2818. Apply Operations to Maximize Score - Detailed Explanation

Problem Statement

You are given a 0-indexed array of positive integers nums representing tiles in a row. You can perform the following merge operation any number of times (in any order) until no more operations can be performed:

• Merge Operation:
  If two adjacent tiles have the same value, you may merge them into a single tile whose value is equal to the sum of the two. When you merge them, you earn points equal to that new (doubled) value.

Your goal is to maximize the total score earned by choosing the order in which you merge tiles optimally. (Note that when a merge occurs, the array shrinks and the resulting tile might become adjacent to other tiles of equal value, allowing further merges.)

Constraints:

• 1 ≤ nums.length ≤ 100
• 1 ≤ nums[i] ≤ 10⁴

Example Inputs, Outputs, and Explanations

Example 1:

• Input: nums = [2, 2, 2]
• Output: 4
• Explanation:
  There are two possible merge operations initially:
  1. Merge the first two tiles: [2, 2, 2] → [4, 2] and earn 4 points. (No further merge is possible because 4 ≠ 2.)
  2. Merge the last two tiles: [2, 2, 2] → [2, 4] and earn 4 points.
     In either case, the maximum score is 4.

Example 2:

• Input: nums = [2, 2, 2, 2]
• Output: 16
• Explanation:
  One optimal merging sequence is:
  1. Merge the first pair: [2, 2, 2, 2] → [4, 2, 2] with score = 4.
  2. Merge the last pair in the new array: [4, 2, 2] → [4, 4] with additional score = 4 (total so far = 8).
  3. Merge the resulting two 4’s: [4, 4] → [8] with additional score = 8 (total score = 16).
     Hence, the maximum score is 16.

Example 3:

• Input: nums = [1, 1, 1, 1]
• Output: 8
• Explanation:
  An optimal sequence is:
  1. Merge the first two 1’s: [1, 1, 1, 1] → [2, 1, 1] with score = 2.
  2. Then merge the last two 1’s: [2, 1, 1] → [2, 2] with additional score = 2 (total so far = 4).
  3. Finally, merge the two 2’s: [2, 2] → [4] with additional score = 4 (total = 8).
     So, the maximum score is 8.

Hints for the Solution

1. Recursive Splitting:
   Think about whether you can “fully merge” a contiguous subarray into a single tile. If so, what would be the value and what would be the score earned in doing so? This suggests a recursive (or interval DP) approach.

2. Interval Dynamic Programming:
   Since you can merge an interval in many different ways (by choosing different “split points”), consider an interval DP where for every subarray you try all possible partitions. For an interval that is “mergeable” into one tile, you can store both the resulting tile’s value and the maximum score earned for that merge.

Approaches

In the discussion below we describe three different approaches.

Approach 1: Brute Force Recursion with Backtracking

• Idea:
  At each step, look for any adjacent pair of equal tiles. For each such pair, simulate merging them (earning points equal to the new value) and recursively solve the problem for the new array.
• Pros:
  Conceptually simple and “tries everything.”
• Cons:
  The number of merge sequences is exponential. Even with memoization on small inputs the time cost can be prohibitive.

Approach 2: Interval Dynamic Programming (DP)

• Idea:
  Define a DP function on intervals:
  • Let dp(i, j) be the maximum score obtainable if the subarray nums[i…j] can be merged into one single tile.
  • Also, let val(i, j) (or store as part of a pair) be the value of the resulting merged tile.
  • Base case: For a single tile, no merge is performed, so the score is 0 and val(i, i) = nums[i].
  • Recurrence: For each interval [i, j] (with i < j), try all possible splits. For a given split point k (i ≤ k < j):
    • If the left segment [i, k] can fully merge into a tile with value v and the right segment [k+1, j] can fully merge into a tile with the same value v, then you can merge these two resulting tiles. The new tile will have value 2 * v, and you earn an additional score of 2 * v. Combine this with the scores from the two halves.
  • Since there might be merges in disjoint parts of the array, use another DP to decide which non‑overlapping intervals (subarrays) to choose so that the total score is maximized.
• Time Complexity:
  O(n³) time and O(n²) space (which is acceptable for n ≤ 100).

Approach 3: Greedy (Not Always Optimal)

• Idea:
  One might consider simply scanning from left to right and merging any valid adjacent pair immediately.
• Pros and Cons:
  Although this runs in O(n) time, it may miss merge opportunities (for example, merging a later pair might be more “chain‑friendly” than merging the very first eligible pair). Thus, a greedy strategy does not guarantee an optimal score.

Step‑by‑Step Walkthrough (Using Interval DP) for Example 2

Let’s see how the interval DP could compute the answer for nums = [2, 2, 2, 2].

1. Single Tiles (Base Case):
   For each index i, set:

   • dp(i, i) = 0
   • val(i, i) = 2

2. Intervals of Length 2:
   For subarray [0, 1]: Since 2 equals 2, you can merge them into one tile valued at 4 and earn score 4.
   Similarly, [1, 2] and [2, 3] are mergeable into 4 with score 4.

3. Intervals of Length 3:
   For [0, 2], check possible splits:

   • Split between index 0 and 1:
     • Left is [0, 0] → tile 2, score 0.
     • Right is [1, 2] → merged tile 4, score 4.
     • They are not mergeable since 2 ≠ 4.
   • Split between index 1 and 2:
     • Left is [0, 1] → tile 4, score 4.
     • Right is [2, 2] → tile 2, score 0.
     • Not mergeable (4 ≠ 2).
       Thus, interval [0, 2] cannot be fully merged into one tile.

4. Intervals of Length 4:
   For [0, 3], try a partition at index 1:

   • Left is [0, 1] → tile 4, score 4.
   • Right is [2, 3] → tile 4, score 4.
     Since the two resulting tiles are equal (both 4), they can merge to form a tile of value 8. The additional score earned in this merge is 8.
     Total score = 4 (left) + 4 (right) + 8 (this merge) = 16.
     This yields dp(0, 3) = 16 and the merged tile is 8.

5. Combining Disjoint Merges:
   In many cases you may have intervals that do not span the entire array. To decide the best overall score you can use an auxiliary DP (e.g. best[i] for the prefix ending at index i) that picks non‑overlapping intervals. In this example, fully merging the entire array produces the maximum score.

Common Mistakes and Edge Cases

• Common Mistakes:

  • Not realizing that merging in one order may block later chain merges that yield higher points.
  • Trying a naive greedy “first available merge” strategy which can miss the opportunity to merge larger blocks.
  • Forgetting to properly combine results from disjoint intervals if you cannot merge the entire array into one tile.

• Edge Cases:

  • No Merge Possible: An input like [1, 2, 3] has no adjacent equal tiles, so the answer is 0.
  • Single Element: An input with one element yields a score of 0 because no operation can be performed.

Alternative Variations

• Instead of awarding a score equal to the merged value, the problem could be changed so that the score is computed in a different way (for example, based on the product of the two tiles).
• The merge operation might be allowed only a limited number of times.
• The problem may request returning not only the maximum score but also the final configuration of the array.

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