Formatted question description: https://leetcode.ca/all/532.html

532. K-diff Pairs in an Array (Easy)

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].

## Solution 1.

Use an `unordered_map<int, int> m` to count the occurrence of each number.

Let `int cnt` be the number of pairs, initially `0`.


For each number `n`,
* If `k == 0` and `m[n] > 1`, increment `cnt`.
* If `k != 0` and `m[n + k] > 0`, increment `cnt`.

```cpp
// OJ: https://leetcode.com/problems/k-diff-pairs-in-an-array
// Time: O(N)
// Space: O(N)
class Solution {
public:
  int findPairs(vector<int>& nums, int k) {
    if (k < 0) return 0;
    unordered_map<int, int> m;
    for (int n : nums) m[n]++;
    int cnt = 0;
    for (auto p : m) {
      if ((!k && p.second > 1)
        || (k && m.count(p.first + k))) ++cnt;
    }
    return cnt;
  }
};

class Solution {
    public int findPairs(int[] nums, int k) {
        if (k < 0 || nums == null || nums.length == 0)
            return 0;
        if (k == 0) {
            Map<Integer, Integer> map = new HashMap<Integer, Integer>();
            int length = nums.length;
            for (int i = 0; i < length; i++) {
                int num = nums[i];
                int count = map.getOrDefault(num, 0);
                count++;
                map.put(num, count);
            }
            int pairs = 0;
            Set<Integer> set = map.keySet();
            for (int num : set) {
                int count = map.get(num);
                if (count > 1)
                    pairs++;
            }
            return pairs;
        } else {
            List<Integer> list = new ArrayList<Integer>();
            for (int num : nums)
                list.add(num);
            Collections.sort(list);
            int pairs = 0;
            int length = list.size();
            int max = list.get(length - 1);
            for (int i = 0; i < length; i++) {
                int num = list.get(i);
                if (i > 0 && list.get(i - 1) == num)
                    continue;
                int nextNum = num + k;
                if (nextNum > max)
                    break;
                if (list.contains(nextNum))
                    pairs++;
            }
            return pairs;
        }
    }
}

############

class Solution {
    public int findPairs(int[] nums, int k) {
        Set<Integer> vis = new HashSet<>();
        Set<Integer> ans = new HashSet<>();
        for (int v : nums) {
            if (vis.contains(v - k)) {
                ans.add(v - k);
            }
            if (vis.contains(v + k)) {
                ans.add(v);
            }
            vis.add(v);
        }
        return ans.size();
    }
}

// OJ: https://leetcode.com/problems/k-diff-pairs-in-an-array
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int findPairs(vector<int>& A, int k) {
        unordered_map<int, int> m;
        for (int n : A) m[n]++;
        int ans = 0;
        for (auto &[n, cnt] : m) {
            ans += k ? m.count(n - k) : cnt > 1;
        }
        return ans;
    }
};

class Solution:
    def findPairs(self, nums: List[int], k: int) -> int:
        vis, ans = set(), set()
        for v in nums:
            if v - k in vis:
                ans.add(v - k)
            if v + k in vis:
                ans.add(v)
            vis.add(v)
        return len(ans)

############

class Solution(object):
  def findPairs(self, nums, k):
    """
    :type nums: List[int]
    :type k: int
    :rtype: int
    """
    if k < 0 or len(nums) < 2:
      return 0
    ans = 0
    nums.sort()
    start, end = 0, 1
    while start < len(nums) and end < len(nums):
      if start > 0 and nums[start - 1] == nums[start]:
        start += 1
        continue
      if nums[end] - nums[start] > k:
        start += 1
      elif start == end or nums[end] - nums[start] < k:
        end += 1
      elif nums[end] - nums[start] == k:
        ans += 1
        end += 1
        while end < len(nums) and nums[end - 1] == nums[end]:
          end += 1
    return ans

func findPairs(nums []int, k int) int {
	vis := map[int]bool{}
	ans := map[int]bool{}
	for _, v := range nums {
		if vis[v-k] {
			ans[v-k] = true
		}
		if vis[v+k] {
			ans[v] = true
		}
		vis[v] = true
	}
	return len(ans)
}

impl Solution {
    pub fn find_pairs(mut nums: Vec<i32>, k: i32) -> i32 {
        nums.sort();
        let n = nums.len();
        let mut res = 0;
        let mut left = 0;
        let mut right = 1;
        while right < n {
            let num = i32::abs(nums[left] - nums[right]);
            if num == k {
                res += 1;
            }
            if num <= k {
                right += 1;
                while right < n && nums[right - 1] == nums[right] {
                    right += 1;
                }
            } else {
                left += 1;
                while left < right && nums[left - 1] == nums[left] {
                    left += 1;
                }
                if left == right {
                    right += 1;
                }
            }
        }
        res
    }
}

532 - K-diff Pairs in an Array

532. K-diff Pairs in an Array (Easy)

All Problems

All Solutions

Welcome to Subscribe On Youtube

532. K-diff Pairs in an Array (Easy)

All Problems

All Solutions