Problem Description

You have an array of integers called nums. You can perform a specific operation on this array multiple times:

• Remove the first two elements from nums and calculate their sum as the "score" of that operation.

The key constraint is that every operation must have the same score. In other words, each time you remove the first two elements, their sum must be identical to the sum from all previous operations.

You need to keep performing this operation as long as:

1. The array has at least two elements remaining
2. The sum of the next two elements equals the required score

Your task is to find the maximum number of operations you can perform while maintaining the same score for all operations.

For example:

• If nums = [3, 2, 1, 4, 5], the first operation removes 3 and 2 with score 5
• The second operation would try to remove 1 and 4, also with score 5
• Since both operations have the same score (5), you can perform 2 operations
• The remaining element 5 cannot form another operation since we need at least 2 elements

The solution determines the required score from the first two elements (s = nums[0] + nums[1]), then iterates through the array in pairs, counting how many consecutive pairs from the beginning have this same sum.

Intuition

The key insight is that once we perform the first operation, the score is fixed - it must be the sum of the first two elements. We have no choice about this since we must start from the beginning of the array.

Since all subsequent operations must have this same score, and we always remove elements from the front of the array, we're essentially asking: "Starting from the beginning, how many consecutive pairs of elements have the same sum as the first pair?"

This leads to a straightforward greedy approach:

1. Calculate the required score as nums[0] + nums[1] - this is set in stone
2. Keep taking pairs from the front and check if their sum equals this required score
3. Stop as soon as we encounter a pair with a different sum, or when we run out of elements

Why does this work? Because:

• We must take elements from the front (as stated in the problem)
• We must maintain the same score for all operations
• Once we encounter a pair that doesn't match the required score, we cannot skip it and continue - we'd have to remove it first, which would violate the "same score" constraint

The solution is optimal because we're maximizing the number of operations by continuing as long as possible. There's no benefit to stopping early when valid pairs are available, and we cannot skip invalid pairs to reach valid ones later.

This is why the implementation simply iterates through the array in steps of 2, checking each pair against the initial sum s, and counting valid operations until we hit a mismatch or run out of elements.

Solution Approach

The implementation uses a simple traversal approach with no additional data structures needed.

Step-by-step walkthrough:

1. Calculate the target score:

   • s = nums[0] + nums[1]
   • This establishes the sum that all operations must match

2. Initialize variables:

   • ans = 0 to count the number of valid operations
   • n = len(nums) to store the array length

3. Traverse the array in pairs:

   • Use a loop with range(0, n, 2) to iterate through indices 0, 2, 4, 6, etc.
   • This ensures we're checking consecutive pairs from the front

4. For each pair, check two conditions:

   • i + 1 == n: Check if we've run out of elements (odd-length array or reached the end)
   • nums[i] + nums[i + 1] != s: Check if the current pair's sum doesn't match our target score

5. Handle the termination:

   • If either condition is true, break the loop immediately
   • Otherwise, increment ans by 1 to count this valid operation

6. Return the result:

   • ans contains the maximum number of operations performed

Time Complexity: O(n) where n is the length of the array, as we traverse the array at most once.

Space Complexity: O(1) as we only use a constant amount of extra space for variables s, ans, and loop counter i.

The algorithm is optimal because it processes each element at most once and stops as soon as an invalid operation is encountered, ensuring we get the maximum number of consecutive valid operations from the beginning of the array.

Example Walkthrough

Let's walk through the solution with nums = [3, 2, 6, 1, 4, 5, 2, 1].

Step 1: Calculate target score

• First two elements: nums[0] = 3, nums[1] = 2
• Target score s = 3 + 2 = 5
• All operations must have sum = 5

Step 2: Process pairs from the front

Operation 1 (i = 0):

• Check pair: nums[0] + nums[1] = 3 + 2 = 5
• Sum equals target score ✓
• Increment ans = 1
• Conceptually remove [3, 2], leaving [6, 1, 4, 5, 2, 1]

Operation 2 (i = 2):

• Check pair: nums[2] + nums[3] = 6 + 1 = 7
• Sum is 7, not 5 ✗
• Break immediately

Result: Maximum operations = 1

Let's try another example: nums = [3, 2, 1, 4, 5, 0]

Step 1: Calculate target score

• s = nums[0] + nums[1] = 3 + 2 = 5

Step 2: Process pairs

Operation 1 (i = 0):

• nums[0] + nums[1] = 3 + 2 = 5 ✓
• ans = 1

Operation 2 (i = 2):

• nums[2] + nums[3] = 1 + 4 = 5 ✓
• ans = 2

Operation 3 (i = 4):

• nums[4] + nums[5] = 5 + 0 = 5 ✓
• ans = 3

Check i = 6:

• i = 6 >= n = 6, no more pairs available
• Break

Result: Maximum operations = 3 (all pairs had sum = 5)

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. The algorithm iterates through the array once with a step size of 2, checking pairs of consecutive elements. In the worst case, it examines all elements in the array (when all pairs have the same sum), resulting in n/2 iterations, which simplifies to O(n).

The space complexity is O(1). The algorithm only uses a constant amount of extra space to store variables s (the target sum), ans (the operation counter), and n (the array length), regardless of the input size.

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

Common Pitfalls

1. Not Handling Edge Cases for Array Length

A common mistake is forgetting to check if the array has fewer than 2 elements at the start. If nums has length 0 or 1, attempting to access nums[0] + nums[1] will cause an IndexError.

Incorrect approach:

def maxOperations(self, nums: List[int]) -> int:
    target_sum = nums[0] + nums[1]  # Crashes if len(nums) < 2
    # ... rest of code

Correct solution:

def maxOperations(self, nums: List[int]) -> int:
    # Guard clause for insufficient elements
    if len(nums) < 2:
        return 0
  
    target_sum = nums[0] + nums[1]
    # ... rest of code

2. Misunderstanding the Problem - Trying All Possible Pairs

Some might incorrectly interpret the problem as finding any pairs in the array that sum to a target, rather than removing consecutive pairs from the front.

Incorrect approach:

def maxOperations(self, nums: List[int]) -> int:
    target_sum = nums[0] + nums[1]
    count = 0
    used = set()
  
    # Wrong: Looking for any pairs that sum to target
    for i in range(len(nums)):
        for j in range(i+1, len(nums)):
            if i not in used and j not in used and nums[i] + nums[j] == target_sum:
                count += 1
                used.add(i)
                used.add(j)
    return count

Correct understanding: Operations must remove the first two elements each time, not any arbitrary pair.

3. Off-by-One Error in Loop Boundary Check

A subtle mistake is checking i + 1 >= n instead of i + 1 == n, which could lead to premature termination or missing the last valid pair.

Incorrect approach:

for i in range(0, n, 2):
    if i + 1 > n:  # Wrong condition
        break
    if nums[i] + nums[i + 1] != target_sum:
        break
    operation_count += 1

Why it's wrong: The condition i + 1 > n will never be true within the loop range, as range(0, n, 2) ensures i never exceeds n-2 for even n or n-1 for odd n.

4. Attempting to Modify the Array During Iteration

Some might try to actually remove elements from the array, which is unnecessary and inefficient.

Inefficient approach:

def maxOperations(self, nums: List[int]) -> int:
    if len(nums) < 2:
        return 0
  
    target_sum = nums[0] + nums[1]
    operation_count = 0
  
    while len(nums) >= 2:
        if nums[0] + nums[1] != target_sum:
            break
        nums.pop(0)  # Inefficient O(n) operation
        nums.pop(0)  # Another O(n) operation
        operation_count += 1
  
    return operation_count

Why it's inefficient: Each pop(0) operation takes O(n) time, making the overall complexity O(n²). The correct approach using indices avoids modifying the array entirely.