# 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.

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 $x - k$.

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

• 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
}
}