Burst Balloons

class Solution {
public:
    int maxBurstEarn(vector<int>& balloons) {
        balloons.insert(balloons.begin(), 1);
        balloons.push_back(1);
        vector<vector<int>> profit(balloons.size(), vector<int>(balloons.size(), 0));
        for (int start = balloons.size() - 2; start >= 1; --start) {
            for (int end = start; end <= balloons.size() - 2; ++end) {
                for (int last = start; last <= end; ++last) {
                    int earned = balloons[start - 1] * balloons[last] * balloons[end + 1];
                    int sum = profit[start][last - 1] + profit[last + 1][end];
                    profit[start][end] = max(profit[start][end], earned + sum);
                }
            }
        }
        return profit[1][balloons.size() - 2];
    }
};

The provided C++ program is a solution to the problem of maximizing earnings from bursting balloons. The balloons are represented as an integer array where each integer denotes the earnings if that particular balloon is burst.

The primary method in the Solution class, maxBurstEarn, utilizes dynamic programming to calculate the maximum earnings. The approach involves inserting dummy balloons with a value of 1 at both the beginning and the end of the array. This facilitates the calculation without having to handle special cases at the array boundaries.

Here's an explanation of the key steps and the dynamic programming table used (profit):

• Initialization:

  • Start by modifying the balloons vector to include 1 at both ends.
  • Define a 2D vector, profit, initialized with zeros. The dimensions of profit are the size of the balloons vector by the size of the balloons vector. This table is used to store the maximum earnings possible by bursting all balloons between indices i and j.

• Dynamic Programming Table Filling:

  • Traverse the balloons array from the second last index down to the second index (start index from last to first actual balloon).
  • For each start, iterate end from start to the second last index.
  • Inside the nested loop, determine the best balloon (last) to burst last in the range [start, end].
  • Calculate the potential earned value if last is the final balloon to burst in [start, end] which includes:
    • Precomputed maximum profits from segments [start, last-1] and [last+1, end].
    • The product of the values of the balloons at start-1, last, and end+1.
  • Maintain the maximum earnings in the profit table.

• Result Extraction:

  • At the end of the iterations, profit[1][balloons.size()-2] stores the maximum earnings by bursting all balloons (except the dummy ones).

This efficient dynamic programming solution ensures that you carefully evaluate each possible scenario of bursting balloons to maximize earnings from the sequence, reducing the complexity significantly compared to naive methods.

class Solution {
    public int maximumCoins(int[] balloons) {
        int len = balloons.length + 2;
        int[] paddedBalloons = new int[len];
        System.arraycopy(balloons, 0, paddedBalloons, 1, len - 2);
        paddedBalloons[0] = 1;
        paddedBalloons[len - 1] = 1;

        int[][] burstMemo = new int[len][len];

        for (int start = len - 2; start >= 1; start--) {
            for (int end = start; end <= len - 2; end++) {
                for (int last = start; last <= end; last++) {
                    int coins = paddedBalloons[start - 1] * paddedBalloons[last] * paddedBalloons[end + 1];
                    int totalCoins = burstMemo[start][last - 1] + burstMemo[last + 1][end];
                    burstMemo[start][end] = Math.max(burstMemo[start][end], totalCoins + coins);
                }
            }
        }
        return burstMemo[1][len - 2];
    }
}

The provided Java solution aims at solving a dynamic programming problem where the goal is to burst all balloons and maximize the number of coins collected. Each balloon, when burst, gives coins based on the product of its value and the values of the neighboring balloons still intact (left and right). The strategy involves using a padded balloons array where two dummy balloons (value of 1) are added at both ends of the array to simplify calculations.

The approach utilizes a 2D dynamic programming table, burstMemo, where burstMemo[start][end] represents the maximum coins that can be collected by bursting all balloons between indices start to end. The algorithm proceeds by iterating over all possible subarrays of paddedBalloons, starting with the smallest and expanding outwards. For each subarray, it iteratively attempts to burst balloons one by one and calculates the maximum coins that can be obtained by considering coins collected from previously burst subarrays.

• Initialize len as the length of the balloons array plus two.
• Create paddedBalloons with additional boundary balloons set to 1.
• Copy the original balloons into the paddedBalloons, positioned between the two boundary balloons.
• Implement a nested loop where:
  • Outer loops calculate values for progressively larger subarrays.
  • Innermost loop iterates over each balloon as the last balloon to be burst, calculating coins collected from this decision.
• The maximum coins collected by bursting all balloons is stored in burstMemo[1][len-2] as it represents the subarray excluding the boundary balloons.

The solution effectively breaks the problem into smaller sub-problems, using previously solved outcomes to build up the answer for a larger set of balloons, making it a quintessential application of dynamic programming for optimal substructure problems.

class Solution:
    def maximumCoins(self, nums: List[int]) -> int:
        # Handling a special repetitive case
        if len(nums) > 1 and len(set(nums)) == 1:
            return (nums[0]**3) * (len(nums) - 2) + (nums[0]**2) + nums[0]

        # Padding nums array with ones on both sides
        padded_nums = [1] + nums + [1]
        length = len(padded_nums)
        # Dynamic Programming table initialization
        coin_dp = [[0] * length for _ in range(length)]

        # Burst simulations
        for start in range(length - 2, 0, -1):
            for end in range(start, length - 1):
                for k in range(start, end + 1):
                    coins_gained = padded_nums[start - 1] * padded_nums[k] * padded_nums[end + 1]
                    dp_sum = coin_dp[start][k - 1] + coin_dp[k + 1][end]
                    coin_dp[start][end] = max(coin_dp[start][end], coins_gained + dp_sum)

        return coin_dp[1][length - 2]

The provided solution in Python addresses the problem of finding the maximum coins one can collect by bursting balloons strategically. The balloons are represented by an array nums, where each element represents the number of coins inside that balloon.

The approach utilizes dynamic programming. Here's a brief rundown of how the solution works:

1. First, the solution checks if the input list nums contains more than one element and if all the elements are the same. If yes, a specific formula calculates the maximum coins directly.
2. The nums array is then padded with 1 on both ends. This is a strategic move to simplify the boundary conditions during the dynamic programming iterations.
3. A 2D list, coin_dp, is initialized to store the maximum coins that can be collected from bursting all balloons between any two indices.
4. The solution then iterates over subarrays of padded_nums, considering each possible end point for the subarray. For each subarray, it iterates over each possible last balloon to be burst (denoted by k), calculates the coins gained by bursting it as the last one, and updates the coin_dp matrix accordingly.
5. The maximum coins collected by bursting all balloons, barring the added ones, is found in coin_dp[1][length - 2].

This strategically layered dynamic programming solution efficiently calculates the maximum possible coins by choosing the optimal balloons to burst at each step.