Easy · Binary search

Binary Search

Given a sorted array and a target, return the index of the target, or -1 if it is not there. Run in O(log n).

nums = [-1, 0, 3, 5, 9, 12], target = 9

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1let lo = 0, hi = arr.length - 1;

3while (lo <= hi) {

4 const mid = (lo + hi) >> 1;

5 if (arr[mid] === target) return mid;

6 arr[mid] < target ? (lo = mid + 1) : (hi = mid - 1);

Binary search and the off-by-one that always gets you

Tunde Alabi0:19