Binary Search

Problem

Given a sorted nums and a target, return the index of target or −1 if not present. Must run in O(log n).

Constraints

• 1 ≤ nums.length ≤ 10⁴
• nums is sorted ascending
• all values unique

Approach — Binary Search

This is a Binary Search problem. The idea: repeatedly halve the search space, discarding the half that can't contain the answer. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(log n) time.

Solution code

Python

class Solution:
    def search(self, nums, target):
        lo, hi = 0, len(nums) - 1
        while lo <= hi:
            mid = (lo + hi) // 2
            if nums[mid] == target:
                return mid
            if nums[mid] < target:
                lo = mid + 1
            else:
                hi = mid - 1
        return -1

Java

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

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.