added partition equal subset problem in cpp

DSA SOLUTIONS (TILL TREE & BASIC GRAPH CONCEPTS)/DP/PartitionEqualSubset.cpp

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+/**
+ * @brief Partition Equal Subset Sum Problem
+ * 
+ * Problem: Determine if an array can be partitioned into two subsets 
+ * with equal sum
+ * 
+ * Key Approach:
+ * - Top-down Dynamic Programming with Memoization
+ * - Recursively explore subset sum possibilities
+ * - Time Complexity: O(n*sum)
+ * - Space Complexity: O(n*sum)
+ */
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    // Recursive helper function with memoization
+    bool f(vector<int> &arr, int i, int sum, vector<vector<int>> &dp) {
+        // Base case: reached first element
+        if (i == 0) {
+            return arr[i] == sum;
+        }
+
+        // Return memoized result if already computed
+        if (dp[i][sum] != -1) {
+            return dp[i][sum];
+        }
+
+        // If target sum is zero, subset found
+        if (sum == 0) {
+            return true;
+        }
+
+        // Invalid sum
+        if (sum < 0) {
+            return false;
+        }
+
+        bool a = false;
+        // Option 1: Include current element if possible
+        if (arr[i] <= sum) {
+            a = f(arr, i-1, sum-arr[i], dp);
+        }
+
+        // Option 2: Exclude current element
+        bool b = f(arr, i-1, sum, dp);
+
+        // Memoize and return result
+        return dp[i][sum] = a || b;
+    }
+
+    // Main function to check partition possibility
+    bool canPartition(vector<int>& nums) {
+        // Calculate total sum
+        int sum = 0;
+        for (int x : nums) {
+            sum += x;
+        }
+
+        // If total sum is odd, cannot be partitioned equally
+        if (sum % 2 != 0) {
+            return false;
+        }
+
+        // Target sum is half of total
+        sum = sum / 2;
+
+        // Create memoization table
+        int n = nums.size();
+        vector<vector<int>> dp(n, vector<int>(sum+1, -1));
+
+        // Call recursive helper
+        return f(nums, n-1, sum, dp);
+    }
+};
+
+// Test cases to demonstrate solution
+int main() {
+    Solution sol;
+
+    // Test Case 1: Partitionable array
+    vector<int> nums1 = {1, 5, 11, 5};
+    cout << "Can Partition 1: " << (sol.canPartition(nums1) ? "True" : "False") << endl;
+
+    // Test Case 2: Non-partitionable array
+    vector<int> nums2 = {1, 2, 3, 5};
+    cout << "Can Partition 2: " << (sol.canPartition(nums2) ? "True" : "False") << endl;
+
+    // Test Case 3: Single element array
+    vector<int> nums3 = {1};
+    cout << "Can Partition 3: " << (sol.canPartition(nums3) ? "True" : "False") << endl;
+
+    return 0;
+}
+
+/* 
+ * Algorithm Walkthrough:
+ * 1. Check if total sum is even (prerequisite for equal partition)
+ * 2. Recursively explore subset sum with memoization
+ * 3. Two choices at each step:
+ *    a) Include current element if possible
+ *    b) Exclude current element
+ * 4. Find if a subset with target sum exists
+ */