LeetCode-in-Java

312. Burst Balloons

Hard

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

public class Solution {
    public int maxCoins(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        int[][] dp = new int[nums.length][nums.length];
        return balloonBurstDp(nums, dp);
    }

    private int balloonBurstDp(int[] nums, int[][] dp) {
        for (int gap = 0; gap < nums.length; gap++) {
            for (int si = 0, ei = gap; ei < nums.length; si++, ei++) {
                int l = (si - 1 == -1) ? 1 : nums[si - 1];
                int r = (ei + 1 == nums.length) ? 1 : nums[ei + 1];
                int maxAns = (int) -1e7;
                for (int cut = si; cut <= ei; cut++) {
                    int leftAns = si == cut ? 0 : dp[si][cut - 1];
                    int rightAns = ei == cut ? 0 : dp[cut + 1][ei];
                    int myAns = leftAns + l * nums[cut] * r + rightAns;
                    maxAns = Math.max(maxAns, myAns);
                }
                dp[si][ei] = maxAns;
            }
        }
        return dp[0][nums.length - 1];
    }
}

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