Merge pull request #289 from AnishTiwari16/anish/burstBalloons

added burst balloons problem on DP

Author: gantavyamalviya

Dynamic Programming/burstBalloons.cpp

@@ -0,0 +1,56 @@
+// this is hard level DP problem on lc
+// https://leetcode.com/problems/burst-balloons/
+
+// both memoization and tabulation approaches
+class Solution
+{
+    // private:
+    //     int solve(int i, int j, vector<int>&nums,vector<vector<int>>&dp){
+    //         //base case
+    //         if(i>j){
+    //             return 0;
+    //         }
+    //         if(dp[i][j]!=-1){
+    //             return dp[i][j];
+    //         }
+    //         int ans = INT_MIN;
+    //         for(int ind = i;ind<=j;ind++){
+    //             //think in terms as we are bursting last balloon
+    //             int cost = nums[ind]*nums[i-1]*nums[j+1] + solve(i,ind-1,nums,dp) + solve(ind+1,j,nums,dp);
+    //             ans = max(ans,cost);
+    //         }
+    //         return dp[i][j] = ans;
+    //     }
+public:
+    int maxCoins(vector<int> &nums)
+    {
+        // mcm as we are given according to ordering and maxi so use dp
+        nums.push_back(1);
+        nums.insert(nums.begin(), 1);
+        //[1,3,1,5,8,1]
+        int n = nums.size();
+        // vector<vector<int>>dp(n,vector<int>(n,-1));
+        // return solve(1,n-2,nums,dp);
+        vector<vector<int>> dp(n + 2, vector<int>(n + 2, 0));
+
+        // think in terms of reverse of memo
+        for (int i = n - 2; i >= 1; i--)
+        {
+            for (int j = 1; j <= n - 2; j++)
+            {
+                if (i > j)
+                {
+                    continue;
+                }
+                int ans = INT_MIN;
+                for (int ind = i; ind <= j; ind++)
+                {
+                    int cost = nums[ind] * nums[i - 1] * nums[j + 1] + dp[i][ind - 1] + dp[ind + 1][j];
+                    ans = max(ans, cost);
+                }
+                dp[i][j] = ans;
+            }
+        }
+        return dp[1][n - 2];
+    }
+};