Comment on page
2364. Count Number of Bad Pairs
Medium
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 <= 10^51 <= nums[i] <= 10^9
Runtime: 119 ms, faster than 100%
Memory Usage: 14.1 MB, less than 16.67%
func countBadPairs(nums []int) int64 {
m := make(map[int]int)
res := (len(nums) * (len(nums) - 1))/2 // 所有可能 C len(nums) 取 2
for i, n := range nums {
m[n - i]++ // 紀錄 nums[i] - i 相同的有多少組
}
for k := range m {
res -= (m[k]*(m[k] - 1))/2 // 扣去 not bad pair 總數
}
return int64(res)
}