Count Number of Bad Pairs - LeetCode solutions

Problem

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 <= 10^5
1 <= nums[i] <= 10^9

Solution (Java)

class Solution {
    public long countBadPairs(int[] nums) {
        HashMap<Integer, Integer> seen = new HashMap<>();
        long count = 0;
        for (int i = 0; i < nums.length; i++) {
            int diff = i - nums[i];
            if (seen.containsKey(diff)) {
                count += (i - seen.get(diff));
            } else {
                count += i;
            }
            seen.put(diff, seen.getOrDefault(diff, 0) + 1);
        }
        return count;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).