Problem Description

You are given a positive integer s. You need to create a 3D array A with dimensions n × n × n, where each element is calculated using a specific formula.

For each element in the array, A[i][j][k] is defined as:

• A[i][j][k] = i * (j OR k), where 0 <= i, j, k < n
• The OR operation is the bitwise OR operation

Your task is to find the maximum possible value of n such that when you sum up all elements in the entire 3D array A, the total sum does not exceed s.

For example, if n = 2, you would have a 2×2×2 array with 8 elements total:

• A[0][0][0] = 0 * (0 OR 0) = 0
• A[0][0][1] = 0 * (0 OR 1) = 0
• A[0][1][0] = 0 * (1 OR 0) = 0
• A[0][1][1] = 0 * (1 OR 1) = 0
• A[1][0][0] = 1 * (0 OR 0) = 0
• A[1][0][1] = 1 * (0 OR 1) = 1
• A[1][1][0] = 1 * (1 OR 0) = 1
• A[1][1][1] = 1 * (1 OR 1) = 1

The sum would be 3, and if s >= 3, then n = 2 would be valid. You need to find the largest such n where the sum stays within the limit s.

Intuition

Let's think about how to calculate the sum efficiently. If we expand the sum of all elements in the 3D array:

Sum = Σ(i=0 to n-1) Σ(j=0 to n-1) Σ(k=0 to n-1) i * (j OR k)

We can rearrange this as: Sum = Σ(i=0 to n-1) i * Σ(j=0 to n-1) Σ(k=0 to n-1) (j OR k)

Notice that the inner double summation Σ(j=0 to n-1) Σ(k=0 to n-1) (j OR k) is independent of i. Let's call this value S_or.

The key insight is that for any fixed value of j OR k, we can count how many (j, k) pairs produce that value. Due to symmetry in the OR operation, we can simplify our calculation.

For efficiency, we can precompute f[n] which represents Σ(i=0 to n-1) Σ(j=0 to i) (i OR j). Why up to i instead of n-1? Because (i OR j) where j <= i captures the essential pattern, and we can use symmetry to handle the full range.

The total sum for a given n can then be expressed as: Total Sum = f[n-1] * (n-1) * n / 2

This formula accounts for the multiplication by each i value and the proper counting of all (j, k) pairs.

Since the sum grows rapidly with n (approximately as n^5), and we want the maximum n where the sum doesn't exceed s, binary search becomes the natural choice. We can estimate that n won't exceed around 1330 based on the growth rate.

By preprocessing the values up to this maximum and then using binary search, we can efficiently find the largest valid n where the total sum stays within the limit s.

Learn more about Binary Search patterns.

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:

def maxSizedArray(self, s: int) -> int:
    l, r = 1, mx
    while l < r:
        m = (l + r + 1) >> 1  # Middle point, biased towards upper half
        if f[m - 1] * (m - 1) * m // 2 <= s:
            l = m  # Sum is within limit, try larger n
        else:
            r = m - 1  # Sum exceeds limit, try smaller n
    return l

The binary search finds the largest n where the total sum remains ≤ s. The expression (l + r + 1) >> 1 ensures we round up when calculating the midpoint, preventing infinite loops when l and r are adjacent values.

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.

Solution Implementation

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. 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

2. Off-by-One Error in Binary Search

Pitfall: The binary search uses (left + right + 1) >> 1 for the midpoint calculation. If you accidentally use (left + right) >> 1 instead, the binary search may enter an infinite loop when left = right - 1 and the condition holds true for mid = left.

Solution: Always use ceiling division when the update rule is left = mid:

mid = (left + right + 1) >> 1  # Correct: ceiling division
# NOT: mid = (left + right) >> 1  # Wrong: can cause infinite loop

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...