Welcome to Subscribe On Youtube

312. Burst Balloons

Description

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 <= 300
• 0 <= nums[i] <= 100

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int maxCoins(int[] nums) {
          int[] vals = new int[nums.length + 2];
          vals[0] = 1;
          vals[vals.length - 1] = 1;
          System.arraycopy(nums, 0, vals, 1, nums.length);
          int n = vals.length;
          int[][] dp = new int[n][n];
          for (int l = 2; l < n; ++l) {
              for (int i = 0; i + l < n; ++i) {
                  int j = i + l;
                  for (int k = i + 1; k < j; ++k) {
                      dp[i][j]
                          = Math.max(dp[i][j], dp[i][k] + dp[k][j] + vals[i] * vals[k] * vals[j]);
                  }
              }
          }
          return dp[0][n - 1];
      }
  }
  

• class Solution {
  public:
      int maxCoins(vector<int>& nums) {
          nums.insert(nums.begin(), 1);
          nums.push_back(1);
          int n = nums.size();
          vector<vector<int>> dp(n, vector<int>(n));
          for (int l = 2; l < n; ++l) {
              for (int i = 0; i + l < n; ++i) {
                  int j = i + l;
                  for (int k = i + 1; k < j; ++k) {
                      dp[i][j] = max(dp[i][j], dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j]);
                  }
              }
          }
          return dp[0][n - 1];
      }
  };
  

• class Solution:
      def maxCoins(self, nums: List[int]) -> int:
          nums = [1] + nums + [1]
          n = len(nums)
          dp = [[0] * n for _ in range(n)]
          for l in range(2, n):
              for i in range(n - l):
                  j = i + l
                  for k in range(i + 1, j):
                      dp[i][j] = max(
                          dp[i][j], dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j]
                      )
          return dp[0][-1]
  
  

• func maxCoins(nums []int) int {
  	vals := make([]int, len(nums)+2)
  	for i := 0; i < len(nums); i++ {
  		vals[i+1] = nums[i]
  	}
  	n := len(vals)
  	vals[0], vals[n-1] = 1, 1
  	dp := make([][]int, n)
  	for i := 0; i < n; i++ {
  		dp[i] = make([]int, n)
  	}
  	for l := 2; l < n; l++ {
  		for i := 0; i+l < n; i++ {
  			j := i + l
  			for k := i + 1; k < j; k++ {
  				dp[i][j] = max(dp[i][j], dp[i][k]+dp[k][j]+vals[i]*vals[k]*vals[j])
  			}
  		}
  	}
  	return dp[0][n-1]
  }
  

• function maxCoins(nums: number[]): number {
      let n = nums.length;
      let dp = Array.from({ length: n + 1 }, v => new Array(n + 2).fill(0));
      nums.unshift(1);
      nums.push(1);
      for (let i = n - 1; i >= 0; --i) {
          for (let j = i + 2; j < n + 2; ++j) {
              for (let k = i + 1; k < j; ++k) {
                  dp[i][j] = Math.max(nums[i] * nums[k] * nums[j] + dp[i][k] + dp[k][j], dp[i][j]);
              }
          }
      }
      return dp[0][n + 1];
  }
  
  

All Problems

All Solutions