532. K-diff Pairs in an Array #

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:

1. The pairs (i, j) and (j, i) count as the same pair.
2. The length of the array won’t exceed 10,000.
3. All the integers in the given input belong to the range: [-1e7, 1e7].

Problem Summary #

Given an array, find how many different pairs in the array have a difference of K.

Solution Approach #

This problem can use a map to record the number of occurrences of each number. Because repeated numbers also correspond to a unique key, there is no need to worry that a certain number will be checked multiple times. Traverse the map once; after adding K to each number, check whether it exists in the dictionary. If it exists, count ++. The case where K = 0 needs to be handled separately: if the frequency of this element in the dictionary is greater than 1, count also needs to ++.

Code #

package leetcode

func findPairs(nums []int, k int) int {
	if k < 0 || len(nums) == 0 {
		return 0
	}
	var count int
	m := make(map[int]int, len(nums))
	for _, value := range nums {
		m[value]++
	}
	for key := range m {
		if k == 0 && m[key] > 1 {
			count++
			continue
		}
		if k > 0 && m[key+k] > 0 {
			count++
		}
	}
	return count
}