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Similar approach as sum of 2 nummber to a given sum in an array. Special case of (k == 0) -> check for frequency of the number >= 2 as only (x-x) will give 0.
*/
classSolution {
publicintfindPairs(int[] nums, intk) {
Map<Integer, Integer> map = newHashMap<>();
for (intn : nums) {
map.put(n, map.getOrDefault(n, 0) + 1);
}
intpairs = 0;
for (Integerkey : map.keySet()) {
intmapVal = map.getOrDefault(key+k, 0);
if (k == 0) {
if (mapVal > 1)
pairs += 1;
}
else {
if (mapVal > 0)
pairs += 1;
}
}
returnpairs;
}
}
// Problem : https://leetcode.com/problems/k-diff-pairs-in-an-array/
// @romitdutta10
// TC : O(n)
// Another Approach with two sets
classSolution {
publicintfindPairs(int[] nums, intk) {
if(nums == null || nums.length == 0) {
return0;
}
Set<String> pairs = newHashSet<>();
Set<Integer> visited = newHashSet<>();
for(intnum : nums) {
intgreater = num + k;
intlesser = num - k;
if(visited.contains(greater)) {
pairs.add(num + " " + greater);
}
if(visited.contains(lesser)) {
pairs.add(lesser + " " + num);
}
visited.add(num);
}
returnpairs.size();
}
}
// Problem : https://leetcode.com/problems/k-diff-pairs-in-an-array/
