Merge pull request #42 from mrid88/feature/div

Solution - #2206 - Mridul/Edited - 23/03/2025

Author: JRS296

Problems/2206 - Divide Array Into Equal Pairs - Easy/Explanation.md

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+# Explanation of the Program
+
+## **Objective**
+The program determines whether it is possible to divide an array of integers into pairs such that each pair consists of identical elements. If this is possible, it returns `true`; otherwise, it returns `false`.
+
+---
+
+## **Step-by-Step Explanation**
+
+### **1. Input and HashMap Initialization**
+- The input is an array `nums` of integers.
+- A `HashMap` is used to count the frequency of each integer in the array.
+  - **Key:** The integer from the array.
+  - **Value:** The number of times the integer appears in the array.
+
+### **2. Counting Frequencies**
+```java
+for (int i = 0; i < n; i++) {
+    map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
+}
+
+Time Complexity
+Building the Frequency Map:
+The loop iterates through the array once, making this O(n), where n is the length of the array.
+Checking Frequencies:
+The loop iterates through the values in the HashMap, which is proportional to the number of distinct integers in the array. In the worst case, there can be up to n distinct integers, so this step is O(n).
+Overall Time Complexity: O(n)
+
+Space Complexity
+HashMap Storage:
+The HashMap stores the frequencies of at most n distinct integers. In the worst case, the space required is O(n).
+Overall Space Complexity: O(n)

Problems/2206 - Divide Array Into Equal Pairs - Easy/Solution.java

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+import java.util.HashMap;
+
+class Solution {
+    public boolean divideArray(int[] nums) {
+        HashMap<Integer,Integer> map = new HashMap<>();
+
+        int n = nums.length;
+
+        for(int i = 0;i<n;i++){
+            map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
+        }
+
+        if(n%2!=0) return false;
+
+        for(int val:map.values()){
+            if(val%2!=0) return false;
+        }
+
+        return true;
+    }
+}