README_EN.md

2364. Count Number of Bad Pairs

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Description

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

Solutions

Python3

class Solution:
    def countBadPairs(self, nums: List[int]) -> int:
        arr = [i - v for i, v in enumerate(nums)]
        cnt = Counter(arr)
        n = len(arr)
        return sum(v * (n - v) for v in cnt.values()) >> 1

Java

class Solution {
    public long countBadPairs(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            nums[i] = i - nums[i];
        }
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int v : nums) {
            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
        }
        long ans = 0;
        for (int v : cnt.values()) {
            ans += v * (n - v);
        }
        return ans >> 1;
    }
}

C++

class Solution {
public:
    long long countBadPairs(vector<int>& nums) {
        int n = nums.size();
        for (int i = 0; i < n; ++i) nums[i] = i - nums[i];
        unordered_map<int, int> cnt;
        for (int v : nums) cnt[v]++;
        long long ans = 0;
        for (auto [_, v] : cnt) ans += 1ll * v * (n - v);
        return ans >> 1;
    }
};

Go

func countBadPairs(nums []int) int64 {
	n := len(nums)
	for i := range nums {
		nums[i] = i - nums[i]
	}
	cnt := map[int]int{}
	for _, v := range nums {
		cnt[v]++
	}
	ans := 0
	for _, v := range cnt {
		ans += v * (n - v)
	}
	ans >>= 1
	return int64(ans)
}

TypeScript

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