Leetcode #2245: Maximum Trailing Zeros in a Cornered Path
In this guide, we solve Leetcode #2245 Maximum Trailing Zeros in a Cornered Path in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 2D integer array grid of size m x n, where each cell contains a positive integer. A cornered path is defined as a set of adjacent cells with at most one turn.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Matrix, Prefix Sum
Intuition
Range queries become simple once we precompute cumulative sums.
We can transform subarray conditions into prefix comparisons.
Approach
Compute prefix sums and use a map to find matching prefixes.
This avoids nested loops while keeping the logic clear.
Steps:
- Compute prefix sums.
- Use a map to find valid ranges.
- Update the answer.
Example
Input: grid = [[23,17,15,3,20],[8,1,20,27,11],[9,4,6,2,21],[40,9,1,10,6],[22,7,4,5,3]]
Output: 3
Explanation: The grid on the left shows a valid cornered path.
It has a product of 15 * 20 * 6 * 1 * 10 = 18000 which has 3 trailing zeros.
It can be shown that this is the maximum trailing zeros in the product of a cornered path.
The grid in the middle is not a cornered path as it has more than one turn.
The grid on the right is not a cornered path as it requires a return to a previously visited cell.
Python Solution
class Solution:
def maxTrailingZeros(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
r2 = [[0] * (n + 1) for _ in range(m + 1)]
c2 = [[0] * (n + 1) for _ in range(m + 1)]
r5 = [[0] * (n + 1) for _ in range(m + 1)]
c5 = [[0] * (n + 1) for _ in range(m + 1)]
for i, row in enumerate(grid, 1):
for j, x in enumerate(row, 1):
s2 = s5 = 0
while x % 2 == 0:
x //= 2
s2 += 1
while x % 5 == 0:
x //= 5
s5 += 1
r2[i][j] = r2[i][j - 1] + s2
c2[i][j] = c2[i - 1][j] + s2
r5[i][j] = r5[i][j - 1] + s5
c5[i][j] = c5[i - 1][j] + s5
ans = 0
for i in range(1, m + 1):
for j in range(1, n + 1):
a = min(r2[i][j] + c2[i - 1][j], r5[i][j] + c5[i - 1][j])
b = min(r2[i][j] + c2[m][j] - c2[i][j], r5[i][j] + c5[m][j] - c5[i][j])
c = min(r2[i][n] - r2[i][j] + c2[i][j], r5[i][n] - r5[i][j] + c5[i][j])
d = min(
r2[i][n] - r2[i][j - 1] + c2[m][j] - c2[i][j],
r5[i][n] - r5[i][j - 1] + c5[m][j] - c5[i][j],
)
ans = max(ans, a, b, c, d)
return ans
Complexity
The time complexity is , and the space complexity is , where and are the number of rows and columns of the grid array, respectively. The space complexity is , where and are the number of rows and columns of the grid array, respectively.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.