequency(String text) {
    java.util.Map<Character, Integer> frequency = new java.util.HashMap<>();

    for (char c : text.toCharArray()) {
        frequency.put(c, frequency.getOrDefault(c, 0) + 1);
    }

    return frequency;
}

// Example usage:
System.out.println(countCharacterFrequency("banana")); // {b=1, a=3, n=2}
System.out.println(countCharacterFrequency("hello")); // {h=1, e=1, l=2, o=1}

Q23. How do you determine if a string is a palindrome using recursion?

Sample Answer: To determine if a string is a palindrome using recursion, we compare the first and last characters while recursively checking the inner substring.  Here is an example of how to determine if a string is a palindrome using recursion: 

boolean isPalindromeRecursive(String text, int left, int right) {
    if (left >= right) return true;
    if (text.charAt(left) != text.charAt(right)) return false;
    return isPalindromeRecursive(text, left + 1, right - 1);
}

// Example usage:
String word = "madam";
System.out.println(isPalindromeRecursive(word, 0, word.length() - 1)); // true

word = "hello";
System.out.println(isPalindromeRecursive(word, 0, word.length() - 1)); // false

Q24. How do you reverse a string using recursion instead of iteration?

Sample Answer: To reverse a string using recursion, we extract the last character and recursively process the rest. Here is an example of a reversed string using recursion instead of iteration:

String reverseStringRecursive(String text) {
    if (text.isEmpty()) return text;
    return reverseStringRecursive(text.substring(1)) + text.charAt(0);
}

// Example usage:
System.out.println(reverseStringRecursive("hello")); // "olleh"
System.out.println(reverseStringRecursive("Java")); // "avaJ"

Q25. How do you find the first non-repeating character in a string?

Sample Answer: To find the first non-repeating character in a string, we use a ‘LinkedHashMap’ to store character counts while preserving order. Here is an example of a first non-repeating character in a string: 

Character firstNonRepeatingCharacter(String text) {
    java.util.Map<Character, Integer> frequency = new java.util.LinkedHashMap<>();

    for (char c : text.toCharArray()) {
        frequency.put(c, frequency.getOrDefault(c, 0) + 1);
    }

    for (char c : frequency.keySet()) {
        if (frequency.get(c) == 1) return c;
    }

    return null; // No unique character found
}

// Example usage:
System.out.println(firstNonRepeatingCharacter("swiss")); // 'w'
System.out.println(firstNonRepeatingCharacter("aabbcc")); // null

Q26. How do you check if one string is a rotation of another?

Sample Answer: To check if one string is a rotation of another, we concatenate the original string with itself and check if the second string exists within it. Here is how you can check if one string is a rotation of another:

boolean areRotations(String str1, String str2) {
    return (str1.length() == str2.length()) && (str1 + str1).contains(str2);
}

// Example usage:
System.out.println(areRotations("waterbottle", "erbottlewat")); // true
System.out.println(areRotations("hello", "elloh")); // false

Q27. How do you find the most frequent word in a sentence?

Sample Answer: We split the sentence into words and count occurrences using a ‘HashMap’ to find the most frequent word. Here is how you can find the most frequent word in a sentence: 

String mostFrequentWord(String sentence) {
    String[] words = sentence.toLowerCase().split("\\s+");
    java.util.Map<String, Integer> frequency = new java.util.HashMap<>();

    for (String word : words) {
        frequency.put(word, frequency.getOrDefault(word, 0) + 1);
    }

    return java.util.Collections.max(frequency.entrySet(), java.util.Map.Entry.comparingByValue()).getKey();
}

// Example usage:
System.out.println(mostFrequentWord("hello world hello Java")); // "hello"
System.out.println(mostFrequentWord("apple banana apple orange banana apple")); // "apple"

Q28. How do you convert a string so that each word starts with an uppercase letter?

Sample Answer: To convert a string to title case, we split the words, capitalize the first letter of each, and reconstruct the string. Here is how you can convert a string so that each word starts with an uppercase letter:

String toTitleCase(String text) {
    String[] words = text.toLowerCase().split("\\s+");
    StringBuilder titleCase = new StringBuilder();

    for (String word : words) {
        titleCase.append(Character.toUpperCase(word.charAt(0))).append(word.substring(1)).append(" ");
    }

    return titleCase.toString().trim();
}

// Example usage:
System.out.println(toTitleCase("java is fun")); // "Java Is Fun"
System.out.println(toTitleCase("hello world")); // "Hello World"

Q29. How do you find the longest word in a given string?

Sample Answer: We split the sentence into words and compare lengths to find the longest word in a string. Here is how you can find the longest word in a given string: 

String longestWord(String sentence) {
    String[] words = sentence.split("\\s+");
    return java.util.Arrays.stream(words).max(java.util.Comparator.comparingInt(String::length)).orElse("");
}

// Example usage:
System.out.println(longestWord("Java is a powerful programming language")); // "programming"
System.out.println(longestWord("Hello world")); // "Hello"

Q30. How do you generate all possible substrings of a given string?

Sample Answer: To generate all possible string substrings, we use nested loops to extract substrings of varying lengths starting from each character position. Here is how you can generate all possible substrings of a given string: 

void findSubstrings(String text) {
    for (int i = 0; i < text.length(); i++) {
        for (int j = i + 1; j <= text.length(); j++) {
            System.out.println(text.substring(i, j));
        }
    }
}

// Example usage:
findSubstrings("abc");
/*
Output:
a
ab
abc
b
bc
c
*/

III. String Programming Interview Questions in Java for Experienced 

For experienced professionals, interviewers often dive deep into advanced string algorithms. These include problems involving substring search, regular expressions, efficient parsing, or even using trie data structures. These string programming interview questions in Java test your conceptual knowledge, optimization, and debugging capabilities. Here is a curated list of 15 Java string program questions for interview for experienced candidates:

Q31. How do you efficiently search for a substring in a larger string using the KMP algorithm?

Sample Answer: The KMP algorithm avoids unnecessary comparisons by preprocessing the pattern (substring to search) using a Longest Prefix Suffix (LPS) array. This array helps determine how many characters can be skipped when a mismatch occurs during the search. You can efficiently search for a substring in a larger string using the KMP algorithm. Here is how you can efficiently search for a substring in a larger string using the KMP algorithm: 

void computeLPSArray(String pattern, int[] lps) {
    int length = 0, i = 1;
    lps[0] = 0;

    while (i < pattern.length()) {
        if (pattern.charAt(i) == pattern.charAt(length)) {
            length++;
            lps[i] = length;
            i++;
        } else {
            if (length != 0) {
                length = lps[length - 1];
            } else {
                lps[i] = 0;
                i++;
            }
        }
    }
}

void KMPSearch(String text, String pattern) {
    int[] lps = new int[pattern.length()];
    computeLPSArray(pattern, lps);
    int i = 0, j = 0;

    while (i < text.length()) {
        if (pattern.charAt(j) == text.charAt(i)) {
            i++;
            j++;
        }
        if (j == pattern.length()) {
            System.out.println("Pattern found at index " + (i - j));
            j = lps[j - 1];
        } else if (i < text.length() && pattern.charAt(j) != text.charAt(i)) {
            j = (j != 0) ? lps[j - 1] : 0;
            i++;
        }
    }
}

// Example usage:
KMPSearch("ababcababaad", "ababa"); // Output: Pattern found at index 2

Q32. How do you find the longest palindromic substring in a given string efficiently?

Sample Answer: To find the longest palindromic substring in a given string efficiently, we can use the expand-around-center technique to find palindromic substrings efficiently. Here is how you can find the longest palindromic substring in a given string efficiently:

String longestPalindromicSubstring(String text) {
    if (text == null || text.length() < 1) return "";
    int start = 0, end = 0;

    for (int i = 0; i < text.length(); i++) {
        int len1 = expandAroundCenter(text, i, i);
        int len2 = expandAroundCenter(text, i, i + 1);
        int maxLen = Math.max(len1, len2);

        if (maxLen > end - start) {
            start = i - (maxLen - 1) / 2;
            end = i + maxLen / 2;
        }
    }

    return text.substring(start, end + 1);
}

int expandAroundCenter(String text, int left, int right) {
    while (left >= 0 && right < text.length() && text.charAt(left) == text.charAt(right)) {
        left--;
        right++;
    }
    Return right - left - 1;
}

// Example usage:
System.out.println(longestPalindromicSubstring("babad")); // "bab" or "aba"
System.out.println(longestPalindromicSubstring("cbbd")); // "bb"

Q33. How do you store and search for prefixes using a Trie data structure?

Sample Answer: A Trie allows efficient prefix-based searching, insertion, and auto-completion functionalities. Here is how we can use a Trie data structure to store and search for prefixes efficiently:

import java.util.HashMap;
import java.util.Map;

class TrieNode {
    Map<Character, TrieNode> children = new HashMap<>();
    boolean isEndOfWord = false;
}

class Trie {
    private final TrieNode root = new TrieNode();

    // Insert a word into the Trie
    void insert(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            node = node.children.computeIfAbsent(c, k -> new TrieNode());
        }
        node.isEndOfWord = true;
    }

    // Search for a complete word in the Trie
    boolean search(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            node = node.children.get(c);
            if (node == null) return false;
        }
        return node.isEndOfWord;
    }

    // Check if any word in the Trie starts with the given prefix
    boolean startsWith(String prefix) {
        TrieNode node = root;
        for (char c : prefix.toCharArray()) {
            node = node.children.get(c);
            if (node == null) return false;
        }
        return true;
    }
}

// Example usage:
public class TrieExample {
    public static void main(String[] args) {
        Trie trie = new Trie();
        trie.insert("apple");
        trie.insert("app");
        
        System.out.println(trie.search("apple"));   // true
        System.out.println(trie.search("app"));     // true
        System.out.println(trie.search("appl"));    // false
        
        System.out.println(trie.startsWith("app")); // true
        System.out.println(trie.startsWith("apl")); // false
    }
}

Q34. How do you find the most frequently occurring substring of length k in a given string?

Sample Answer: To find the most frequently occurring substring of length k in a given string, we need to use a HashMap to count occurrences of substrings of size k. Here is how you can find the most frequently occurring substring of length k in a given string: 

import java.util.HashMap;
import java.util.Map;

public class FrequentSubstring {

    public static String mostFrequentSubstring(String text, int k) {
        Map<String, Integer> frequencyMap = new HashMap<>();
        String maxSubstring = "";
        int maxCount = 0;

        for (int i = 0; i <= text.length() - k; i++) {
            String sub = text.substring(i, i + k);
            frequencyMap.put(sub, frequencyMap.getOrDefault(sub, 0) + 1);

            if (frequencyMap.get(sub) > maxCount) {
                maxCount = frequencyMap.get(sub);
                maxSubstring = sub;
            }
        }

        return maxSubstring;
    }

    public static void main(String[] args) {
        System.out.println(mostFrequentSubstring("abcabcabc", 3)); // Output: abc
        System.out.println(mostFrequentSubstring("banana", 2));    // Output: an
    }
}

Q35. How do you validate an email address using Java regular expressions?

Sample Answer: To validate an email address using Java regular expressions, we need to use Java’s ‘Pattern’ class with a regex pattern to validate email addresses. Here is how you can validate an email address using Java regular expressions:

boolean isValidEmail(String email) {
    // Regex requires a valid domain with at least one dot and 2-6 letter extension
    String regex = "^[A-Za-z0-9+_.-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,6}$";
    return java.util.regex.Pattern.matches(regex, email);
}

// Example usage:
System.out.println(isValidEmail("test@example.com"));   // true
System.out.println(isValidEmail("invalid-email@com"));  // false
System.out.println(isValidEmail("user@domain.co.in"));  // true
System.out.println(isValidEmail("user@domain."));       // false

Pro Tip: Explore more concepts to enhance your string programming in Java interview questions preparation. Check out our blog on Java coding interview questions and answers. 

Q36. How do you generate all possible permutations of a given string?

Sample Answer: We use recursion and backtracking to generate all possible permutations of a given string. Here is how you can generate all possible permutations of a given string:

public class StringPermutations {

    // Recursive method to generate permutations
    public static void generatePermutations(String str, String ans) {
        if (str.isEmpty()) {
            System.out.println(ans);
            return;
        }

        for (int i = 0; i < str.length(); i++) {
            // Choose the character at index i and recurse for remaining characters
            generatePermutations(str.substring(0, i) + str.substring(i + 1), ans + str.charAt(i));
        }
    }

    public static void main(String[] args) {
        generatePermutations("abc", "");
    }
}


/*
Output:
abc
acb
bac
bca
cab
cba
*/

Q37. How do you find the smallest substring that contains all characters of a given pattern?

Sample Answer: To find the smallest substring that contains all characters of a given pattern, we need to use a sliding window approach for efficient substring searching. Here is how you can  find the smallest substring that contains all characters of a given pattern: 

import java.util.HashMap;
import java.util.Map;

public class MinWindowSubstring {

    public static String minWindowSubstring(String text, String pattern) {
        Map<Character, Integer> map = new HashMap<>();
        for (char c : pattern.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }

        int left = 0, right = 0;
        int minStart = 0, minLen = Integer.MAX_VALUE;
        int count = pattern.length();

        while (right < text.length()) {
            char c = text.charAt(right);
            if (map.containsKey(c)) {
                if (map.get(c) > 0) count--;
                map.put(c, map.get(c) - 1);
            }
            right++;

            while (count == 0) {
                if (right - left < minLen) {
                    minLen = right - left;
                    minStart = left;
                }
                char leftChar = text.charAt(left);
                if (map.containsKey(leftChar)) {
                    map.put(leftChar, map.get(leftChar) + 1);
                    if (map.get(leftChar) > 0) count++;
                }
                left++;
            }
        }

        return minLen == Integer.MAX_VALUE ? "" : text.substring(minStart, minStart + minLen);
    }

    public static void main(String[] args) {
        System.out.println(minWindowSubstring("ADOBECODEBANCD", "A

Q38. How do you find the longest repeating substring in a given string?

Sample Answer: We must use suffix arrays or dynamic programming for efficient pattern detection to find the longest repeating substring in a given string. Here is how you can find the longest repeating substring in a given string:

public class LongestRepeatingSubstring {

    public static String longestRepeatingSubstring(String text) {
        int n = text.length();
        int[][] dp = new int[n + 1][n + 1];
        int length = 0, index = 0;

        for (int i = 1; i <= n; i++) {
            for (int j = i + 1; j <= n; j++) {
                if (text.charAt(i - 1) == text.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    if (dp[i][j] > length) {
                        length = dp[i][j];
                        index = i;
                    }
                } else {
                    dp[i][j] = 0;
                }
            }
        }

        return text.substring(index - length, index);
    }

    public static void main(String[] args) {
        System.out.println(longestRepeatingSubstring("banana"));   // Output: ana
        System.out.println(longestRepeatingSubstring("abcdefg"));  // Output: (empty string)
        System.out.println(longestRepeatingSubstring("abcabcabc")); // Output: abc
    }
}

Q39. How do you implement regular expression pattern matching in Java?

Sample Answer: To implement regular expression pattern matching in Java, we must use recursion and dynamic programming to match ‘.’ and ‘* ‘patterns effectively. Here is how you can  implement regular expression pattern matching in Java:

public class RegexPatternMatching {

    // Recursive regex pattern matching method
    public static boolean isMatch(String text, String pattern) {
        if (pattern.isEmpty()) return text.isEmpty();

        boolean firstMatch = (!text.isEmpty() &&
                             (text.charAt(0) == pattern.charAt(0) || pattern.charAt(0) == '.'));

        if (pattern.length() >= 2 && pattern.charAt(1) == '*') {
            // Case 1: skip "x*" in the pattern
            // Case 2: consume a matching character and use "*" again
            return isMatch(text, pattern.substring(2)) ||
                   (firstMatch && isMatch(text.substring(1), pattern));
        } else {
            return firstMatch && isMatch(text.substring(1), pattern.substring(1));
        }
    }

    public static void main(String[] args) {
        System.out.println(isMatch("aa", "a*"));              // true
        System.out.println(isMatch("mississippi", "mis*is*p*.")); // false
        System.out.println(isMatch("ab", ".*"));              // true
        System.out.println(isMatch("aab", "c*a*b"));          // true
        System.out.println(isMatch("aaa", "a*a"));            // true
    }
}

Q40. How do you check if one string is an interleaving of two other strings?

Sample Answer: To check if one string is an interleaving of two other strings, we need to use dynamic programming to check for interleaving sequences efficiently. Here is how you can check if one string is an interleaving of two other strings:

public class InterleavingChecker {

    public static boolean isInterleaving(String s1, String s2, String s3) {
        if (s1.length() + s2.length() != s3.length()) return false;
        boolean[][] dp = new boolean[s1.length() + 1][s2.length() + 1];

        for (int i = 0; i <= s1.length(); i++) {
            for (int j = 0; j <= s2.length(); j++) {
                if (i == 0 && j == 0) {
                    dp[i][j] = true;
                } else if (i == 0) {
                    dp[i][j] = dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
                } else if (j == 0) {
                    dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
                } else {
                    dp[i][j] = (dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) ||
                               (dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));
                }
            }
        }

        return dp[s1.length()][s2.length()];
    }

    public static void main(String[] args) {
        System.out.println(isInterleaving("abc", "def", "adbcef"));  // true
        System.out.println(isInterleaving("abc", "def", "abcdef"));  // true
        System.out.println(isInterleaving("abc", "def", "abdecf"));  // false
    }
}

Q41. How do you find all palindromic substrings in a given string?

Sample Answer: We use the expand-around-center approach to find all palindromic substrings efficiently. Here is how you can find all palindromic substrings in a given string: 

void findPalindromicSubstrings(String text) {
    for (int i = 0; i < text.length(); i++) {
        expandAndPrint(text, i, i); // Odd-length palindromes
        expandAndPrint(text, i, i + 1); // Even-length palindromes
    }
}

void expandAndPrint(String text, int left, int right) {
    while (left >= 0 && right < text.length() && text.charAt(left) == text.charAt(right)) {
        System.out.println(text.substring(left, right + 1));
        left--;
        right++;
    }
}

// Example usage:
findPalindromicSubstrings("abba");
/*
Output:
a
b
bb
abba
b
a
*/

Q42. How do you implement a custom string tokenizer that splits a string based on multiple delimiters?

Sample Answer: We use regular expressions with ‘split()’ to tokenize a string with multiple delimiters. Here is how you can implement a custom string tokenizer that splits a string based on multiple delimiters: 

public class CustomStringTokenizer {

    // Method to split a string based on multiple delimiters
    public static String[] customTokenizer(String text, String delimiters) {
        return text.split("[" + delimiters + "]+");
    }

    // Example usage
    public static void main(String[] args) {
        String text = "Hello, World! Welcome-to Java.";
        String delimiters = " ,!-";

        String[] tokens = customTokenizer(text, delimiters);

        for (String token : tokens) {
            System.out.println(token);
        }
    }
}


Output:
Hello
World
Welcome
to
Java

Q43. How do you check if a string contains properly balanced parentheses?

Sample Answer: We use a stack to track opening and closing brackets to validate balanced parentheses. Here is how you can check if a string contains properly balanced parentheses:

import java.util.Stack;

public class ParenthesesChecker {

    // Method to check if the parentheses are balanced
    public static boolean areParenthesesBalanced(String text) {
        Stack<Character> stack = new Stack<>();
        for (char c : text.toCharArray()) {
            if (c == '(' || c == '{' || c == '[') {
                stack.push(c);
            } else if (c == ')') {
                if (stack.isEmpty() || stack.pop() != '(') return false;
            } else if (c == '}') {
                if (stack.isEmpty() || stack.pop() != '{') return false;
            } else if (c == ']') {
                if (stack.isEmpty() || stack.pop() != '[') return false;
            }
        }
        return stack.isEmpty();
    }

    // Example usage
    public static void main(String[] args) {
        String[] testCases = {
            "{[()]}",    // true
            "{[(])}",    // false
            "({[]})",    // true
            "(()",       // false
            "",          // true
            "[{()}]",    // true
            "[(])",      // false
            "({[)]}"     // false
        };

        for (String test : testCases) {
            System.out.println("Input: " + test + " -> Balanced: " + areParenthesesBalanced(test));
        }
    }
}

Output:
Input: {[()]} -> Balanced: true  
Input: {[(])} -> Balanced: false  
Input: ({[]}) -> Balanced: true  
Input: (() -> Balanced: false  
Input:  -> Balanced: true  
Input: [{()}] -> Balanced: true  
Input: [(]) -> Balanced: false  
Input: ({[)]} -> Balanced: false  

Q44. How do you find the most extended sequence of connected words where each word starts with the last letter of the previous word

Sample Answer: To find the longest sequence of connected words where each word starts with the last letter of the previous word, we need to sort the words and build the longest possible sequence using backtracking. Here is how you can find the most extended sequence of connected words where each word starts with the last letter of the previous word:

import java.util.*;

public class WordChainFinder {

    private static List<String> longestChain = new ArrayList<>();

    public static List<String> longestWordChain(String[] words) {
        // Build a map where each key is a starting character and the value is a list of words starting with it
        Map<Character, List<String>> wordMap = new HashMap<>();
        for (String word : words) {
            char key = word.charAt(0);
            wordMap.computeIfAbsent(key, k -> new ArrayList<>()).add(word);
        }

        longestChain.clear();
        for (String word : words) {
            List<String> currentChain = new ArrayList<>();
            Set<String> visited = new HashSet<>();
            buildChain(word, currentChain, visited, wordMap);
        }

        return longestChain;
    }

    private static void buildChain(String word, List<String> currentChain, Set<String> visited,
                                   Map<Character, List<String>> wordMap) {
        currentChain.add(word);
        visited.add(word);

        if (currentChain.size() > longestChain.size()) {
            longestChain = new ArrayList<>(currentChain);
        }

        char lastChar = word.charAt(word.length() - 1);
        if (wordMap.containsKey(lastChar)) {
            for (String nextWord : wordMap.get(lastChar)) {
                if (!visited.contains(nextWord)) {
                    buildChain(nextWord, currentChain, visited, wordMap);
                }
            }
        }

        // Backtrack
        currentChain.remove(currentChain.size() - 1);
        visited.remove(word);
    }

    public static void main(String[] args) {
        String[] words = {"apple", "elephant", "tiger", "rabbit", "tree", "egg"};
        List<String> result = longestWordChain(words);
        System.out.println(result);
    }
}


// Example usage:
String[] words = {"apple", "elephant", "tiger", "rabbit", "tree", "egg"};
System.out.println(longestWordChain(words));
/*
Output: [elephant, tiger, rabbit, tree, egg]
*/

Q45. How do you convert a Roman numeral string into its integer representation?

Sample Answer: To convert a Roman numeral string into its integer representation, we need to iterate through the string and add/subtract values based on Roman numeral rules. Here is how you can convert a Roman numeral string into its integer representation: 

import java.util.Map;

public class RomanConverter {
    
    // Method to convert Roman numeral to integer
    public static int romanToInteger(String s) {
        Map<Character, Integer> map = Map.of(
            'I', 1, 'V', 5, 'X', 10, 'L', 50,
            'C', 100, 'D', 500, 'M', 1000
        );

        int result = 0, prev = 0;
        for (int i = s.length() - 1; i >= 0; i--) {
            int value = map.get(s.charAt(i));
            if (value < prev) {
                result -= value;
            } else {
                result += value;
            }
            prev = value;
        }
        return result;
    }

    // Example usage
    public static void main(String[] args) {
        String[] examples = { "III", "IV", "IX", "LVIII", "MCMXCIV" };

        for (String roman : examples) {
            int number = romanToInteger(roman);
            System.out.println("Roman: " + roman + " => Integer: " + number);
        }
    }
}

How to Prepare for String Programming Interview Questions in Java?

Preparing for string programming interview questions in Java requires a strong understanding of string manipulation, optimization techniques, and problem-solving strategies. Since strings are widely used in coding challenges, developing expertise in handling them efficiently is crucial for technical interviews. Here are key steps to help you prepare effectively:

• Master String Fundamentals: Understand immutable and mutable strings, string pool concepts, and commonly used methods like substring(), replace(), split(), and concat().
• Practice Common Coding Problems: Solve string-based problems related to reversing, anagram checking, substring searching, and pattern matching to improve problem-solving skills.
• Learn Time Complexity Optimization: To optimize performance, focus on efficient algorithms for string manipulation, such as Sliding Window, Hashing, and Dynamic Programming.
• Explore Real-World Applications: Gain knowledge about how strings are used in database queries, network communication, encryption, and text processing in Java applications.
• Attempt Mock Interviews: Practice coding challenges and interview questions on platforms like Internshala blogs to build confidence. You can also take practice courses, like how to ace coding interviews, by Internshala, to boost your preparation.
• Review Frequently Asked Questions: Study commonly asked string programming questions in Java for freshers, intermediate candidates, and experienced professionals.

Conclusion

Mastering string programming interview questions in Java is essential for developers at all levels, whether you are a fresher, an intermediate coder, or an experienced professional. Understanding what a string is, its applications, and how to prepare for interview questions effectively can significantly boost your confidence and performance.

By familiarizing yourself with core string concepts, practicing common coding problems, and applying structured problem-solving techniques, you can improve your chances of excelling in technical interviews. Keep refining your knowledge, explore different approaches to string manipulation, and stay updated with best practices to strengthen your programming skills.

Want to ace your next interview? Start with the Capgemini Java developer questions to master string concepts.

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