532. K-diff Pairs in an Array

• Difficulty: Easy

• Topics: Array, Two Pointers

• Similar Questions:

  • Minimum Absolute Difference in BST

Problem:

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2:

Input:[1, 2, 3, 4, 5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

Solutions:

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int count = 0;
        int j = 0;
        for (int i = 1; i < nums.size(); ++i) {
            while (j < i && nums[i] - nums[j] > k) {
                ++j;
            }
            if (j < i && nums[i] - nums[j] == k) {
                ++count;
                while (j + 1 < nums.size() && nums[j] == nums[j+1]) ++j; // this essential to deduplication.
                ++j;
            }
        }

        return count;
    }
};