Solution Approach

The solution uses the binary search template to find the maximum value at nums[index]. The key insight is to search for the first value that exceeds the budget, then return the value just before it.

Helper Function: calculateSum(peakValue, count)

This function calculates the sum of a decreasing sequence starting from peakValue with count elements:

• Case 1: When peakValue >= count - All elements can decrease by 1 without reaching below 1

  • The sequence is: [peakValue, peakValue-1, ..., peakValue-count+1]
  • Sum = (peakValue + peakValue - count + 1) * count / 2

• Case 2: When peakValue < count - The sequence reaches 1 before using all count positions

  • Sum = (peakValue + 1) * peakValue / 2 + (count - peakValue)

Feasible Function: exceedsBudget(mid)

The feasibility condition checks if placing value mid at the index exceeds maxSum:

def exceeds_budget(mid: int) -> bool:  total_sum = calculate_sum(mid - 1, index) + calculate_sum(mid, n - index)  return total_sum > maxSum

This creates a pattern: false, false, ..., false, true, true, ...

• Values 1, 2, 3, ... are valid (don't exceed budget) → return false
• At some point, values become too large → return true

Binary Search Template

We use the standard template to find the first value where exceeds_budget returns true:

left, right = 1, maxSum + 1 first_true_index = -1  while left <= right:  mid = (left + right) // 2  if exceeds_budget(mid):  first_true_index = mid  right = mid - 1  else:  left = mid + 1  return first_true_index - 1 if first_true_index != -1 else maxSum

The answer is first_true_index - 1 because we want the last valid value (the one just before the first invalid value).

The time complexity is O(log(maxSum)) for the binary search, with O(1) for each validation check. Space complexity is O(1).

Example Walkthrough

Let's walk through a concrete example with n = 6, index = 2, and maxSum = 9.

We want to find the maximum value we can place at position 2 in an array of length 6, where the total sum cannot exceed 9.

Step 1: Set up binary search bounds

• left = 1 (minimum possible value)
• right = 9 (theoretical maximum)

Step 2: First iteration - try mid = 5

• Place 5 at index 2
• Build the array following the "hill" pattern:
  • Position 2: value = 5
  • Position 1: value = 4 (decrease by 1)
  • Position 0: value = 3 (decrease by 1)
  • Position 3: value = 4 (decrease by 1)
  • Position 4: value = 3 (decrease by 1)
  • Position 5: value = 2 (decrease by 1)
• Array: [3, 4, 5, 4, 3, 2]
• Calculate sum using helper function:
  • Left side (2 elements): sum(4, 2) = 4 + 3 = 7
  • Right side (4 elements): sum(5, 4) = 5 + 4 + 3 + 2 = 14
  • Total = 7 + 14 = 21
• Since 21 > 9, this is too large. Update right = 4

Step 3: Second iteration - try mid = 3

• Place 3 at index 2
• Build the array:
  • Position 2: value = 3
  • Position 1: value = 2
  • Position 0: value = 1
  • Position 3: value = 2
  • Position 4: value = 1
  • Position 5: value = 1 (can't go below 1)
• Array: [1, 2, 3, 2, 1, 1]
• Calculate sum:
  • Left side: sum(2, 2) = 2 + 1 = 3
  • Right side: sum(3, 4) = 3 + 2 + 1 + 1 = 7
  • Total = 3 + 7 = 10
• Since 10 > 9, still too large. Update right = 2

Step 4: Third iteration - try mid = 2

• Place 2 at index 2
• Build the array:
  • Position 2: value = 2
  • Position 1: value = 1
  • Position 0: value = 1 (can't go below 1)
  • Position 3: value = 1
  • Position 4: value = 1
  • Position 5: value = 1
• Array: [1, 1, 2, 1, 1, 1]
• Calculate sum:
  • Left side: sum(1, 2) = 1 + 1 = 2
  • Right side: sum(2, 4) = 2 + 1 + 1 + 1 = 5
  • Total = 2 + 5 = 7
• Since 7 ≤ 9, this is valid! Update left = 2

Step 5: Check convergence

• Now left = 2 and right = 2
• Binary search converges

Answer: 2

The maximum value we can place at index 2 is 2, resulting in the array [1, 1, 2, 1, 1, 1] with sum 7, which satisfies all constraints.

Time and Space Complexity

The time complexity of this algorithm is O(log maxSum). The binary search operates on the range [1, maxSum], where left = 1 and right = maxSum. In each iteration, the search space is halved by updating either left = mid or right = mid - 1. The number of iterations required is therefore log₂(maxSum), which gives us O(log maxSum) time complexity.

The sum helper function performs constant time operations with arithmetic calculations that don't depend on the input size. Each call to sum within the binary search loop takes O(1) time.

The space complexity is O(1). The algorithm only uses a fixed number of variables (left, right, mid) regardless of the input size. The sum function also uses only local variables (x, cnt) and performs calculations without requiring additional space that scales with the input. No recursive calls are made, and no additional data structures are allocated, resulting in constant space usage.

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

Pitfall: Using the while left < right with left = mid pattern instead of the standard template. This can cause infinite loops and is harder to reason about.

Wrong Implementation:

while left < right:  mid = (left + right + 1) >> 1  if total_sum <= maxSum:  left = mid  else:  right = mid - 1 return left

Correct Implementation (using template):

left, right = 1, maxSum + 1 first_true_index = -1  while left <= right:  mid = (left + right) // 2  if exceeds_budget(mid):  first_true_index = mid  right = mid - 1  else:  left = mid + 1  return first_true_index - 1 if first_true_index != -1 else maxSum

2. Incorrect Sum Calculation for the Sides

Pitfall: Incorrectly calculating the sum for the left and right sides of the index.

Solution: Remember that we're building a "hill" where:

• The peak is at position index with value mid
• Left side has index elements decreasing from mid - 1
• Right side has n - index elements starting from mid (including the peak)

3. Integer Overflow in Arithmetic Sequence Formula

Pitfall: When calculating the sum of an arithmetic sequence, the multiplication can cause integer overflow for large values.

Solution: In Python, this isn't typically an issue due to arbitrary precision integers, but in other languages, use long/int64 data types.

4. Forgetting Edge Cases with Zero Elements

Pitfall: Not handling cases where one side has zero elements (when index = 0 or index = n - 1).

Solution: Ensure your helper function properly handles count = 0 case by returning 0.

5. Misunderstanding the Problem Constraints

Pitfall: Assuming the array can have zeros.

Solution: Always use 1 as the minimum value for any array position. The optimal strategy is to decrease by 1 from the peak until reaching 1, then fill remaining positions with 1s.