Solution Approach

The solution consists of two main parts: preprocessing and binary search.

Preprocessing Phase:

First, we establish an upper bound for n. Since the sum grows approximately as (n-1)^5 / 4, and we need (n-1)^5 / 4 ≤ s, we can determine that n won't exceed approximately 1330 for reasonable values of s.

We precompute an array f where f[i] represents the sum Σ(j=0 to i) Σ(k=0 to j) (j OR k). The preprocessing code:

mx = 1330
f = [0] * mx
for i in range(1, mx):
    f[i] = f[i - 1] + i  # Add the case where j OR k = i (when j = k = i)
    for j in range(i):
        f[i] += 2 * (i | j)  # Count both (i, j) and (j, i) cases

The formula f[i] = f[i-1] + i + 2 * Σ(j=0 to i-1) (i OR j) builds upon the previous value and adds new contributions from index i.

Binary Search Phase:

Once we have the preprocessed values, we use binary search to find the maximum n such that the total sum doesn't exceed s.

For a given n, the total sum of the 3D array is calculated as: Total Sum = f[n-1] * (n-1) * n / 2

This formula accounts for:

• f[n-1]: The sum pattern for OR operations
• (n-1): Multiplication by each value of i from 0 to n-1
• n / 2: Proper scaling factor for the 3D array structure

The binary search implementation uses the standard template. Since we want the maximum valid n, we find the first n where sum exceeds s:

def maxSizedArray(self, s: int) -> int:
    left, right = 1, mx
    first_exceed_index = -1  # First n where sum > s

    while left <= right:
        mid = (left + right) // 2
        cost = f[mid - 1] * (mid - 1) * mid // 2

        if cost > s:  # Feasible: sum exceeds s
            first_exceed_index = mid
            right = mid - 1  # Look for earlier exceed
        else:
            left = mid + 1  # Sum still valid, try larger

    # If never exceeded, return max; otherwise return one less than first exceed
    if first_exceed_index == -1:
        return mx
    return first_exceed_index - 1

The binary search finds the first n where the sum exceeds s, then returns first_exceed_index - 1 as the maximum valid n.

Example Walkthrough

Let's walk through finding the maximum n for s = 10.

Step 1: Understanding the 3D Array

For n = 2, we calculate all elements:

• When i = 0: All elements are 0 (since 0 * anything = 0)
• When i = 1:
  • A[1][0][0] = 1 * (0 OR 0) = 1 * 0 = 0
  • A[1][0][1] = 1 * (0 OR 1) = 1 * 1 = 1
  • A[1][1][0] = 1 * (1 OR 0) = 1 * 1 = 1
  • A[1][1][1] = 1 * (1 OR 1) = 1 * 1 = 1

Total sum = 0 + 0 + 0 + 0 + 0 + 1 + 1 + 1 = 3

Step 2: Preprocessing Values

We build our f array:

• f[0] = 0
• f[1] = f[0] + 1 + 2*(1 OR 0) = 0 + 1 + 2*1 = 3
• f[2] = f[1] + 2 + 2*(2 OR 0) + 2*(2 OR 1) = 3 + 2 + 2*2 + 2*3 = 3 + 2 + 4 + 6 = 15

Step 3: Binary Search

Now we search for maximum n where the sum ≤ 10:

Initial range: l = 1, r = 1330

First iteration:

• m = (1 + 1330 + 1) / 2 = 666
• Calculate sum for n = 666: This would be enormous, much greater than 10
• Update: r = 665

After several iterations, we narrow down to:

• When checking n = 2: Sum = f[1] * 1 * 2 / 2 = 3 * 1 * 1 = 3 ✓ (≤ 10)
• When checking n = 3: Sum = f[2] * 2 * 3 / 2 = 15 * 2 * 3 / 2 = 45 ✗ (> 10)

Therefore, the maximum n = 2.

Verification: For n = 2, the total sum is 3, which is indeed ≤ 10. For n = 3, we would have a 3×3×3 array where the sum would be 45, exceeding our limit. Thus, n = 2 is the correct answer.

Time and Space Complexity

The time complexity consists of two parts:

1. Preprocessing phase: The outer loop runs from 1 to mx-1 (i.e., O(mx) iterations). For each iteration i, the inner loop runs i times, performing a bitwise OR operation and arithmetic operations. This results in 1 + 2 + 3 + ... + (mx-1) = (mx-1) * mx / 2 operations, giving us O(mx^2) time complexity for preprocessing.

2. Binary search phase: The maxSizedArray method performs a binary search between 1 and mx. Each iteration of the while loop reduces the search space by half, resulting in O(log mx) time complexity.

Therefore, the total time complexity is O(mx^2 + log mx). Since mx = 1330 is a constant, we can also express this as O(n^2 + log n) where n represents the upper bound of the search range.

The space complexity is O(mx) or O(n), as we allocate an array f of size mx to store the precomputed values. No additional space that scales with input is used during the binary search phase.

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

Pitfall: Using while left < right with left = mid for finding the maximum valid n:

# WRONG: Non-standard template for finding maximum
while left < right:
    mid = (left + right + 1) >> 1  # Need ceiling division
    if cost <= s:
        left = mid
    else:
        right = mid - 1
return left

Solution: Use the standard template by finding the first n where cost exceeds s:

# CORRECT: Standard template
first_exceed_index = -1
while left <= right:
    mid = (left + right) // 2
    if cost > s:  # Feasible: exceeds limit
        first_exceed_index = mid
        right = mid - 1
    else:
        left = mid + 1
return first_exceed_index - 1 if first_exceed_index != -1 else MAX_SIZE - 1

2. Integer Overflow in Cost Calculation

Pitfall: The most critical issue in this solution is potential integer overflow when calculating the cost:

cost = cumulative_or_sum[mid - 1] * (mid - 1) * mid // 2

For large values of mid (approaching 1330), the multiplication cumulative_or_sum[mid - 1] * (mid - 1) * mid can exceed Python's integer limits in other languages or cause performance issues even in Python with very large numbers.

Solution: Rearrange the calculation to perform division earlier or check for overflow:

# Option 1: Divide earlier to reduce intermediate values
cost = cumulative_or_sum[mid - 1] * ((mid - 1) * mid // 2)

# Option 2: Check if multiplication would exceed s before computing
if cumulative_or_sum[mid - 1] > s * 2 // ((mid - 1) * mid):
    right = mid - 1
    continue

3. Incorrect Precomputation Formula

Pitfall: The precomputation loop might be implemented incorrectly:

# Wrong: Missing the factor of 2 for symmetric pairs
for array_size in range(1, MAX_SIZE):
    cumulative_or_sum[array_size] = cumulative_or_sum[array_size - 1] + array_size
    for element in range(array_size):
        cumulative_or_sum[array_size] += (array_size | element)  # Missing factor of 2!

This misses that for each unique OR value, we need to count both (j, k) and (k, j) pairs when j ≠ k.

Solution: Ensure the factor of 2 is included for non-diagonal elements:

for array_size in range(1, MAX_SIZE):
    cumulative_or_sum[array_size] = cumulative_or_sum[array_size - 1] + array_size
    for element in range(array_size):
        cumulative_or_sum[array_size] += 2 * (array_size | element)  # Correct

4. Hardcoded Maximum Size Without Validation

Pitfall: Using a fixed MAX_SIZE = 1330 without considering the actual input range. If s is extremely large (beyond what 1330 can handle), the solution would return an incorrect answer.

Solution: Dynamically determine the upper bound based on s:

def maxSizedArray(self, s: int) -> int:
    # Dynamically calculate upper bound
    # Since sum ~ n^5/4, we have n ~ (4*s)^(1/5)
    import math
    max_possible = min(MAX_SIZE, int((4 * s) ** 0.2) + 10)  # Add buffer

    left, right = 1, max_possible
    # ... rest of binary search

5. Not Handling Edge Cases

Pitfall: Not properly handling small values of s:

• When s = 0, the answer should be 1 (array of size 1 has sum 0)
• When s < 0, the input is invalid

Solution: Add edge case handling:

def maxSizedArray(self, s: int) -> int:
    if s == 0:
        return 1
    if s < 0:
        raise ValueError("s must be non-negative")

    # Continue with normal binary search...