[LeetCode] 1395. Count Number of Teams

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

统计作战单位数。

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/count-number-of-teams
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这道题不涉及算法。我这里提供一个N平方的思路。这道题是在求一个三元组，满足下标i < j < k同时三个下标对应的rating是单调增或者是单调减。暴力解是O(n^3)的，需要对每一种i，j，k的组合进行比较看看是否符合条件。

优化的思路是O(n^2)的。这里我们对每个下标 j 记录他的左边有多少比他小的元素和他右边有多少比他大的元素。这样，包含 ratings[j] 的组合数 = j左边比他小的元素个数 * j右边比他大的元素个数。这是单调增的情况。

同时单调减的组合数 = j左边比他大的元素个数 * j右边比他小的元素个数。

时间O(n^2)

空间O(1)

Java实现

class Solution {
    public int numTeams(int[] rating) {
        int res = 0;
        for (int i = 1; i < rating.length - 1; i++) {
            int leftBigger = 0;
            int leftSmaller = 0;
            for (int j = 0; j < i; j++) {
                if (rating[j] > rating[i]) {
                    leftBigger++;
                }
                if (rating[j] < rating[i]) {
                    leftSmaller++;
                }
            }

            int rightBigger = 0;
            int rightSmaller = 0;
            for (int j = i + 1; j < rating.length; j++) {
                if (rating[j] > rating[i]) {
                    rightBigger++;
                }
                if (rating[j] < rating[i]) {
                    rightSmaller++;
                }
            }
            res += leftBigger * rightSmaller + leftSmaller * rightBigger;
        }
        return res;
    }
}

 

LeetCode 题目总结