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Problem solution · C++

Number of Ways to Reach Destination in the Grid

Number of Ways to Reach Destination in the Grid: 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
39 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 Number of Ways to Reach Destination in the Grid, 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

  • 39 lines of C++ from the credited upstream file 2912-2.cpp.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 codeNumber of Ways to Reach Destination in the Grid · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int numberOfWays(int n, int m, int k, vector<int>& source,                   vector<int>& dest) {    constexpr int kMod = 1'000'000'007;    // the number of ways of `source` to `dest` using steps so far    int ans = source[0] == dest[0] && source[1] == dest[1];    // the number of ways of `source` to dest's row using steps so far    int row = source[0] == dest[0] && source[1] != dest[1];    // the number of ways of `source` to dest's col using steps so far    int col = source[0] != dest[0] && source[1] == dest[1];    // the number of ways of `source` to others using steps so far    int others = source[0] != dest[0] && source[1] != dest[1];     for (int i = 0; i < k; ++i) {      const int nextAns = (row + col) % kMod;      const int nextRow = static_cast<int>((ans * (m - 1L) +  // -self                                            row * (m - 2L) +  //-self, -center                                            others) %                                           kMod);      const int nextCol = static_cast<int>((ans * (n - 1L) +  // -self                                            col * (n - 2L) +  // -self, -center                                            others) %                                           kMod);      const int nextOthers =          static_cast<int>((row * (n - 1L) +              // -self                            col * (m - 1L) +              // -self                            others * (m + n - 1 - 3L)) %  // -self, -row, -col                           kMod);      ans = nextAns;      row = nextRow;      col = nextCol;      others = nextOthers;    }     return ans;  }};

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