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1512. Number of Good Pairs

Description

Given an array of integers nums, return the number of good pairs.

A pair (i, j) is called good if nums[i] == nums[j] and i < j.

 

Example 1:

Input: nums = [1,2,3,1,1,3]
Output: 4
Explanation: There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed.

Example 2:

Input: nums = [1,1,1,1]
Output: 6
Explanation: Each pair in the array are good.

Example 3:

Input: nums = [1,2,3]
Output: 0

 

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 100

Solutions

Solution 1: Counting

Traverse the array, and for each element $x$, count how many elements before it are equal to $x$. This count represents the number of good pairs formed by $x$ and the previous elements. After traversing the entire array, we obtain the answer.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the array, and $C$ is the range of values in the array. In this problem, $C = 101$.

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class Solution:
    def numIdenticalPairs(self, nums: List[int]) -> int:
        ans = 0
        cnt = Counter()
        for x in nums:
            ans += cnt[x]
            cnt[x] += 1
        return ans

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class Solution {
    public int numIdenticalPairs(int[] nums) {
        int ans = 0;
        int[] cnt = new int[101];
        for (int x : nums) {
            ans += cnt[x]++;
        }
        return ans;
    }
}

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class Solution {
public:
    int numIdenticalPairs(vector<int>& nums) {
        int ans = 0;
        int cnt[101]{};
        for (int& x : nums) {
            ans += cnt[x]++;
        }
        return ans;
    }
};

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func numIdenticalPairs(nums []int) (ans int) {
    cnt := [101]int{}
    for _, x := range nums {
        ans += cnt[x]
        cnt[x]++
    }
    return
}

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function numIdenticalPairs(nums: number[]): number {
    const cnt: number[] = Array(101).fill(0);
    let ans = 0;
    for (const x of nums) {
        ans += cnt[x]++;
    }
    return ans;
}

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impl Solution {
    pub fn num_identical_pairs(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut cnt = [0; 101];
        for &x in nums.iter() {
            ans += cnt[x as usize];
            cnt[x as usize] += 1;
        }
        ans
    }
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var numIdenticalPairs = function (nums) {
    const cnt = Array(101).fill(0);
    let ans = 0;
    for (const x of nums) {
        ans += cnt[x]++;
    }
    return ans;
};

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class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function numIdenticalPairs($nums) {
        $ans = 0;
        $cnt = array_fill(0, 101, 0);
        foreach ($nums as $x) {
            $ans += $cnt[$x]++;
        }
        return $ans;
    }
}

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int numIdenticalPairs(int* nums, int numsSize) {
    int cnt[101] = {0};
    int ans = 0;
    for (int i = 0; i < numsSize; i++) {
        ans += cnt[nums[i]]++;
    }
    return ans;
}

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