| 3 | // Space Complexity : O(V) |
| 4 | |
| 5 | class Solution { |
| 6 | public: |
| 7 | vector<int> del_i = {1, 0, -1, 0}; |
| 8 | vector<int> del_j = {0, 1, 0, -1}; |
| 9 | |
| 10 | bool checkWithinBounds(int i, int j, int row, int col) { |
| 11 | if (i < 0 || j < 0 || i >= row || j >= col) return false; |
| 12 | return true; |
| 13 | } |
| 14 | |
| 15 | void markIsland(vector<vector<int>>& A, int i, int j) { |
| 16 | A[i][j] = -1; |
| 17 | |
| 18 | for (int k = 0; k < 4; ++k) |
| 19 | { |
| 20 | int next_i = i + del_i[k]; |
| 21 | int next_j = j + del_j[k]; |
| 22 | |
| 23 | if (checkWithinBounds(next_i, next_j, A.size(), A[0].size()) && A[next_i][next_j] == 1) |
| 24 | markIsland(A, next_i, next_j); |
| 25 | } |
| 26 | |
| 27 | } |
| 28 | |
| 29 | int shortestBridge(vector<vector<int>>& A) { |
| 30 | int r = A.size(); |
| 31 | int c = A[0].size(); |
| 32 | |
| 33 | // mark island of one group as -1 |
| 34 | int flag = 0; |
| 35 | for (int i = 0; i < r; ++i) |
| 36 | { |
| 37 | for (int j = 0; j < c; ++j) |
| 38 | { |
| 39 | if (A[i][j] == 1) { |
| 40 | flag = 1; |
| 41 | markIsland(A, i, j); |
| 42 | break; |
| 43 | } |
| 44 | } |
| 45 | |
| 46 | if (flag) break; |
| 47 | } |
| 48 | |
| 49 | // queue that represent flips on a particular path |
| 50 | queue<pair<int, int>> q; |
| 51 | |
| 52 | for (int i = 0; i < r; ++i) |
| 53 | { |
| 54 | for (int j = 0; j < c; ++j) |
| 55 | { |
| 56 | if (A[i][j] == -1) { |
| 57 | q.push({i * r + j, 0}); |
| 58 | } |
| 59 | } |
| 60 | } |
| 61 | |
| 62 | while (!q.empty()) { |
nothing calls this directly
no outgoing calls
no test coverage detected