Leetcode - 704. Binary Search

Easy

Dec 22, 2025

#binary-search

704. Binary Search

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

1 <= nums.length <= 104
-104 < nums[i], target < 104
All the integers in nums are unique.
nums is sorted in ascending order.

Constraints:

Notes

Intuition

In a sorted array, we can eliminate half the search space with each comparison. If the middle element is too small, the target must be in the right half; if too large, it's in the left half.

Implementation

Maintain left and right pointers. While l ≤ r, calculate the midpoint. If nums[mid] equals the target, return mid. If it's less than the target, search the right half by setting l = mid + 1. If greater, search the left half by setting r = mid - 1. If the loop exits, the target doesn't exist—return -1.

Edge-cases

The condition must be l ≤ r (not l < r) to handle the case where the target is the only remaining element when l == r.

Time: O(log n) — halve the search space each iteration
Space: O(1) — only tracking pointers

Complexity

Solution

from typing import List

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        l, r = 0, len(nums) - 1

        while l <= r:
            m = (l + r) // 2
            if nums[m] == target:
                return m
            elif nums[m] < target:
                l = m + 1
            else:
                r = m - 1
        
        return -1



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()