Top 25 Binary Search Interview Questions and Answers (2026)

1. Introduction to Binary Search Interview Questions

Binary Search is one of the most efficient techniques in computer science, halving the search space at each step. Below are 25 essential Binary Search questions with full implementations in C++, Java, Python, and JavaScript.

2. Core Binary Search Questions

Q1. Classic Binary Search

Question: Given a sorted array of integers nums and a target, return its index or -1.

cpp
int search(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return mid;
        else if (nums[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}

java
public int search(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return mid;
        else if (nums[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}

python
def search(nums: list, target: int) -> int:
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = left + (right - left) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
    return -1

javascript
function search(nums, target) {
    let left = 0, right = nums.length - 1;
    while (left <= right) {
        let mid = Math.floor(left + (right - left) / 2);
        if (nums[mid] === target) return mid;
        else if (nums[mid] < target) left = mid + 1;
        else right = mid - 1;
    }
    return -1;
}

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

Q2. Search in Rotated Sorted Array

cpp
int searchRotated(vector<int>& nums, int target) {
    int left = 0, right = nums.size() - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return mid;
        if (nums[left] <= nums[mid]) {
            if (nums[left] <= target && target < nums[mid]) right = mid - 1;
            else left = mid + 1;
        } else {
            if (nums[mid] < target && target <= nums[right]) left = mid + 1;
            else right = mid - 1;
        }
    }
    return -1;
}

java
public int searchRotated(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return mid;
        if (nums[left] <= nums[mid]) {
            if (nums[left] <= target && target < nums[mid]) right = mid - 1;
            else left = mid + 1;
        } else {
            if (nums[mid] < target && target <= nums[right]) left = mid + 1;
            else right = mid - 1;
        }
    }
    return -1;
}

python
def searchRotated(nums: list, target: int) -> int:
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = left + (right - left) // 2
        if nums[mid] == target: return mid
        if nums[left] <= nums[mid]:
            if nums[left] <= target < nums[mid]: right = mid - 1
            else: left = mid + 1
        else:
            if nums[mid] < target <= nums[right]: left = mid + 1
            else: right = mid - 1
    return -1

javascript
function searchRotated(nums, target) {
    let left = 0, right = nums.length - 1;
    while (left <= right) {
        let mid = Math.floor(left + (right - left) / 2);
        if (nums[mid] === target) return mid;
        if (nums[left] <= nums[mid]) {
            if (nums[left] <= target && target < nums[mid]) right = mid - 1;
            else left = mid + 1;
        } else {
            if (nums[mid] < target && target <= nums[right]) left = mid + 1;
            else right = mid - 1;
        }
    }
    return -1;
}

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

3. Summary Table

Practice all binary search problems on DSAMaster's practice platform.