added maximum subarray problem in cpp

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

@@ -0,0 +1,66 @@
+/**
+ * @brief Maximum Product Subarray Problem Solution
+ * 
+ * Problem: Find the contiguous subarray with the largest product
+ * 
+ * Key Approach:
+ * - Dynamic Programming with simultaneous tracking of max and min products
+ * - Handles both positive and negative numbers
+ * - O(n) time complexity, O(1) space complexity
+ */
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int maxProduct(vector<int>& nums) {
+    // Handle empty input
+    if (nums.empty()) return 0;
+
+    // Initialize variables to track max and min products
+    int max_prod = nums[0];  // Current max product
+    int min_prod = nums[0];  // Current min product (crucial for negative numbers)
+    int result = nums[0];    // Overall max product
+
+    // Iterate through array starting from second element
+    for (int i = 1; i < nums.size(); i++) {
+        // Swap max and min if current number is negative 
+        // This helps handle multiplications with negative numbers
+        if (nums[i] < 0) {
+            std::swap(max_prod, min_prod);
+        }
+
+        // Update max and min products
+        max_prod = std::max(nums[i], max_prod * nums[i]);
+        min_prod = std::min(nums[i], min_prod * nums[i]);
+
+        // Update overall result
+        result = std::max(result, max_prod);
+    }
+
+    return result;
+}
+
+// Test cases to demonstrate solution
+int main() {
+    // Test Case 1: Mixed positive and negative numbers
+    vector<int> nums1 = {2, 3, -2, 4};
+    cout << "Max Product 1: " << maxProduct(nums1) << endl;  // Expected: 6
+
+    // Test Case 2: All negative numbers
+    vector<int> nums2 = {-2, -3, -4};
+    cout << "Max Product 2: " << maxProduct(nums2) << endl;  // Expected: 12
+
+    // Test Case 3: Single element
+    vector<int> nums3 = {-2};
+    cout << "Max Product 3: " << maxProduct(nums3) << endl;  // Expected: -2
+
+    return 0;
+}
+
+/* 
+ * Algorithm Explanation:
+ * 1. Track both max and min products at each step
+ * 2. Swap max/min for negative numbers to handle sign changes
+ * 3. Choose between current number and product with previous elements
+ * 4. Continuously update overall max product
+ */