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

Python Code
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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

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

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

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

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

PHP Code
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class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function numIdenticalPairs($nums) {
        $arr = array_values(array_unique($nums));
        for ($i = 0; $i < count($nums); $i++) {
            $v[$nums[$i]] += 1;
        }
        $rs = 0;
        for ($j = 0; $j < count($arr); $j++) {
            $rs += ($v[$arr[$j]] * ($v[$arr[$j]] - 1)) / 2;
        }
        return $rs;
    }
}

C Code
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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;
}

Solution 2

Python Code
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class Solution:
    def numIdenticalPairs(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        return sum(v * (v - 1) for v in cnt.values()) >> 1

Java Code
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class Solution {
    public int numIdenticalPairs(int[] nums) {
        int[] cnt = new int[101];
        for (int x : nums) {
            ++cnt[x];
        }
        int ans = 0;
        for (int v : cnt) {
            ans += v * (v - 1) / 2;
        }
        return ans;
    }
}

C++ Code
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class Solution {
public:
    int numIdenticalPairs(vector<int>& nums) {
        int cnt[101]{};
        for (int& x : nums) {
            ++cnt[x];
        }
        int ans = 0;
        for (int v : cnt) {
            ans += v * (v - 1) / 2;
        }
        return ans;
    }
};

Go Code
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func numIdenticalPairs(nums []int) (ans int) {
	cnt := [101]int{}
	for _, x := range nums {
		cnt[x]++
	}
	for _, v := range cnt {
		ans += v * (v - 1) / 2
	}
	return
}

TypeScript Code
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function numIdenticalPairs(nums: number[]): number {
    const cnt = new Array(101).fill(0);
    for (const x of nums) {
        ++cnt[x];
    }
    let ans = 0;
    for (const v of cnt) {
        ans += v * (v - 1);
    }
    return ans >> 1;
}

Rust Code
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impl Solution {
    pub fn num_identical_pairs(nums: Vec<i32>) -> i32 {
        let mut cnt = [0; 101];
        for &num in nums.iter() {
            cnt[num as usize] += 1;
        }
        let mut ans = 0;
        for &v in cnt.iter() {
            ans += (v * (v - 1)) / 2;
        }
        ans
    }
}

C Code
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int numIdenticalPairs(int* nums, int numsSize) {
    int cnt[101] = {0};
    for (int i = 0; i < numsSize; i++) {
        cnt[nums[i]]++;
    }
    int ans = 0;
    for (int i = 0; i < 101; ++i) {
        ans += cnt[i] * (cnt[i] - 1) / 2;
    }
    return ans;
}