Question
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).
Code
class Solution {
public int findPairs(int[] nums, int k) {
if(k<0) return 0;
Map<Integer, Set<Integer>> map = new HashMap<>();
Set<Integer> visited = new HashSet<>();
for(int n:nums) {
if(visited.contains(n-k)) {
Set<Integer> set = map.getOrDefault(n-k, new HashSet<Integer>());
set.add(n);
map.put(n-k, set);
}
if(visited.contains(n+k)) {
Set<Integer> set = map.getOrDefault(n, new HashSet<Integer>());
set.add(n+k);
map.put(n, set);
}
visited.add(n);
}
int count = 0;
for(Map.Entry<Integer, Set<Integer>> entry : map.entrySet()) {
count += entry.getValue().size();
}
return count;
}
}
class Solution {
public int findPairs(int[] nums, int k) {
if(k<0) return 0;
Set<Integer> starts = new HashSet<>();
Set<Integer> uniqs = new HashSet<>();
int count = 0;
for(int n:nums) {
if(uniqs.contains(n+k)) starts.add(n);
if(uniqs.contains(n-k)) starts.add(n-k);
uniqs.add(n);
}
return starts.size();
}
}
Performance
O(n)