Add problem 532

Author: avhn

algorithms/532/README.md

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+## 532. K-diff Pairs in an Array
+
+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:**
+<pre>
+<b>Input:</b> [3, 1, 4, 1, 5], k = 2
+<b>Output:</b> 2
+<b>Explanation:</b> 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.
+</pre>
+
+**Example 2:**
+<pre>
+<b>Input:</b> [1, 2, 3, 4, 5], k = 1
+<b>Output:</b> 4
+<b>Explanation:</b> There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and
+(4, 5).
+</pre>
+
+**Example 3:**
+<pre>
+<b>Input:</b> [1, 3, 1, 5, 4], k = 0
+<b>Output:</b> 1
+<b>Explanation:</b> There is one 0-diff pair in the array, (1, 1).
+</pre>
+
+**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].

algorithms/532/solution.go

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+package main
+
+// Time and space complexity, O(n) where n is size of the array
+func findPairs(nums []int, k int) int {
+if k < 0 {
+return 0
+}
+countMap := make(map[int]int, len(nums))
+for _, n := range nums {
+countMap[n]++
+}
+var c int
+for n, count := range countMap {
+if 0 < k {
+if 0 < countMap[n+k] {
+c++
+}
+} else if 1 < count {
+c++
+}
+}
+return c
+}