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

Solution: Dynamic Programming

• Time complexity: O(n³)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const maxCoins = function (nums) {
  const n = nums.length;
  const coins = [1, ...nums, 1];
  const dp = new Array(n + 2)
    .fill('')
    .map(_ => new Array(n + 2).fill(0));

  for (let left = n; left >= 1; left--) {
    for (let right = left; right <= n; right++) {
      let max = 0;

      for (let index = left; index <= right; index++) {
        const current = coins[left - 1] * coins[index] * coins[right + 1];

        max = Math.max(max, current + dp[left][index - 1] + dp[index + 1][right]);
      }
      dp[left][right] = max;
    }
  }
  return dp[1][n];
};