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Problem solution · C++

Maximum Trailing Zeros in a Cornered Path

Maximum Trailing Zeros in a Cornered Path: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
67 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Maximum Trailing Zeros in a Cornered Path, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 67 lines of C++ from the credited upstream file 2245.cpp.
  • The implementation visibly relies on sequence storage.
  • 7 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum Trailing Zeros in a Cornered Path · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int maxTrailingZeros(vector<vector<int>>& grid) {    const int m = grid.size();    const int n = grid[0].size();    // leftPrefix2[i][j] := the number of 2 in grid[i][0..j]    // leftPrefix5[i][j] := the number of 5 in grid[i][0..j]    // topPrefix2[i][j] := the number of 2 in grid[0..i][j]    // topPrefix5[i][j] := the number of 5 in grid[0..i][j]    vector<vector<int>> leftPrefix2(m, vector<int>(n));    vector<vector<int>> leftPrefix5(m, vector<int>(n));    vector<vector<int>> topPrefix2(m, vector<int>(n));    vector<vector<int>> topPrefix5(m, vector<int>(n));     for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j) {        leftPrefix2[i][j] = getCount(grid[i][j], 2);        leftPrefix5[i][j] = getCount(grid[i][j], 5);        if (j > 0) {          leftPrefix2[i][j] += leftPrefix2[i][j - 1];          leftPrefix5[i][j] += leftPrefix5[i][j - 1];        }      }     for (int j = 0; j < n; ++j)      for (int i = 0; i < m; ++i) {        topPrefix2[i][j] = getCount(grid[i][j], 2);        topPrefix5[i][j] = getCount(grid[i][j], 5);        if (i > 0) {          topPrefix2[i][j] += topPrefix2[i - 1][j];          topPrefix5[i][j] += topPrefix5[i - 1][j];        }      }     int ans = 0;    for (int i = 0; i < m; ++i)      for (int j = 0; j < n; ++j) {        const int curr2 = getCount(grid[i][j], 2);        const int curr5 = getCount(grid[i][j], 5);        const int l2 = leftPrefix2[i][j];        const int l5 = leftPrefix5[i][j];        const int r2 = leftPrefix2[i][n - 1] - (j ? leftPrefix2[i][j - 1] : 0);        const int r5 = leftPrefix5[i][n - 1] - (j ? leftPrefix5[i][j - 1] : 0);        const int t2 = topPrefix2[i][j];        const int t5 = topPrefix5[i][j];        const int d2 = topPrefix2[m - 1][j] - (i ? topPrefix2[i - 1][j] : 0);        const int d5 = topPrefix5[m - 1][j] - (i ? topPrefix5[i - 1][j] : 0);        ans = max({ans, min(l2 + t2 - curr2, l5 + t5 - curr5),                   min(r2 + t2 - curr2, r5 + t5 - curr5),                   min(l2 + d2 - curr2, l5 + d5 - curr5),                   min(r2 + d2 - curr2, r5 + d5 - curr5)});      }     return ans;  }  private:  int getCount(int num, int factor) {    int count = 0;    while (num % factor == 0) {      num /= factor;      ++count;    }    return count;  }};

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