Approach: Count by value

def numIdenticalPairs(nums):
    from collections import Counter
    return sum(c*(c-1)//2 for c in Counter(nums).values())

Complexity: O(n) time, O(n) space.

Full Solution — all languages

class Solution:
    def numIdenticalPairs(self, nums):
        cnt = {}
        total = 0
        for x in nums:
            total += cnt.get(x, 0)
            cnt[x] = cnt.get(x, 0) + 1
        return total

import java.util.*;
class Solution {
    public int numIdenticalPairs(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        int total = 0;
        for (int x : nums) {
            int c = cnt.getOrDefault(x, 0);
            total += c;
            cnt.put(x, c + 1);
        }
        return total;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int numIdenticalPairs(vector<int>& nums) {
        unordered_map<int, int> cnt;
        int total = 0;
        for (int x : nums) {
            total += cnt[x];
            cnt[x]++;
        }
        return total;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var numIdenticalPairs = function(nums) {
    const cnt = {};
    let total = 0;
    for (const x of nums) {
        total += (cnt[x] || 0);
        cnt[x] = (cnt[x] || 0) + 1;
    }
    return total;
};

Number of Good Pairs

Example 1

Example 2

Constraints