How do you solve Maximum Trailing Zeros in a Cornered Path on LeetCode?

Maximum Trailing Zeros in a Cornered Path (LeetCode #2245) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n²) time complexity and O(m * n) space complexity. Key insight: Trailing zeros depend on the minimum of the count of factors of 2 and 5.

What is the brute force solution for Maximum Trailing Zeros in a Cornered Path?

The brute force approach for Maximum Trailing Zeros in a Cornered Path is Brute Force, with O(m * n * (m + n)) time complexity and O(1) space complexity. The brute force approach involves checking every possible cornered path in the grid. For each path, we calculate the product of the numbers and count the trailing zeros based on the factors of 2 and 5. The optimal Optimal Solution approach improves this to O(m * n²).

What algorithm or data structure is used to solve Maximum Trailing Zeros in a Cornered Path?

Maximum Trailing Zeros in a Cornered Path uses the following concepts: Array, Matrix, Prefix Sum. The recommended patterns to study are: Prefix Sum, Dynamic Programming.

What are the interview tips for Maximum Trailing Zeros in a Cornered Path?

Clearly explain your thought process while coding. Consider edge cases, such as single-row or single-column grids. Practice calculating trailing zeros for small numbers to build intuition.

What are common mistakes when solving Maximum Trailing Zeros in a Cornered Path?

Not considering all possible cornered paths. Failing to account for previously visited cells.

#2245

Maximum Trailing Zeros in a Cornered Path

Medium

ArrayMatrixPrefix SumPrefix SumDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * (m + n))	O(m * n²)
Space	O(1)	O(m * n)

💡

Intuition

Time O(m * n²)Space O(m * n)

To optimize, we can precompute the number of factors of 2 and 5 for each cell. This allows us to quickly calculate the number of trailing zeros for any cornered path without recalculating products.

⚙️

Algorithm

3 steps

1Step 1: Create two matrices to store the count of factors of 2 and 5 for each cell in the grid.
2Step 2: For each cell, calculate the cumulative sum of factors of 2 and 5 in both horizontal and vertical directions.
3Step 3: Iterate through each cell, and for each possible cornered path, compute the number of trailing zeros using the precomputed matrices.

solution.py29 lines

1def count_factors(n, factor):
2    count = 0
3    while n > 0 and n % factor == 0:
4        count += 1
5        n //= factor
6    return count
7
8def max_trailing_zeros(grid):
9    m, n = len(grid), len(grid[0])
10    count2 = [[0] * n for _ in range(m)]
11    count5 = [[0] * n for _ in range(m)]
12
13    for i in range(m):
14        for j in range(n):
15            count2[i][j] = count_factors(grid[i][j], 2)
16            count5[i][j] = count_factors(grid[i][j], 5)
17
18    max_zeros = 0
19    for i in range(m):
20        for j in range(n):
21            for k in range(j, n):  # horizontal path
22                zeros2 = sum(count2[i][x] for x in range(j, k + 1))
23                zeros5 = sum(count5[i][x] for x in range(j, k + 1))
24                max_zeros = max(max_zeros, min(zeros2, zeros5))
25                for l in range(i + 1, m):  # vertical path
26                    zeros2 += count2[l][k]
27                    zeros5 += count5[l][k]
28                    max_zeros = max(max_zeros, min(zeros2, zeros5))
29    return max_zeros

ℹ

Complexity note: The time complexity is O(m * n²) because for each cell, we may iterate through all columns in the worst case. The space complexity is O(m * n) due to the storage of the factor counts.

1Trailing zeros depend on the minimum of the count of factors of 2 and 5.
2Precomputation of factors allows for faster calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.