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목록Count of Smaller Numbers After Self (1)
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LeetCode: 315. Count of Smaller Numbers After Self
Several solutions and short analysis of each one. 1. List + Binary Search The most straightforward solution, however, time complexity is O(N**2). class Solution: def countSmaller(self, nums: List[int]) -> List[int]: from bisect import insort, bisect_left arr = [] count = [] for i in range(len(nums) - 1, -1, -1): num = nums[i] idx = bisect_left(arr, num) count.append(idx) insort(arr, num) return ..
Computer Science/Coding Test
2022. 2. 10. 11:22