03. Number of Good Pairs

The problem can be found at the following link: Question Link

My Approach

1. Initialize an unordered map of int.

2. Iterate through the given values in the array. For each piece in the iteration:

   • Check if the current value is present in the map or not If yes, increment count as it is a valid pair.

   • If it is not present in the map then insert the value in the map

3. Return count which stores the count of all valid pairs .

Time and Auxiliary Space Complexity

• Time Complexity: O(n)

• Auxiliary Space Complexity: O(n)

Code (C++)

class Solution {
public:
    int numIdenticalPairs(vector<int>& a) {
        int i,count=0;

        unordered_map<int,int>map;

        for(i=0;i<a.size();i++){
            if(map.find(a[i])!=map.end())
            count +=map[a[i]];
                
            map[a[i]]++;
        }
        return count;
    }
};

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