This problem looks busy at first, but the accepted solution is built around one steady invariant. For 130. Surrounded Regions, the solution in this repository is mainly a union-find solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: union-find, graph traversal.
The notes already sitting in the source point us in the right direction:
- bfs
Guide
When?
This pattern shows up when the brute force version has too many repeated checks, too many possible branches, or too much bookkeeping to do by hand. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are solve, changeNeighbor, unite, isConnected, node.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- The queue gives the solution a level-by-level or frontier-style traversal.
- The two-dimensional vector is the memory of the solution: grid state, DP state, or adjacency shape.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Start from the smallest reliable state.
- Expand one legal move at a time.
- Cache, count, or merge information as soon as it becomes settled.
- Let the final stored value answer the original question.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01//bfs
02//Runtime: 80 ms, faster than 5.44% of C++ online submissions for Surrounded Regions.
03//Memory Usage: 19.1 MB, less than 8.24% of C++ online submissions for Surrounded Regions.
04class Solution {
05public:
06 void solve(vector<vector<char>>& board) {
07 int m = board.size();
08 if(m == 0) return;
09 int n = board[0].size();
10 if(n == 0) return;
11 vector<vector<bool>> visited(m, vector<bool>(n, false));
12 vector<vector<int>> dirs = {
13 {0,1},
14 {0,-1},
15 {1,0},
16 {-1,0}
17 };
18 vector<vector<int>> component;
19 vector<int> cur;
20
21 for(int i = 0; i < m; ++i){
22 for(int j = 0; j < n; ++j){
23 bool surrounded = true;
24 component.clear();
25 if(board[i][j] == 'O' && !visited[i][j]){
26 queue<vector<int>> q;
27 q.push({i, j});
28 visited[i][j] = true;
29
30 while(!q.empty()){
31 cur = q.front(); q.pop();
32 if(cur[0] == 0 || cur[0] == m-1 || cur[1] == 0 || cur[1] == n-1){
33 surrounded = false;
34 }
35 component.push_back(cur);
36
37 for(vector<int>& dir : dirs){
38 int ni = cur[0] + dir[0];
39 int nj = cur[1] + dir[1];
40 if(ni >= 0 && ni < m && nj >= 0 && nj < n && board[ni][nj] == 'O' && !visited[ni][nj]){
41 visited[ni][nj] = true;
42 q.push({ni, nj});
43 }
44 }
45 }
46 }
47
48 if(surrounded && !component.empty()){
49 for(vector<int>& cur : component){
50 board[cur[0]][cur[1]] = 'X';
51 }
52 }
53 }
54 }
55 }
56};
57
58//bfs
59//https://leetcode.com/problems/surrounded-regions/discuss/41612/A-really-simple-and-readable-C%2B%2B-solutionuff0conly-cost-12ms
60//Runtime: 32 ms, faster than 54.14% of C++ online submissions for Surrounded Regions.
61//Memory Usage: 11 MB, less than 44.85% of C++ online submissions for Surrounded Regions.
62class Solution {
63public:
64 int m, n;
65 vector<vector<bool>> visited;
66 vector<vector<int>> dirs;
67 vector<int> cur;
68
69 void changeNeighbor(vector<vector<char>>& board, int i, int j, char src, char dst){
70 queue<vector<int>> q;
71 q.push({i, j});
72 visited[i][j] = true;
73
74 while(!q.empty()){
75 cur = q.front(); q.pop();
76 board[cur[0]][cur[1]] = dst;
77
78 for(vector<int>& dir : dirs){
79 int ni = cur[0] + dir[0];
80 int nj = cur[1] + dir[1];
81 if(ni >= 0 && ni < m && nj >= 0 && nj < n && board[ni][nj] == 'O' && !visited[ni][nj]){
82 visited[ni][nj] = true;
83 q.push({ni, nj});
84 }
85 }
86 }
87 };
88
89 void solve(vector<vector<char>>& board) {
90 m = board.size();
91 if(m == 0) return;
92 n = board[0].size();
93 if(n == 0) return;
94 visited = vector<vector<bool>>(m, vector<bool>(n, false));
95 dirs = {
96 {0,1},
97 {0,-1},
98 {1,0},
99 {-1,0}
100 };
101
102 for(int j = 0; j < n; ++j){
103 for(int i : {0, m-1}){
104 if(board[i][j] == 'O' && !visited[i][j]){
105 changeNeighbor(board, i, j, 'O', '1');
106 }
107 }
108 }
109
110 for(int i = 0; i < m; ++i){
111 for(int j : {0, n-1}){
112 if(board[i][j] == 'O' && !visited[i][j]){
113 changeNeighbor(board, i, j, 'O', '1');
114 }
115 }
116 }
117
118 for(int i = 0; i < m; ++i){
119 for(int j = 0; j < n; ++j){
120 if(board[i][j] == 'O'){
121 board[i][j] = 'X';
122 }else if(board[i][j] == '1'){
123 board[i][j] = 'O';
124 }
125 }
126 }
127 }
128};
129
130//union find
131//https://leetcode.com/problems/surrounded-regions/discuss/41617/Solve-it-using-Union-Find
132//Runtime: 36 ms, faster than 41.79% of C++ online submissions for Surrounded Regions.
133//Memory Usage: 10.4 MB, less than 63.29% of C++ online submissions for Surrounded Regions.
134class DSU{
135public:
136 vector<int> parent;
137
138 DSU(int N){
139 parent = vector<int>(N);
140 iota(parent.begin(), parent.end(), 0);
141 };
142
143 //wrong implementation
144 // int find(int x){
145 // if(parent[x] == x){
146 // return x;
147 // }
148 // return parent[x] = find(parent[x]);
149 // };
150
151 int find(int x){
152 if(parent[x] != x){ //initial state
153 parent[x] = find(parent[x]);
154 }
155 return parent[x];
156 }
157
158 void unite(int x, int y){
159 parent[find(x)] = find(y);
160 };
161
162 bool isConnected(int x, int y){
163 return find(x) == find(y);
164 }
165};
166
167class Solution {
168public:
169 int m, n;
170
171 int node(int i, int j){
172 return i*n+j;
173 };
174
175 void solve(vector<vector<char>>& board) {
176 m = board.size();
177 if(m == 0) return;
178 n = board[0].size();
179 if(n == 0) return;
180
181 vector<vector<int>> dirs = {
182 {0,1},
183 {0,-1},
184 {1,0},
185 {-1,0}
186 };
187
188 //[0, m*n-1] for each cell, m*n for dummy cell
189 DSU dsu(m*n+1);
190 int dummyNode = m*n;
191
192 for(int i = 0; i < m; ++i){
193 for(int j = 0; j < n; ++j){
194 if(board[i][j] != 'O') continue;
195 if(i == 0 || i == m-1 || j == 0 || j == n-1){
196 dsu.unite(node(i, j), dummyNode);
197 }else{
198 for(vector<int>& dir : dirs){
199 int ni = i + dir[0];
200 int nj = j + dir[1];
201 if(ni >= 0 && ni < m && nj >= 0 && nj < n && board[ni][nj] == 'O'){
202 dsu.unite(node(ni, nj), node(i, j));
203 }
204 }
205 //another implementation
206 // if(i > 0 && board[i-1][j] == 'O') dsu.unite(node(i,j), node(i-1,j));
207 // if(i < m-1 && board[i+1][j] == 'O') dsu.unite(node(i,j), node(i+1,j));
208 // if(j > 0 && board[i][j-1] == 'O') dsu.unite(node(i,j), node(i, j-1));
209 // if(j < n-1 && board[i][j+1] == 'O') dsu.unite(node(i,j), node(i, j+1));
210 }
211 }
212 }
213
214 for(int i = 0; i < m; ++i){
215 for(int j = 0; j < n; ++j){
216 if(dsu.isConnected(node(i, j), dummyNode)){
217 board[i][j] = 'O';
218 }else{
219 board[i][j] = 'X';
220 }
221 }
222 }
223 }
224};Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.