Q46: Binary Search on Sorted Boundaries. - DSA Patterns (Multi-Lang)

In Simple Words: Standard binary search template: mid = l + Math.floor((r - l) / 2). For lower bound (first element $\ge target$), keep moving left r = mid - 1 when nums[mid] >= target.

Time: $O(\log N)$ | Space: $O(1)$

Multi-Language SolutionsPython 3 • JavaScript • TypeScript • Java • C++

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