LeetCode 2006: Count Number of Pairs With Absolute Difference K Solution

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2006. Count Number of Pairs With Absolute Difference K

Easy

Array Hash Table Counting

Problem Explanation

Explanation

To solve this problem, we can iterate through the array and for each element nums[i], we check if there exists any element nums[j] such that |nums[i] - nums[j]| == k. We can achieve this by maintaining a count of the occurrences of each number in a HashMap.

Initialize a HashMap to store the count of each number in the input array.
Iterate through the array and for each element:
- Check if nums[i] - k or nums[i] + k exists in the HashMap. If either exists, increment the count by the value in the HashMap.
- Update the count of nums[i] in the HashMap.
Return the total count of pairs found.

The time complexity of this approach is O(n) where n is the size of the input array nums. The space complexity is also O(n) to store the count of each number in the HashMap.

Solution Code

import java.util.HashMap;

class Solution {
    public int countKDifference(int[] nums, int k) {
        HashMap<Integer, Integer> countMap = new HashMap<>();
        int pairsCount = 0;

        for (int num : nums) {
            pairsCount += countMap.getOrDefault(num - k, 0);
            pairsCount += countMap.getOrDefault(num + k, 0);
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
        }

        return pairsCount;
    }
}

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