Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.
A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:
0 <= i, j < nums.length
i != j
|nums[i] - nums[j]| == k
Notice that |val| denotes the absolute value of val.
Example 1:
Input: nums = [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: nums = [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: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
Example 4:
Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2
Example 5:
Input: nums = [-1,-2,-3], k = 1
Output: 2
Constraints:
1 <= nums.length <= 104
-107 <= nums[i] <= 107
0 <= k <= 107
Solution:
class Solution {
public int findPairs(int[] nums, int k) {
Set<Integer> vals = new HashSet();
Set<List<Integer>> set = new HashSet();
int res = 0;
// |i - j| = k
// i >= j, i - j == k, j = i - k
// i < j, j - i == k, j - i + k
for (int i = 0; i < nums.length; i ++) {
int val = nums[i];
if (vals.contains(val + k) && set.add(Arrays.asList(val + k, val))) {
set.add(Arrays.asList(val, val + k));
res ++;
}
if (vals.contains(val - k) && set.add(Arrays.asList(val - k, val))) {
set.add(Arrays.asList(val, val - k));
res ++;
}
vals.add(val);
}
return res;
}
}
class Solution {
public int findPairs(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<>();
int result = 0;
for (int i : nums) {
if (map.containsKey(i)) {
if (k == 0 && map.get(i) == 1) {
result++;
}
map.put(i, map.get(i) + 1);
} else {
if (map.containsKey(i - k)) {
result++;
}
if (map.containsKey(i + k)) {
result++;
}
map.put(i, 1);
}
}
return result;
}
}