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 $nums$.

Python3JavaC++GoTypeScriptRust

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)
        )

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;
    }
};

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;
}