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Problem solution · C++

Surroundedregions

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

Technique
Breadth-first search
Source
Kamyu LeetCode Solutions
Length
78 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

Breadth-first search

For Surroundedregions, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 78 lines of C++ from the credited upstream file surroundedRegions.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 7 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

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 codeSurroundedregions · C++C++
Use this to learn the idea, then write your own version.
// LeetCode, Surrounded Regions// BFS method// Complexity://     O(n) time, O(n) space class Solution {public:    void solve(vector<vector<char>> &board) {        if (board.empty()) return;        const int m = board.size();        const int n = board[0].size();         // do BFS from up and down boundary        for (int i = 0; i < n; i++) {            bfs(board, 0, i);            bfs(board, m - 1, i);        }         // do BFS from left and right boundary        for (int j = 1; j < m - 1; j++) {            bfs(board, j, 0);            bfs(board, j, n - 1);        }         for (int i = 0; i < m; i++)        	for (int j = 0; j < n; j++)        		// mark the BFS-unvisited node with 'X'			    if (board[i][j] == 'O')			        board[i][j] = 'X';			    // mark the BFS-visited node with 'O' 			    else if (board[i][j] == '+')			        board[i][j] = 'O';	}  private:	void bfs(vector<vector<char>> &board, int i, int j) {	    typedef pair<int, int> state_t;	    queue<state_t> q;	    const int m = board.size();	    const int n = board[0].size(); 	    auto is_valid = [&](const state_t &s) {	        const int x = s.first;	        const int y = s.second;	        if (x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'O')	            return false;	        return true;	    }; 	    auto state_extend = [&](const state_t &s) {	        vector<state_t> result;	        const int x = s.first;	        const int y = s.second;	        const state_t new_states[4] = {{x-1,y}, {x+1,y},{x,y-1}, {x,y+1}};	        for(int k=0;k<4; ++k){	        	if (is_valid(new_states[k])) {	        		// mark	        		board[new_states[k].first][new_states[k].second] = '+';	        		result.push_back(new_states[k]);	        	} 	        }	        return result;	    };                state_t start = { i, j };        if (is_valid(start)) {            board[i][j] = '+';            q.push(start);        }                while (!q.empty()) {            auto cur = q.front();            q.pop();            auto new_states = state_extend(cur);            for (auto s : new_states) q.push(s);        }     }};

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