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2006. Count Number of Pairs With Absolute Difference K

Description

Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.

The value of |x| is defined as:

• x if x >= 0.
• -x if x < 0.

 

Example 1:

Input: nums = [1,2,2,1], k = 1
Output: 4
Explanation: The pairs with an absolute difference of 1 are:
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]

Example 2:

Input: nums = [1,3], k = 3
Output: 0
Explanation: There are no pairs with an absolute difference of 3.

Example 3:

Input: nums = [3,2,1,5,4], k = 2
Output: 3
Explanation: The pairs with an absolute difference of 2 are:
- [3,2,1,5,4]
- [3,2,1,5,4]
- [3,2,1,5,4]

 

Constraints:

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100
• 1 <= k <= 99

Solutions

Solution 1: Brute Force Enumeration

We notice that the length of the array $nums$ does not exceed $200$, so we can enumerate all pairs $(i, j)$, where $i < j$, and check if $

nums[i] - nums[j]

$ equals $k$. If it does, we increment the answer by one.

Finally, we return the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

Solution 2: Hash Table or Array

We can use a hash table or array to record the occurrence count of each number in the array $nums$. Then, we enumerate each number $x$ in the array $nums$, and check if $x + k$ and $x - k$ are in the array $nums$. If they are, we increment the answer by the sum of the occurrence counts of $x + k$ and $x - k$.

Finally, we return the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int countKDifference(int[] nums, int k) {
          int ans = 0;
          for (int i = 0, n = nums.length; i < n; ++i) {
              for (int j = i + 1; j < n; ++j) {
                  if (Math.abs(nums[i] - nums[j]) == k) {
                      ++ans;
                  }
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int countKDifference(vector<int>& nums, int k) {
          int n = nums.size();
          int ans = 0;
          for (int i = 0; i < n; ++i) {
              for (int j = i + 1; j < n; ++j) {
                  ans += abs(nums[i] - nums[j]) == k;
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def countKDifference(self, nums: List[int], k: int) -> int:
          n = len(nums)
          return sum(
              abs(nums[i] - nums[j]) == k for i in range(n) for j in range(i + 1, n)
          )
  
  

• func countKDifference(nums []int, k int) int {
  	n := len(nums)
  	ans := 0
  	for i := 0; i < n; i++ {
  		for j := i + 1; j < n; j++ {
  			if abs(nums[i]-nums[j]) == k {
  				ans++
  			}
  		}
  	}
  	return ans
  }
  
  func abs(x int) int {
  	if x > 0 {
  		return x
  	}
  	return -x
  }
  

• function countKDifference(nums: number[], k: number): number {
      let ans = 0;
      let cnt = new Map();
      for (let num of nums) {
          ans += (cnt.get(num - k) || 0) + (cnt.get(num + k) || 0);
          cnt.set(num, (cnt.get(num) || 0) + 1);
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn count_k_difference(nums: Vec<i32>, k: i32) -> i32 {
          let mut res = 0;
          let n = nums.len();
          for i in 0..n - 1 {
              for j in i..n {
                  if (nums[i] - nums[j]).abs() == k {
                      res += 1;
                  }
              }
          }
          res
      }
  }
  
  

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