2022-01-01 Daily-Challenge

Today I have done leetcode's January LeetCoding Challenge with cpp.

January LeetCoding Challenge 1

Description

Burst Balloons

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

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

Return the maximum coins you can collect by bursting the balloons wisely.

 

Example 1:

Input: nums = [3,1,5,8]
Output: 167
Explanation:
nums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> []
coins =  3*1*5    +   3*5*8   +  1*3*8  + 1*8*1 = 167

Example 2:

Input: nums = [1,5]
Output: 10

 

Constraints:

  • n == nums.length
  • 1 <= n <= 500
  • 0 <= nums[i] <= 100

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

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

Return the maximum coins you can collect by bursting the balloons wisely.

 

Example 1:

Input: nums = [3,1,5,8]
Output: 167
Explanation:
nums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> []
coins =  3*1*5    +   3*5*8   +  1*3*8  + 1*8*1 = 167

Example 2:

Input: nums = [1,5]
Output: 10

 

Constraints:

  • n == nums.length
  • 1 <= n <= 500
  • 0 <= nums[i] <= 100

Solution

int a[502];
int dp[502][502];
class Solution {
public:
  int maxCoins(vector<int>& nums) {
    memset(dp, 0, sizeof(dp));
    int n = nums.size();
    
    a[0] = 1;
    a[n+1] = 1;
    for(int i = 0; i < n; i++) {
      a[i+1] = nums[i];
    }
    
    for(int l = 3; l <= n+2; l++) {
      for(int i = 0; i + l - 1 <= n + 1; i++){
        for(int j = 1; j <= l - 2; j++){
          dp[i][i+l-1] = max(dp[i][i + j] + dp[i + j][i+l-1] + a[i] * a[i+j] * a[i + l - 1], dp[i][i+l-1]);
        }
      }
    }
    return dp[0][n+1];
  }
};

// Accepted
// 70/70 cases passed (128 ms)
// Your runtime beats 99.93 % of cpp submissions
// Your memory usage beats 95.76 % of cpp submissions (8.9 MB)