Problem Description

You have a 0-indexed array nums containing positive integers. You can perform a specific merge operation on adjacent elements any number of times.

The operation works as follows:

• Select an index i where 0 <= i < nums.length - 1
• Check if nums[i] <= nums[i + 1]
• If the condition is met, replace nums[i + 1] with the sum nums[i] + nums[i + 1]
• Remove nums[i] from the array

Your goal is to find the largest possible value that can be obtained in the final array after performing any number of these operations.

For example, if you have [2, 3, 7, 9, 3]:

• You could merge indices 0 and 1: [2, 3] becomes [5], resulting in [5, 7, 9, 3]
• Then merge indices 0 and 1 again: [5, 7] becomes [12], resulting in [12, 9, 3]
• You cannot merge index 0 and 1 now since 12 > 9
• But you could have started differently to get a larger result

The key insight is that to maximize the final value, you should process the array from right to left. This greedy approach ensures that elements on the right become as large as possible, enabling more merge operations and ultimately producing the maximum possible value.

The solution traverses the array in reverse order (from index n-2 to 0). At each position i, if nums[i] <= nums[i + 1], it adds nums[i] to nums[i + 1] and effectively "removes" nums[i] by storing the sum at position i. After processing all possible merges, the maximum value in the modified array is returned as the answer.

Intuition

When we can merge adjacent elements where the left element is smaller or equal to the right element, we want to create the largest possible value. The key observation is that merging creates a larger number that might enable more merges.

Consider what happens when we merge from left to right versus right to left. If we start from the left, we might create a large number early on that prevents further merges. For instance, with [2, 3, 4], merging left to right gives us [5, 4], and we're stuck because 5 > 4. But if we merge from right to left, we get [2, 7], then [9], achieving a larger result.

The crucial insight is that we want to build up the largest possible "accumulator" value that can keep absorbing elements. Since we can only merge when nums[i] <= nums[i + 1], we want nums[i + 1] to be as large as possible to allow more merges with elements to its left.

Think of it like a snowball rolling down a hill. If we roll from right to left (processing in reverse), the snowball at position i + 1 has already accumulated all possible values from the right. When we check if nums[i] can merge with it, we're checking against the maximum possible value at that position. If it can merge, we add to the snowball; if not, we start a new snowball.

This greedy approach works because once we've processed all elements to the right of position i, we've already found the optimal way to merge them. The decision at position i only depends on whether we can merge with the accumulated result to the right, not on any future decisions to the left. By processing right to left, we ensure each merge decision is made with complete information about what's possible on the right side.

Learn more about Greedy patterns.

Solution Approach

The implementation follows a greedy strategy of merging elements from right to left. Here's how the algorithm works:

Algorithm Steps:

1. Traverse in Reverse: Start from index len(nums) - 2 and move towards index 0. We start at n-2 because we need to compare each element with its right neighbor.

2. Check Merge Condition: At each position i, check if nums[i] <= nums[i + 1]. This is the condition that allows us to perform the merge operation.

3. Perform Merge: If the condition is satisfied, update nums[i] to nums[i] + nums[i + 1]. This effectively merges the two elements and stores the result at position i. We don't actually delete nums[i + 1] from the array; instead, we'll ignore it in future iterations.

4. Find Maximum: After processing all positions, return the maximum value in the array using max(nums).

Implementation Details:

class Solution:
    def maxArrayValue(self, nums: List[int]) -> int:
        for i in range(len(nums) - 2, -1, -1):
            if nums[i] <= nums[i + 1]:
                nums[i] += nums[i + 1]
        return max(nums)

The key insight is that when we update nums[i] = nums[i] + nums[i + 1], the value at nums[i + 1] becomes irrelevant for future iterations. Each element nums[i] will either:

• Absorb its right neighbor if nums[i] <= nums[i + 1], becoming the new accumulated value
• Remain unchanged if nums[i] > nums[i + 1], acting as a barrier that prevents further merges

Time and Space Complexity:

• Time Complexity: O(n) where n is the length of the array. We traverse the array once and then find the maximum in another pass.
• Space Complexity: O(1) as we modify the array in-place without using additional data structures.

This approach guarantees the maximum possible value because at each step, we're making the locally optimal choice that enables the most future merges, and processing from right to left ensures we build up the largest possible accumulated values.

Example Walkthrough

Let's walk through the solution with the array [2, 3, 7, 9, 3].

Initial state: nums = [2, 3, 7, 9, 3]

We traverse from right to left, starting at index 3 (second-to-last element).

Step 1: i = 3

• Compare nums[3] with nums[4]: Is 9 <= 3? No.
• Since 9 > 3, we cannot merge. No change.
• Array remains: [2, 3, 7, 9, 3]

Step 2: i = 2

• Compare nums[2] with nums[3]: Is 7 <= 9? Yes.
• We can merge! Update nums[2] = 7 + 9 = 16
• Array becomes: [2, 3, 16, 9, 3] (note: nums[3] is now ignored)

Step 3: i = 1

• Compare nums[1] with nums[2]: Is 3 <= 16? Yes.
• We can merge! Update nums[1] = 3 + 16 = 19
• Array becomes: [2, 19, 16, 9, 3] (note: nums[2] is now ignored)

Step 4: i = 0

• Compare nums[0] with nums[1]: Is 2 <= 19? Yes.
• We can merge! Update nums[0] = 2 + 19 = 21
• Array becomes: [21, 19, 16, 9, 3] (note: nums[1] is now ignored)

Final step: Find the maximum value in the array.

• max([21, 19, 16, 9, 3]) = 21

The algorithm returns 21 as the maximum possible value.

Notice how the right-to-left traversal allowed us to build up a large accumulated value (16) at position 2, which then enabled further merges with elements to its left. If we had started from the left, we might have created barriers that prevented optimal merging.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array. The algorithm iterates through the array once from right to left (from index len(nums) - 2 to 0), performing constant-time operations (comparison and addition) at each step. Even though max(nums) is called at the end, which also takes O(n) time, the overall complexity remains O(n) since O(n) + O(n) = O(n).

The space complexity is O(1). The algorithm modifies the input array in-place and only uses a constant amount of extra space for the loop variable i. No additional data structures are created that depend on the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding the Direction of Traversal

The Problem: A common mistake is attempting to traverse the array from left to right instead of right to left. This greedy approach would fail to produce the maximum value.

Why It Fails: Consider the array [2, 3, 7, 9, 3]:

• Left-to-right approach:

  • Merge [2, 3] → [5, 7, 9, 3]
  • Merge [5, 7] → [12, 9, 3]
  • Cannot merge 12 and 9 (12 > 9)
  • Maximum value: 12

• Right-to-left approach:

  • Merge [9, 3] → [2, 3, 7, 12]
  • Merge [7, 12] → [2, 3, 19]
  • Merge [3, 19] → [2, 22]
  • Merge [2, 22] → [24]
  • Maximum value: 24

Solution: Always traverse from right to left to build up larger values that can absorb more elements:

for i in range(len(nums) - 2, -1, -1):  # Correct: right to left
    # NOT: for i in range(len(nums) - 1):  # Wrong: left to right

Pitfall 2: Modifying the Array While Tracking Indices

The Problem: Some might try to actually remove elements from the array during the merge operation, which would cause index management issues and complexity.

Why It Fails:

# Incorrect approach - actually removing elements
i = len(nums) - 2
while i >= 0:
    if nums[i] <= nums[i + 1]:
        nums[i] = nums[i] + nums[i + 1]
        nums.pop(i + 1)  # This changes array size!
    i -= 1  # Index becomes invalid after pop

Solution: Keep the array size constant and simply overwrite values. The unused values will be ignored when finding the maximum:

for i in range(len(nums) - 2, -1, -1):
    if nums[i] <= nums[i + 1]:
        nums[i] += nums[i + 1]  # Just update, don't remove

Pitfall 3: Edge Case Handling

The Problem: Forgetting to handle arrays with only one element or assuming the array has at least two elements.

Why It Fails: If nums = [5], the loop range(len(nums) - 2, -1, -1) would be range(-1, -1, -1), which executes zero iterations. However, max(nums) still correctly returns 5.

Solution: The current implementation actually handles this correctly! The loop naturally handles edge cases:

• Single element: Loop doesn't execute, max([5]) returns 5
• Two elements: Loop executes once for index 0
• Empty array would need separate handling if allowed by constraints

Pitfall 4: Attempting to Track Multiple Merge Paths

The Problem: Some might think they need dynamic programming or need to explore all possible merge sequences to find the optimal solution.

Why It Fails (in terms of complexity): While this would work, it's unnecessarily complex. The greedy right-to-left approach is provably optimal and much simpler.

Solution: Trust the greedy approach. Processing from right to left guarantees the maximum value because:

1. Making elements on the right as large as possible enables more merges
2. A larger right element can absorb more left elements
3. There's no benefit to keeping smaller separate values when they could be merged