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This approach involves using a recursive function to try bursting each balloon and calculating the coins collected, while storing results of already computed subproblems to avoid redundant calculations. We pad the nums array with 1 at both ends to handle boundary conditions easily. For each subproblem defined by indices left and right, we choose a balloon i to burst last and calculate the coins.
Time Complexity: O(n^3), where n is the total number of balloons including the padding ones.
Space Complexity: O(n^2) due to memoization table.
1public class Solution {
2 public int MaxCoins(int[] nums) {
3 int n = nums.Length;
4 int[] numsWithBoundaries = new int[n + 2];
5 System.Array.Copy(nums, 0, numsWithBoundaries, 1, n);
6 numsWithBoundaries[0] = numsWithBoundaries[n + 1] = 1;
7
8 int[,] memo = new int[n + 2, n + 2];
9
10 return DP(0, n + 1, numsWithBoundaries, memo);
11 }
12
13 private int DP(int left, int right, int[] nums, int[,] memo) {
14 if (left + 1 == right) return 0;
15 if (memo[left, right] > 0) return memo[left, right];
16 int ans = 0;
17 for (int i = left + 1; i < right; i++) {
18 ans = Math.Max(ans, nums[left] * nums[i] * nums[right] + DP(left, i, nums, memo) + DP(i, right, nums, memo));
19 }
20 memo[left, right] = ans;
21 return ans;
22 }
23}C# implementation uses a similar logic pattern. A 2D array is used for memoization, and an auxiliary array is prepared with boundaries. The recursive function DP performs the calculations.
In this approach, we use a DP table to systematically build up solutions from smaller problems to a larger one without recursion. We create a dynamic programming matrix dp where dp[left][right] indicates the max coins to be obtained from subarray nums[left:right]. For each subarray defined by len, we consider bursting each possible k as the last balloon and compute the result using previously computed results.
Time Complexity: O(n^3)
Space Complexity: O(n^2)
1def
The Python solution uses a bottom-up dynamic programming table to iterate through all subarray lengths from 3 upwards since we initially padded the nums array. This way, we are able to systematically determine the most coins by considering each k element in this sub-array as the last balloon to burst and combining results from smaller subproblems.