Burst Balloons LeetCode Solution

Problem – 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 <= 300
• 0 <= nums[i] <= 100

Burst Balloons LeetCode Solution in Java

public int maxCoins(int[] iNums) {
    int[] nums = new int[iNums.length + 2];
    int n = 1;
    for (int x : iNums) if (x > 0) nums[n++] = x;
    nums[0] = nums[n++] = 1;


    int[][] memo = new int[n][n];
    return burst(memo, nums, 0, n - 1);
}

public int burst(int[][] memo, int[] nums, int left, int right) {
    if (left + 1 == right) return 0;
    if (memo[left][right] > 0) return memo[left][right];
    int ans = 0;
    for (int i = left + 1; i < right; ++i)
        ans = Math.max(ans, nums[left] * nums[i] * nums[right] 
        + burst(memo, nums, left, i) + burst(memo, nums, i, right));
    memo[left][right] = ans;
    return ans;
}
// 12 ms

Burst Balloons LeetCode Solution in Python

class Solution:
    def maxCoins(self, nums):
        A = [1] + nums + [1]
        
        @lru_cache(None)
        def dfs(i, j):
            return max([A[i]*A[k]*A[j] + dfs(i,k) + dfs(k,j) for k in range(i+1, j)] or [0])
        
        return dfs(0, len(A) - 1)

Burst Balloons LeetCode Solution in C++

int maxCoins(vector<int>& nums) {
    int N = nums.size();
    nums.insert(nums.begin(), 1);
    nums.insert(nums.end(), 1);

    // rangeValues[i][j] is the maximum # of coins that can be obtained
    // by popping balloons only in the range [i,j]
    vector<vector<int>> rangeValues(nums.size(), vector<int>(nums.size(), 0));
    
    // build up from shorter ranges to longer ranges
    for (int len = 1; len <= N; ++len) {
        for (int start = 1; start <= N - len + 1; ++start) {
            int end = start + len - 1;
            // calculate the max # of coins that can be obtained by
            // popping balloons only in the range [start,end].
            // consider all possible choices of final balloon to pop
            int bestCoins = 0;
            for (int final = start; final <= end; ++final) {
                int coins = rangeValues[start][final-1] + rangeValues[final+1][end]; // coins from popping subranges
                coins += nums[start-1] * nums[final] * nums[end+1]; // coins from final pop
                if (coins > bestCoins) bestCoins = coins;
            }
            rangeValues[start][end] = bestCoins;
        }
    }
    return rangeValues[1][N];
}

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