Top 25 String Interview Questions and Answers (2026)

1. Introduction to String Interview Questions

Strings are one of the most universally tested topics in software engineering interviews. Unlike arrays, strings are immutable in many languages (like Python, Java, and JavaScript), requiring careful memory awareness.

Below are 25 essential string interview questions with complete solutions in C++, Java, Python, and JavaScript.

2. Easy String Questions

Q1. Reverse a String In-Place

Question: Given a character array s, reverse it in-place.

cpp
void reverseString(vector<char>& s) {
    int left = 0, right = s.size() - 1;
    while (left < right) swap(s[left++], s[right--]);
}

java
public void reverseString(char[] s) {
    int left = 0, right = s.length - 1;
    while (left < right) {
        char temp = s[left];
        s[left++] = s[right];
        s[right--] = temp;
    }
}

python
def reverseString(s: list) -> None:
    left, right = 0, len(s) - 1
    while left < right:
        s[left], s[right] = s[right], s[left]
        left += 1; right -= 1

javascript
function reverseString(s) {
    let left = 0, right = s.length - 1;
    while (left < right) {
        let temp = s[left];
        s[left++] = s[right];
        s[right--] = temp;
    }
}

Time Complexity: O(n) | Space Complexity: O(1)

Q2. Valid Palindrome

Question: Return true if a string is a palindrome, considering only alphanumeric characters and ignoring cases.

cpp
bool isPalindrome(string s) {
    int left = 0, right = s.size() - 1;
    while (left < right) {
        while (left < right && !isalnum(s[left])) left++;
        while (left < right && !isalnum(s[right])) right--;
        if (tolower(s[left]) != tolower(s[right])) return false;
        left++; right--;
    }
    return true;
}

java
public boolean isPalindrome(String s) {
    int left = 0, right = s.length() - 1;
    while (left < right) {
        while (left < right && !Character.isLetterOrDigit(s.charAt(left))) left++;
        while (left < right && !Character.isLetterOrDigit(s.charAt(right))) right--;
        if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) return false;
        left++; right--;
    }
    return true;
}

python
def isPalindrome(s: str) -> bool:
    left, right = 0, len(s) - 1
    while left < right:
        while left < right and not s[left].isalnum(): left += 1
        while left < right and not s[right].isalnum(): right -= 1
        if s[left].lower() != s[right].lower(): return False
        left += 1; right -= 1
    return True

javascript
function isPalindrome(s) {
    let left = 0, right = s.length - 1;
    while (left < right) {
        while (left < right && !/[a-zA-Z0-9]/.test(s[left])) left++;
        while (left < right && !/[a-zA-Z0-9]/.test(s[right])) right--;
        if (s[left].toLowerCase() !== s[right].toLowerCase()) return false;
        left++; right--;
    }
    return true;
}

Time Complexity: O(n) | Space Complexity: O(1)

Q3. Valid Anagram

Question: Return true if t is an anagram of s.

cpp
bool isAnagram(string s, string t) {
    if (s.size() != t.size()) return false;
    int count[26] = {0};
    for (char c : s) count[c - 'a']++;
    for (char c : t) if (--count[c - 'a'] < 0) return false;
    return true;
}

java
public boolean isAnagram(String s, String t) {
    if (s.length() != t.length()) return false;
    int[] count = new int[26];
    for (char c : s.toCharArray()) count[c - 'a']++;
    for (char c : t.toCharArray()) if (--count[c - 'a'] < 0) return false;
    return true;
}

python
from collections import Counter
def isAnagram(s: str, t: str) -> bool:
    return Counter(s) == Counter(t)

javascript
function isAnagram(s, t) {
    if (s.length !== t.length) return false;
    let count = new Array(26).fill(0);
    for (let c of s) count[c.charCodeAt(0) - 97]++;
    for (let c of t) {
        let idx = c.charCodeAt(0) - 97;
        count[idx]--;
        if (count[idx] < 0) return false;
    }
    return true;
}

Time Complexity: O(n) | Space Complexity: O(1)

Q4. Longest Common Prefix

cpp
string longestCommonPrefix(vector<string>& strs) {
    if (strs.empty()) return "";
    string prefix = strs[0];
    for (int i = 1; i < strs.size(); i++) {
        while (strs[i].find(prefix) != 0) {
            prefix = prefix.substr(0, prefix.size() - 1);
            if (prefix.empty()) return "";
        }
    }
    return prefix;
}

java
public String longestCommonPrefix(String[] strs) {
    if (strs == null || 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;
}

python
def longestCommonPrefix(strs: list) -> str:
    if not strs: return ""
    prefix = strs[0]
    for s in strs[1:]:
        while not s.startswith(prefix):
            prefix = prefix[:-1]
            if not prefix: return ""
    return prefix

javascript
function longestCommonPrefix(strs) {
    if (!strs || strs.length === 0) return "";
    let prefix = strs[0];
    for (let i = 1; i < strs.length; i++) {
        while (strs[i].indexOf(prefix) !== 0) {
            prefix = prefix.substring(0, prefix.length - 1);
            if (!prefix) return "";
        }
    }
    return prefix;
}

Time Complexity: O(S) | Space Complexity: O(1)

3. Summary Table

Practice all string problems on DSAMaster's practice platform.