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Problem solution · C++

Minimum Moves to Capture The Queen

Minimum Moves to Capture The Queen: 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
24 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 Minimum Moves to Capture The Queen, 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

  • 24 lines of C++ from the credited upstream file 3001.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit 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 codeMinimum Moves to Capture The Queen · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minMovesToCaptureTheQueen(int a, int b, int c, int d, int e, int f) {    // The rook is in the same row as the queen.    if (a == e)      // The bishop blocks the rook or not.      return (c == a && (b < d && d < f || b > d && d > f)) ? 2 : 1;    // The rook is in the same column as the queen.    if (b == f)      // The bishop blocks the rook or not.      return (d == f && (a < c && c < e || a > c && c > e)) ? 2 : 1;    // The bishop is in the same up-diagonal as the queen.    if (c + d == e + f)      // The rook blocks the bishop or not.      return (a + b == c + d && (c < a && a < e || c > a && a > e)) ? 2 : 1;    // The bishop is in the same down-diagonal as the queen.    if (c - d == e - f)      // The rook blocks the bishop or not.      return (a - b == c - d && (c < a && a < e || c > a && a > e)) ? 2 : 1;    // The rook can always get the green in two steps.    return 2;  }};

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