✅ Top Coding Interview Questions with Answers: Part-1 💻🧠

1️⃣ Reverse a String
Q: Write a function to reverse a string.

Python:

def reverse_string(s):
    return s[::-1]

C++:

string reverseString(string s) {
    reverse(s.begin(), s.end());
    return s;
}

Java:

String reverseString(String s) {
    return new StringBuilder(s).reverse().toString();
}

2️⃣ Check for Palindrome
Q: Check if a string is a palindrome.

Python:

def is_palindrome(s):
    s = s.lower().replace(" ", "")
    return s == s[::-1]

C++:

bool isPalindrome(string s) {
    transform(s.begin(), s.end(), s.begin(), ::tolower);
    s.erase(remove(s.begin(), s.end(), ' '), s.end());
    return s == string(s.rbegin(), s.rend());
}

Java:

boolean isPalindrome(String s) {
    s = s.toLowerCase().replaceAll(" ", "");
    return s.equals(new StringBuilder(s).reverse().toString());
}

3️⃣ Count Vowels in a String
Q: Count number of vowels in a string.

Python:

def count_vowels(s):
    return sum(1 for c in s.lower() if c in "aeiou")

C++:

int countVowels(string s) {
    int count = 0;
    for (char c: s) {
        c = tolower(c);
        if (string("aeiou").find(c)!= string::npos)
            count++;
    }
    return count;
}

Java:

int countVowels(String s) {
    int count = 0;
    s = s.toLowerCase();
    for (char c : s.toCharArray()) {
        if ("aeiou".indexOf(c) != -1)
            count++;
    }
    return count;
}

4️⃣ Find Factorial (Recursion)
Q: Find factorial using recursion.

Python:

def factorial(n):
    return 1 if n <= 1 else n * factorial(n - 1)

C++:

int factorial(int n) {
    return (n <= 1) ? 1 : n * factorial(n - 1);
}

Java:

int factorial(int n) {
    return (n <= 1) ? 1 : n * factorial(n - 1);
}

5️⃣ Find Duplicate Elements in List/Array
Q: Print all duplicates from a list.

Python:

from collections import Counter
def find_duplicates(lst):
    return [k for k, v in Counter(lst).items() if v > 1]

C++:

vector<int> findDuplicates(vector<int>& nums) {
    unordered_map<int, int> freq;
    vector<int> res;
    for (int n : nums) freq[n]++;
    for (auto& p : freq)
        if (p.second > 1) res.push_back(p.first);
    return res;
}

Java:

List<Integer> findDuplicates(int[] nums) {
    Map<Integer, Integer> map = new HashMap<>();
    List<Integer> result = new ArrayList<>();
    for (int n : nums) map.put(n, map.getOrDefault(n, 0) + 1);
    for (Map.Entry<Integer, Integer> entry : map.entrySet())
        if (entry.getValue() > 1) result.add(entry.getKey());
    return result;
}

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