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Problem solution · C++

The Knights Tour

The Knights Tour: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
87 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

Depth-first search

For The Knights Tour, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 87 lines of C++ from the credited upstream file the-knights-tour.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeThe Knights Tour · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(m * n)// Space: O(1) // backtracking, greedy, warnsdorff's ruleclass Solution {public:    vector<vector<int>> tourOfKnight(int m, int n, int r, int c) {        static const vector<pair<int, int>> DIRECTIONS = {{1, 2}, {-1, 2}, {1, -2}, {-1, -2},                                                          {2, 1}, {-2, 1}, {2, -1}, {-2, -1}};         vector<vector<int>> result(m, vector<int>(n, -1));        result[r][c] = 0;        const function<bool(int, int, int)> backtracking = [&](int r, int c, int i) {            const auto& degree = [&](const auto& x) {                int cnt = 0;                const auto& [r, c] = x;                for (const auto& [dr, dc] : DIRECTIONS) {                    const int nr = r + dr, nc = c + dc;                    if (0 <= nr && nr < m && 0 <= nc && nc < n && result[nr][nc] == -1) {                        ++cnt;                    }                }                return cnt;            };             if (i == m * n) {                return true;            }            vector<pair<int, int>> candidates;            for (const auto& [dr, dc] : DIRECTIONS) {                const int nr = r + dr, nc = c + dc;                if (0 <= nr && nr < m && 0 <= nc && nc < n && result[nr][nc] == -1) {                    candidates.emplace_back(nr, nc);                }            }            sort(begin(candidates), end(candidates), [&](const auto& a, const auto& b) {                return degree(a) < degree(b);            });            for (const auto& [nr, nc] : candidates) {  // warnsdorff's rule                result[nr][nc] = i;                if (backtracking(nr, nc, i + 1)) {                    return true;                }                result[nr][nc] = -1;            }            return false;        };         backtracking(r, c, 1);        return result;    }}; // Time:  O(8^(m * n - 1))// Space: O(m * n)// backtrackingclass Solution2 {public:    vector<vector<int>> tourOfKnight(int m, int n, int r, int c) {        static const vector<pair<int, int>> DIRECTIONS = {{1, 2}, {-1, 2}, {1, -2}, {-1, -2},                                                          {2, 1}, {-2, 1}, {2, -1}, {-2, -1}};         vector<vector<int>> result(m, vector<int>(n, -1));        result[r][c] = 0;        const function<bool(int, int, int)> backtracking = [&](int r, int c, int i) {            if (i == m * n) {                return true;            }            for (const auto& [dr, dc] : DIRECTIONS) {                const int nr = r + dr, nc = c + dc;                if (!(0 <= nr && nr < m && 0 <= nc && nc < n && result[nr][nc] == -1)) {                    continue;                }                result[nr][nc] = i;                if (backtracking(nr, nc, i + 1)) {                    return true;                }                result[nr][nc] = -1;            }            return false;        };         backtracking(r, c, 1);        return result;    }};

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