Most Beautiful Item for Each Query — binary search
Halve the sorted range until the target is found or the range collapses. Built for LeetCode 2070. Most Beautiful Item for Each Query.
Because the slice is sorted, comparing nums[mid] to target tells you which half can be thrown away. That is the whole algorithm.
Algorithm
- While
lo <= hi, pickmid. - Equal → return
mid. Too small → search right. Too big → search left. - Empty range → not found (
-1).
- Time
- O(log n)
- Space
- O(1)
class Solution: def mostBeautifulItemForEachQuery(self, nums: list[int], target: int) -> int: 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