K-diff Pairs in an Array – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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).

Constraints:

• 1 <= nums.length <= 104
• -107 <= nums[i] <= 107
• 0 <= k <= 107

C++ K-diff Pairs in an Array LeetCode Solution

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end());
        int ans=0,i=0,j=1;
        for(i,j;i<nums.size() and j<nums.size();)
        {
            if(i==j or nums[j]-nums[i]<k)
                j++;
            else 
            {
                if(nums[j]-nums[i]==k)
                {
                    ans++;
                    j++;
                    for(;j<nums.size();j++)
                        if(nums[j]!=nums[j-1])
                            break;
                    if(j==nums.size())
                    return ans;
                    j--;                  
                }
                i++;
                while(i<j and nums[i]==nums[i-1])
                    i++;
            }
        }
        return ans;
    }
};

Java K-diff Pairs in an Array LeetCode Solution

public class Solution {
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length == 0 || k < 0)   return 0;
        
        Map<Integer, Integer> map = new HashMap<>();
        int count = 0;
        for (int i : nums) {
            map.put(i, map.getOrDefault(i, 0) + 1);
        }
        
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (k == 0) {
                //count how many elements in the array that appear more than twice.
                if (entry.getValue() >= 2) {
                    count++;
                } 
            } else {
                if (map.containsKey(entry.getKey() + k)) {
                    count++;
                }
            }
        }
        
        return count;
    }
}
 

Python 3 K-diff Pairs in an Array LeetCode Solution

def findPairs(self, nums, k):
        res = 0
        c = collections.Counter(nums)
        for i in c:
            if k > 0 and i + k in c or k == 0 and c[i] > 1:
                res += 1
        return res

 
which equals to:
 
def findPairs(self, nums, k):
        c = collections.Counter(nums)
        return  sum(k > 0 and i + k in c or k == 0 and c[i] > 1 for i in c)

Array-1180
String-562
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Tree-195

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Stack-133
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Backtracking-92
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Sliding Window-73
Linked List-69
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Interactive-18
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