<p>You are given <code>n</code> balloons, indexed from <code>0</code> to <code>n - 1</code>. Each balloon is painted with a number on it represented by an array <code>nums</code>. You are asked to burst all the balloons.</p>

<p>If you burst the <code>i<sup>th</sup></code> balloon, you will get <code>nums[i - 1] * nums[i] * nums[i + 1]</code> coins. If <code>i - 1</code> or <code>i + 1</code> goes out of bounds of the array, then treat it as if there is a balloon with a <code>1</code> painted on it.</p>

<p>Return <em>the maximum coins you can collect by bursting the balloons wisely</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,5,8]
<strong>Output:</strong> 167
<strong>Explanation:</strong>
nums = [3,1,5,8] --&gt; [3,5,8] --&gt; [3,8] --&gt; [8] --&gt; []
coins =  3*1*5    +   3*5*8   +  1*3*8  + 1*8*1 = 167</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5]
<strong>Output:</strong> 10
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 300</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
  
  //MCM +memo taking every kth balloon bursting in the last;
  
  
  int memo[501][501];
  int solve(vector<int> &nums,int i ,int j){
    if(i>j)return 0;
    if(memo[i][j]!=-1)return memo[i][j];
    
    int max_score=INT_MIN;
    for(int k=i;k<=j;k++){
      int curr_score=nums[i-1]*nums[k]*nums[j+1];
      int temp_score=curr_score+solve(nums,i,k-1)+solve(nums,k+1,j);
      if(temp_score>max_score)max_score=temp_score;
    }
    return memo[i][j]=max_score;
  }
    int maxCoins(vector<int>& nums) {
      memset(memo,-1,sizeof(memo));
      //-1 th position pe && nth pos pe 1 score milega (by example)
      nums.insert(nums.begin(),1);
      nums.push_back(1);
      
      return solve(nums,1,nums.size()-2);
    }
};
10800