LeetCode | 1395. Count Number of Teams統計作戰單位數【Python】

LeetCode 1395. Count Number of Teams統計作戰單位數【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

問題

力扣

n 名士兵站成一排。每個士兵都有一個 獨一無二 的評分 rating 。

每 3 個士兵可以組成一個作戰單位，分組規則如下：

• 從隊伍中選出下標分別爲 i、j、k 的 3 名士兵，他們的評分分別爲 rating[i]、rating[j]、rating[k]
• 作戰單位需滿足： rating[i] < rating[j] < rating[k] 或者 rating[i] > rating[j] > rating[k] ，其中 0 <= i < j < k < n

請你返回按上述條件可以組建的作戰單位數量。每個士兵都可以是多個作戰單位的一部分。

示例 1：

輸入：rating = [2,5,3,4,1]
輸出：3
解釋：我們可以組建三個作戰單位 (2,3,4)、(5,4,1)、(5,3,1) 。

示例 2：

輸入：rating = [2,1,3]
輸出：0
解釋：根據題目條件，我們無法組建作戰單位。

示例 3：

輸入：rating = [1,2,3,4]
輸出：4

提示：

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

思路

暴力

三層循環，暴力求解。
因爲數據 n 是 [1, 200]，所以不會 LTE。

時間複雜度: O(n^3)
空間複雜度: O(1)

Python3代碼

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鏈接

Python