Problem Description

The task is to determine the largest size for a three-dimensional array A with dimensions n x n x n such that the sum of all its elements does not exceed a given positive integer s. The array is constructed with the elements A[i][j][k] defined by the formula i * (j OR k), where 0 <= i, j, k < n. The challenge is to find the maximum integer n that allows the sum of all these elements in A to stay within the limit of s.

Intuition

To approach the solution, the problem can be tackled in two major parts: preprocessing and binary search. The preprocessing step involves precomputing possible sums that can be constructed from various sizes of n. This is done by evaluating the expression f[n] = sum(i * (j OR k)) for all valid indices and storing these cumulative sums until a safe estimate of n is reached.

By considering the mathematical behavior of the operations involved, especially the bitwise OR, we can deduce that the sum grows approximately as a fifth power of n, i.e., (n-1)^5 / 4. This approximation suggests that n should not exceed approximately 1320 for feasible computation within most reasonable constraints of s.

Once precomputation is complete, a binary search is employed to efficiently find the largest n such that the cumulative sum for an (n-1)-sized array is within the allowed limit s. Binary searching between 1 and the estimated maximum ensures that the optimal n is found quickly and accurately. The binary search leverages the cumulative sums precomputed in f to decide whether increasing n is safe with respect to the sum condition.

Learn more about Binary Search patterns.

Solution Approach

The solution employs a combination of preprocessing and binary search to determine the largest possible n for which the sum of the 3D array A does not exceed the given integer s.

Preprocessing

1. Precompute Possible Sums:
   Allocate an array f to store cumulative sums. The array is initialized with a size based on an estimated maximum n, specifically up to 1330. This allows the computation of potential sums without reevaluating them during each binary search step.

2. Calculate Sums Using Bitwise Operations:
   For each possible i, iterate through possible j values up to i, and compute the sum using the formula 2 * (i | j) for each combination. Accumulate these values to build f which represents the cumulative sum for possible sizes of n.

   mx = 1330
   f = [0] * mx
   for i in range(1, mx):
       f[i] = f[i - 1] + i
       for j in range(i):
           f[i] += 2 * (i | j)

Binary Search

3. Initialize Search Bounds:
   Set l (lower bound) to 1 and r (upper bound) to 1330. These bounds encompass all possible n values derived from the preprocessing step.

4. Perform Binary Search:
   The goal is to find the largest n where f[n-1] * (n-1) * n / 2 <= s. Begin with the middle point m and compute whether the constraint is satisfied. Adjust l and r based on whether the condition holds, effectively narrowing down to the maximum feasible n.

   class Solution:
       def maxSizedArray(self, s: int) -> int:
           l, r = 1, mx
           while l < r:
               m = (l + r + 1) >> 1
               if f[m - 1] * (m - 1) * m // 2 <= s:
                   l = m
               else:
                   r = m - 1
           return l

Through these steps, the algorithm efficiently determines the largest dimension n that satisfies the sum constraint, combining both pre-calculation and dynamic searching strategies.

Example Walkthrough

To understand the solution approach, let's take a small example where s = 100. We want to find the largest n for a 3D array A such that the sum of all its elements does not exceed 100.

Step 1: Preprocessing

• Precompute Possible Sums:
  We create an array f to store cumulative sums for potential dimensions n from 1 to 1330. This requires calculating sums based on the formula A[i][j][k] = i * (j OR k).

  Consider n = 3:

  • For i = 0, 1, 2, compute sums using:
    • i = 1: 1 * (j OR k) when j, k < 3 gives combinations like 1*(0 OR 0), 1*(0 OR 1), 1*(0 OR 2), 1*(1 OR 0), 1*(1 OR 1), 1*(1 OR 2), ..., all computed to accumulate the sum.
    • Similar calculations follow for i = 2.

  Using this approach, the cumulative sum for n = 3 could be evaluated and stored in the array f.

Step 2: Binary Search

• Initialize Search Bounds:
  Set l = 1 and r = 1330. We search for the largest n where the computed sum does not exceed s = 100.

• Perform Binary Search:
  Check mid-point m of the current interval [l, r]. Calculate if f[m-1] * (m-1) * m / 2 <= 100:

  • For a smaller m, such as m = 3, check if the condition holds.
  • If it does, increase l to explore larger n; if not, decrease r.

• Continue this process by selecting the midpoint and checking the feasible n until the largest valid n is determined.

Using this method, you identify the largest n—let's say n = 2—where all conditions are satisfied.

This walkthrough demonstrates how preprocessing coupled with a binary search can efficiently solve the problem of determining the largest possible dimension n for the given sum condition s.

Solution Implementation

Time and Space Complexity

The time complexity of the code involves two main parts:

1. Preprocessing Loop: This loop initializes the array f and has a nested loop inside it. The outer loop runs mx times and the inner loop runs up to i times for each i. This results in a total complexity of [ \sum_{i=1}^{mx-1} i = \frac{(mx-1) \times mx}{2} ] Hence, the preprocessing has a time complexity of (O(n^2)), where n corresponds to mx.

2. Binary Search: This part involves searching within the bounds of 1 to mx, making the time complexity of the binary search (O(\log n)).

Therefore, the total time complexity of the algorithm is (O(n^2 + \log n)).

For space complexity:

• The space used is mainly for the array f, which is of size mx.

Therefore, the space complexity is (O(n)).

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