[Solved] Count Number of Bad Pairs LeetCode Contest

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.

Count Number of Bad Pairs LeetCode Contest

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

Solution

class Solution {
    public long countBadPairs(int[] nums) {
        
        HashMap<Integer, Integer> fMap = new HashMap<>();
        int n = nums.length;
        long N = 1L * (n-1);
        long tp = (N*(N+1))/2;
        long vp = 0;
        
        for(int i = 0; i < n ; i++){
            int val = nums[i] - i;
            fMap.put(val, fMap.getOrDefault(val,0)+1);
        }
        
        
        for(int i = 0; i < n ; i++){
            
            Integer key = nums[i] - i;
            Integer value = fMap.get(key);

            if(value >= 2){
                vp += value-1;

                fMap.put(key, value-1);
            }
        }
        
        
        return tp-vp;
    }
}

long long countBadPairs(vector<int>& nums) {
    long long ans = 0;
    unordered_map<int, int> m;
    for (int i = 0; i < nums.size(); i++) {
        ans += i - m[nums[i] - i];
        m[nums[i] - i] += 1;
    }
    return ans;
}

def countBadPairs(self, A: List[int]) -> int:
    m, ans = defaultdict(int), 0
    for i in range(len(A)):
        ans += i - m[A[i] - i]
        m[A[i] - i] += 1
    return ans

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