Question

1512. Number of Good Pairs

中文文档

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

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

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

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

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

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;
}

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
  }
}

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;
  }
}

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

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

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;
  }
}

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;
  }
};

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
}

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;
}

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
  }
}

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;
}