312. Burst Balloons - Detailed Explanation

Problem Statement

You are given an array of integers, where each element represents the number on a balloon. When you burst a balloon at index i, you gain coins equal to the product of the numbers on the balloon at index i and its adjacent balloons. After bursting a balloon, it is removed from the array, and its adjacent balloons become neighbors. To simplify edge cases, you can imagine that there are two extra balloons with a value of 1 at both ends of the array (i.e., nums[-1] = nums[n] = 1, where n is the length of the original array).

The goal is to burst all the balloons in an order that maximizes the total number of coins collected. Return the maximum coins you can collect by bursting the balloons in an optimal order.

Example 1

• Input: nums = [3,1,5,8]
• Output: 167
• Explanation:
  One optimal order is:
  1. Burst balloon 1 (value 1):
     • Coins gained = 3 * 1 * 5 = 15
     • New array becomes [3,5,8]
  2. Burst balloon 5 (value 5):
     • Coins gained = 3 * 5 * 8 = 120
     • New array becomes [3,8]
  3. Burst balloon 3 (value 3):
     • Coins gained = 1 * 3 * 8 = 24
     • New array becomes [8]
  4. Burst balloon 8 (value 8):
     • Coins gained = 1 * 8 * 1 = 8
  • Total coins = 15 + 120 + 24 + 8 = 167

Example 2

• Input: nums = [1,5]
• Output: 10
• Explanation:
  With the imaginary boundaries, the array becomes [1,1,5,1].
  • If you burst balloon 1 (the one in the middle with value 1):
    Coins = 1 * 1 * 5 = 5, new array: [1,5,1].
  • Then burst balloon 5:
    Coins = 1 * 5 * 1 = 5, new array: [1,1].
  • Finally burst the last balloon:
    Coins = 1 * 1 * 1 = 1, but note that there might be a better order to reach maximum. The optimal sequence, however, gives a total of 10 coins.

Hints

1. Add Boundary Balloons:
   To avoid handling edge cases separately, add two virtual balloons with value 1 at both ends of the array. This transforms the problem so that every balloon burst is now "internal" to the augmented array.

2. Interval Dynamic Programming:
   Think in terms of intervals. Define a DP function that represents the maximum coins that can be obtained by bursting all the balloons in a specific interval (excluding the boundaries of the interval).

3. Choose the Last Balloon to Burst:
   Instead of considering the order in which to burst the balloons, think in reverse: choose which balloon in an interval will be burst last. When a balloon is burst last, its adjacent balloons are fixed and come from the boundaries of the interval.

Approach: Dynamic Programming (Interval DP)

Concept

• DP Definition:
  Let dp[i][j] be the maximum coins you can collect by bursting all the balloons between indices i and j in the augmented array (not including i and j themselves).
  • The augmented array is defined as:
    [ nums' = [1] + nums + [1] ]
• Recurrence Relation:
  For an interval (i, j), consider every possible index k where i < k < j. If balloon k is the last to be burst in that interval, then: [ dp[i][j] = \max_{i < k < j} \{ dp[i][k] + dp[k][j] + nums'[i] \times nums'[k] \times nums'[j] \} ] Here, nums'[i] and nums'[j] are the boundaries of the interval and are not burst until the end.

Steps

1. Augment the Array:
   Create a new array nums' by adding 1 at the beginning and the end of nums.

2. Initialize DP Table:
   Create a 2D dp array of size n x n, where n is the length of the augmented array, and initialize all values to 0.
   Base cases: When the interval is empty (i.e., when j = i+1), dp[i][j] = 0.

3. Fill the DP Table:

   • Iterate over the length of intervals starting from 2 (since an interval with two elements contains no balloon to burst) up to n.
   • For each interval (i, j) of the current length, iterate over all possible k (i < k < j) and update: [ dp[i][j] = \max(dp[i][j], dp[i][k] + dp[k][j] + nums'[i] \times nums'[k] \times nums'[j]) ]

4. Final Answer:
   The final answer is stored in dp[0][n-1], representing the maximum coins obtained from bursting all balloons in the full augmented array.

Complexity Analysis

• Time Complexity:
  The dp table has O(n²) entries, and for each entry, we iterate over O(n) possible choices for k. Thus, the total time complexity is O(n³).

• Space Complexity:
  The dp table uses O(n²) space.

Step-by-Step Walkthrough with Example

Consider nums = [3,1,5,8].
Augmented array: nums' = [1, 3, 1, 5, 8, 1].
Let n = 6.

• Base Cases:
  For every i, dp[i][i+1] = 0 since there are no balloons between i and i+1.

• Interval Length 2:
  For interval (0,2): Only k=1 is valid.
  dp[0][2] = dp[0][1] + dp[1][2] + (1 * 3 * 1) = 0 + 0 + 3 = 3
  Similarly, fill for other intervals of length 2.

• Increasing Interval Lengths:
  For each interval (i, j) with j-i >= 2, consider each k in (i, j) and update dp[i][j] accordingly.

• Final Answer:
  dp[0][5] will hold the maximum coins obtainable, which in this example is 167.

Python Code

Java Code

Step-by-Step Walkthrough and Visual Examples

1. Augment the Array:
   Original nums: [3, 1, 5, 8]
   Augmented nums: [1, 3, 1, 5, 8, 1]

2. DP Table Initialization:
   Create a 6x6 dp table with all values initialized to 0.

3. Filling the DP Table:

   • For intervals of length 2, dp[i][i+2] is computed using the single balloon between i and i+2.
   • For longer intervals, consider every possible last balloon to burst within the interval.
   • For example, to compute dp[0][5] (full interval), try bursting balloon at each k from 1 to 4:
     • If k = 1: dp[0][5] = dp[0][1] + dp[1][5] + 1 * 3 * 1
     • If k = 2: dp[0][5] = dp[0][2] + dp[2][5] + 1 * 1 * 1
     • And so on, taking the maximum value among these.

4. Final Answer:
   After filling the dp table, dp[0][5] contains the maximum coins obtainable from bursting all balloons optimally, which in this case is 167.

Common Pitfalls

• Not Augmenting the Array:
  Failing to add the boundary balloons (1's) complicates the implementation because of edge conditions.

• Incorrect DP Boundaries:
  Remember that dp[i][j] represents the maximum coins for bursting all balloons in (i, j) — the boundaries i and j are not burst until the final step.

• Iteration Order:
  It is crucial to iterate the intervals by increasing length to ensure that smaller subproblems are solved before solving larger intervals.

Alternative Variations

• Greedy Approaches:
  There are no greedy algorithms that solve this problem optimally due to the complex interdependencies between balloon bursts.

• Memoization:
  Instead of bottom-up dp, one can also use top-down recursion with memoization to solve the same recurrence relation.

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