# 2364. Count Number of Bad Pairs

## 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

• class Solution {
Map<Integer, Integer> cnt = new HashMap<>();
long ans = 0;
for (int i = 0; i < nums.length; ++i) {
int x = i - nums[i];
ans += i - cnt.getOrDefault(x, 0);
cnt.merge(x, 1, Integer::sum);
}
return ans;
}
}

• class Solution {
public:
unordered_map<int, int> cnt;
long long ans = 0;
for (int i = 0; i < nums.size(); ++i) {
int x = i - nums[i];
ans += i - cnt[x];
++cnt[x];
}
return ans;
}
};

• class Solution:
def countBadPairs(self, nums: List[int]) -> int:
cnt = Counter()
ans = 0
for i, x in enumerate(nums):
ans += i - cnt[i - x]
cnt[i - x] += 1
return ans


• func countBadPairs(nums []int) (ans int64) {
cnt := map[int]int{}
for i, x := range nums {
x = i - x
ans += int64(i - cnt[x])
cnt[x]++
}
return
}

• function countBadPairs(nums: number[]): number {
const cnt = new Map<number, number>();
let ans = 0;
for (let i = 0; i < nums.length; ++i) {
const x = i - nums[i];
ans += i - (cnt.get(x) ?? 0);
cnt.set(x, (cnt.get(x) ?? 0) + 1);
}
return ans;
}