Binary Search

Problem Link

Approach

Maintain left and right pointers on the sorted array.

Repeatedly check the middle index:

• If nums[mid] == target, return mid.
• If nums[mid] < target, search the right half.
• Otherwise search the left half.

Stop when the search space is empty.

Linear scan is O(n). Binary search halves the range each step for O(log n) time.

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

Code

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

        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                return mid
            if nums[mid] < target:
                left = mid + 1
            else:
                right = mid - 1

        return -1