LeetCode 1395. Count Number of Teams统计作战单位数【Medium】【Python】【暴力】

Problem

LeetCode

There are n soldiers standing in a line. Each soldier is assigned a unique rating value.

You have to form a team of 3 soldiers amongst them under the following rules:

  • Choose 3 soldiers with index (i, j, k) with rating (rating[i], rating[j], rating[k]).
  • A team is valid if: (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k]) where (0 <= i < j < k < n).

Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).

Example 1:

Input: rating = [2,5,3,4,1]
Output: 3
Explanation: We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1).

Example 2:

Input: rating = [2,1,3]
Output: 0
Explanation: We can't form any team given the conditions.

Example 3:

Input: rating = [1,2,3,4]
Output: 4

Constraints:

  • n == rating.length
  • 1 <= n <= 200
  • 1 <= rating[i] <= 10^5

问题

力扣

n 名士兵站成一排。每个士兵都有一个 独一无二 的评分 rating 。

每 3 个士兵可以组成一个作战单位，分组规则如下：

  • 从队伍中选出下标分别为 i、j、k 的 3 名士兵，他们的评分分别为 rating[i]、rating[j]、rating[k]
  • 作战单位需满足： rating[i] < rating[j] < rating[k] 或者 rating[i] > rating[j] > rating[k] ，其中 0 <= i < j < k < n

请你返回按上述条件可以组建的作战单位数量。每个士兵都可以是多个作战单位的一部分。

示例 1：

输入：rating = [2,5,3,4,1]
输出：3
解释：我们可以组建三个作战单位 (2,3,4)、(5,4,1)、(5,3,1) 。

示例 2：

输入：rating = [2,1,3]
输出：0
解释：根据题目条件，我们无法组建作战单位。

示例 3：

输入：rating = [1,2,3,4]
输出：4

提示：

  • n == rating.length
  • 1 <= n <= 200
  • 1 <= rating[i] <= 10^5

思路

暴力

三层循环，暴力求解。
因为数据 n 是 [1, 200]，所以不会 LTE。

时间复杂度: O(n^3)
空间复杂度: O(1)

Python3代码
from typing import List

class Solution:
    def numTeams(self, rating: List[int]) -> int:
        n = len(rating)
        count = 0
        for i in range(n-2):
            for j in range(i+1, n-1):
                for k in range(j+1, n):
                    if rating[i] < rating[j]:
                        if rating[j] < rating[k]:
                            count += 1
                    elif rating[i] > rating[j]:
                        if rating[j] > rating[k]:
                            count += 1
        return count

GitHub链接

Python