LeetCode每日一题(2364. Count Number of Bad Pairs)

最新推荐文章于 2023-01-10 12:07:20 发布

wangjun861205

最新推荐文章于 2023-01-10 12:07:20 发布

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给定一个整数数组nums，返回其中坏对的总数。坏对定义为i < j且j - i != nums[j] - nums[i]。例如，输入nums = [4,1,3,3]，存在5个坏对。通过计算diff[i] = nums[i] - i，将数组分组并计算每组的非坏对，从而得出答案。" 134278289,10526265,Qt应用：Linux程序监控、QScreen截屏与GIF动画自适应,"['Qt开发', 'Linux编程', 'GUI设计', '图像处理', '动画']

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You are given a 0-indexed integer array nums. A pair of indices (i, j) is a bad pair if i < j and j - i != nums[j] - nums[i].

Return the total number of bad pairs in nums.

Example 1:

Input: nums = [4,1,3,3]
Output: 5

Explanation:
The pair (0, 1) is a bad pair since 1 - 0 != 1 - 4.
The pair (0, 2) is a bad pair since 2 - 0 != 3 - 4, 2 != -1.
The pair (0, 3) is a bad pair since 3 - 0 != 3 - 4, 3 != -1.
The pair (1, 2) is a bad pair since 2 - 1 != 3 - 1, 1 != 2.
The pair (2, 3) is a bad pair since 3 - 2 != 3 - 3, 1 != 0.
There are a total of 5 bad pairs, so we return 5.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 0

Explanation: There are no bad pairs.

Constraints:

1 <= nums.length <= 105
1 <= nums[i] <= 109

bad_pairs = total_pairs - not_bad_pairs
j - i != nums[j] - nums[i] => nums[i] - i != nums[j] - j

以上两条是解这题的核心, 假设 diff[i] = nums[i] - i, 那如果 diff[i] == diff[j]则 nums[i]和 nums[j]一定可以组成not bad pair, 所以任何两个有相同的 diff 的 nums 相互之间一定是可以组成not bad pair 的，这样我们可以根据 diff 来对 nums 进行分组，在求出每一组所能组成的 pairs 的数量，用 nums 所能组成的所有 pairs 的数量减去每组的 pairs 的数量就是最终答案


use std::collections::HashMap;

impl Solution {
    fn pairs(count: i64) -> i64 {
        let mut ans = 0;
        for i in 1..count {
            ans += i;
        }
        ans
    }
    pub fn count_bad_pairs(nums: Vec<i32>) -> i64 {
        let mut total = Solution::pairs(nums.len() as i64);
        let mut diffs = HashMap::new();
        for (i, n) in nums.into_iter().enumerate() {
            *diffs.entry(n - i as i32).or_insert(0i64) += 1;
        }
        for &d in diffs.values() {
            total -= Solution::pairs(d);
        }
        total
    }
}