Merge pull request #396 from Noob-coder-007/main

Added Burst_Balloons.cpp

Author: Pradeepsingh61

CPP/dynamic_programming/Burst_Balloons.cpp

@@ -0,0 +1,77 @@
+/*
+ * Burst Balloons (Dynamic Programming)
+ *
+ * Description:
+ * You are given n balloons, each balloon has a number on it represented by an array nums[].
+ * When you burst a balloon i, you gain coins equal to:
+ *      nums[left] * nums[i] * nums[right]
+ * where left and right are the adjacent balloons still remaining.
+ * After bursting, balloon i is removed and the array shrinks.
+ * Return the maximum coins you can collect by bursting balloons wisely.
+ *
+ * Approach:
+ * - Add virtual balloons with value 1 at both ends for easier handling.
+ * - Use Dynamic Programming (DP) on intervals.
+ *   dp[i][j] = maximum coins obtainable by bursting balloons between index i and j (exclusive).
+ * - For each range (i, j), consider every balloon k between i and j as the last balloon to burst:
+ *       dp[i][j] = max(dp[i][j], nums[i]*nums[k]*nums[j] + dp[i][k] + dp[k][j])
+ * - Final answer = dp[0][n-1]
+ *
+ * Time Complexity: O(n^3)
+ * Space Complexity: O(n^2)
+ */
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    int maxCoins(vector<int>& nums) {
+        int n = nums.size();
+        nums.insert(nums.begin(), 1);
+        nums.push_back(1);
+        vector<vector<int>> dp(n + 2, vector<int>(n + 2, 0));
+
+        // len is the distance between i and j
+        for (int len = 2; len < n + 2; len++) {
+            for (int i = 0; i + len < n + 2; i++) {
+                int j = i + len;
+                for (int k = i + 1; k < j; k++) {
+                    dp[i][j] = max(dp[i][j], nums[i]*nums[k]*nums[j] + dp[i][k] + dp[k][j]);
+                }
+            }
+        }
+
+        return dp[0][n + 1];
+    }
+};
+
+int main() {
+    Solution sol;
+
+    // Test Case 1
+    cout << "Test Case 1:" << endl;
+    vector<int> nums1 = {3, 1, 5, 8};
+    cout << "Balloons: [3, 1, 5, 8]" << endl;
+    cout << "Max Coins: " << sol.maxCoins(nums1) << endl << endl; // 167
+
+    // Test Case 2
+    cout << "Test Case 2:" << endl;
+    vector<int> nums2 = {1, 5};
+    cout << "Balloons: [1, 5]" << endl;
+    cout << "Max Coins: " << sol.maxCoins(nums2) << endl << endl; // 10
+
+    // Test Case 3
+    cout << "Test Case 3:" << endl;
+    vector<int> nums3 = {9, 76, 64, 21};
+    cout << "Balloons: [9, 76, 64, 21]" << endl;
+    cout << "Max Coins: " << sol.maxCoins(nums3) << endl << endl;
+
+    // Test Case 4 (Single balloon)
+    cout << "Test Case 4:" << endl;
+    vector<int> nums4 = {7};
+    cout << "Balloons: [7]" << endl;
+    cout << "Max Coins: " << sol.maxCoins(nums4) << endl << endl; // 7
+
+    return 0;
+}