Check if cells numbered 1 to K in a grid can be connected after removal of atmost one blocked cell
Given a grid A of size N*M consisting of K cells denoted by values in range [1, K], some blocked cells denoted by -1 and remaining unblocked cells denoted by 0, the task is to check if it is possible to connect those K cells, directly or indirectly, by unblocking atmost one cell. It is possible to move only to the adjacent horizontal and adjacent vertical cells.
Examples
Input:
A = {{0, 5, 6, 0},
{3, -1, -1, 4},
{-1, 2, 1, -1},
{-1, -1, -1, -1}},
K = 6
Output: Yes
Explanation:
Unblocking cell (2, 2) or (2, 3) or (3, 1) or
(3, 4) would make all the K cells connected.
Input:
A = {{-1, -1, 3, -1},
{1, 0, -1, -1},
{-1, -1, -1, 0},
{-1, 0, 2, -1}},
K = 3
Output: No
Explanation:
Atleast two cells need to be unblocked.
Approach: Perform BFS from the cells numbered 1 to K and mark every cell by the component to which it belongs. Check if there is any blocked cell having adjacent cells belonging to different components. If there exists any, then it is possible to connect by unblocking that cell. Otherwise, it is not possible.
Example:
After performing BFS and labeling the cells by their no of components, the array appears as follows:
A={{1, 1, 1, 1}, {1, -1, -1, 1}, {-1, 2, 2, -1}, {-1, -1, -1, -1}}
The number of different label around the cell (2, 2) is 2.
Hence, unblocking it will connect the K cells.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; #define pairs pair<int, int> void check(int k, vector<vector<int> > a, int n, int m) { int cells[k][2]; bool visited[n][m]; int count = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] != 0 && a[i][j] != -1) { cells[count][0] = i; cells[count][1] = j; count++; } visited[i][j] = false; } } // Arrays to make grid traversals easier int dx[] = { 0, 0, 1, -1 }; int dy[] = { 1, -1, 0, 0 }; // Store number of components int component = 0; // Perform BFS and maark every cell // by the component in which it belongs for (int i = 0; i < k; i++) { int x = cells[i][0], y = cells[i][1]; if (visited[x][y]) continue; component++; queue<pairs> cells; cells.push(make_pair(x, y)); visited[x][y] = true; while (!cells.empty()) { pairs z = cells.front(); cells.pop(); a[z.first][z.second] = component; for (int j = 0; j < 4; j++) { int new_x = z.first + dx[j]; int new_y = z.second + dy[j]; if (new_x < 0 || new_x >= n || new_y < 0 || new_y >= m) continue; if (visited[new_x][new_y] || a[new_x][new_y] == -1) continue; cells.push(make_pair(new_x, new_y)); visited[new_x][new_y] = true; } } } int maximum = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] == -1) { unordered_set<int> set; for (int kk = 0; kk < 4; kk++) { int xx = i + dx[kk]; int yy = j + dy[kk]; if (xx < 0 || xx >= n || yy < 0 || yy >= m) continue; // if the cell doesn't // belong to any component if (a[xx][yy] <= 0) continue; set.insert(a[xx][yy]); } int s = set.size(); maximum = max(s, maximum); } } } if (maximum == component) { cout << "Yes\n"; } else { cout << "No\n"; } } int main() { int k = 6; int n = 4, m = 4; vector<vector<int> > a = { { 0, 5, 6, 0 }, { 3, -1, -1, 4 }, { -1, 2, 1, -1 }, { -1, -1, -1, -1 } }; check(k, a, n, m); return 0; } |
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Java
// Java implementation of the above approach import java.util.*; class GFG{ static class pair { int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } static void check(int k, int [][]a, int n, int m) { int [][]cell = new int[k][2]; boolean [][]visited = new boolean[n][m]; int count = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] != 0 && a[i][j] != -1) { cell[count][0] = i; cell[count][1] = j; count++; } visited[i][j] = false; } } // Arrays to make grid traversals easier int dx[] = { 0, 0, 1, -1 }; int dy[] = { 1, -1, 0, 0 }; // Store number of components int component = 0; // Perform BFS and maark every cell // by the component in which it belongs for (int i = 0; i < k; i++) { int x = cell[i][0], y = cell[i][1]; if (visited[x][y]) continue; component++; Queue<pair> cells = new LinkedList<>(); cells.add(new pair(x, y)); visited[x][y] = true; while (!cells.isEmpty()) { pair z = cells.peek(); cells.remove(); a[z.first][z.second] = component; for (int j = 0; j < 4; j++) { int new_x = z.first + dx[j]; int new_y = z.second + dy[j]; if (new_x < 0 || new_x >= n || new_y < 0 || new_y >= m) continue; if (visited[new_x][new_y] || a[new_x][new_y] == -1) continue; cells.add(new pair(new_x, new_y)); visited[new_x][new_y] = true; } } } int maximum = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] == -1) { HashSet<Integer> set = new HashSet<Integer>(); for (int kk = 0; kk < 4; kk++) { int xx = i + dx[kk]; int yy = j + dy[kk]; if (xx < 0 || xx >= n || yy < 0 || yy >= m) continue; // if the cell doesn't // belong to any component if (a[xx][yy] <= 0) continue; set.add(a[xx][yy]); } int s = set.size(); maximum = Math.max(s, maximum); } } } if (maximum == component) { System.out.print("Yes\n"); } else { System.out.print("No\n"); } } public static void main(String[] args) { int k = 6; int n = 4, m = 4; int [][]a = { { 0, 5, 6, 0 }, { 3, -1, -1, 4 }, { -1, 2, 1, -1 }, { -1, -1, -1, -1 } }; check(k, a, n, m); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the above approach from collections import deque as queue def check(k, a, n, m): cells = [[0 for i in range(2)] for i in range(k)] visited = [[0 for i in range(m)] for i in range(n)] count = 0 for i in range(n): for j in range(m): if (a[i][j] != 0 and a[i][j] != -1): cells[count][0] = i cells[count][1] = j count += 1 visited[i][j] = False # Arrays to make grid traversals easier dx = [0, 0, 1, -1] dy = [1, -1, 0, 0] # Store number of components component = 0 # Perform BFS and maark every cell # by the component in which it belongs for i in range(k): x = cells[i][0] y = cells[i][1] if (visited[x][y]): continue component += 1 cell = queue() cell.append([x, y]) visited[x][y] = True while (len(cell) > 0): z = cell.popleft() a[z[0]][z[1]] = component for j in range(4): new_x = z[0] + dx[j] new_y = z[1] + dy[j] if (new_x < 0 or new_x >= n or new_y < 0 or new_y >= m): continue if (visited[new_x][new_y] or a[new_x][new_y] == -1): continue cell.append([new_x, new_y]) visited[new_x][new_y] = True maximum = 0 for i in range(n): for j in range(m): if (a[i][j] == -1): se = dict() for kk in range(4): xx = i + dx[kk] yy = j + dy[kk] if (xx < 0 or xx >= n or yy < 0 or yy >= m): continue # if the cell doesn't # belong to any component if (a[xx][yy] <= 0): continue se[a[xx][yy]] = 1 s = len(se) maximum = max(s, maximum) if (maximum == component): print("Yes\n") else: print("No\n") # Driver code if __name__ == '__main__': k = 6 n = 4 m = 4 a=[[0, 5, 6, 0 ], [3, -1, -1, 4], [-1, 2, 1, -1], [-1, -1,-1,-1]] check(k, a, n, m) # This code is contributed by mohit kumar 29 |
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C#
// C# implementation of the above approach using System; using System.Collections.Generic; class GFG{ class pair { public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } static void check(int k, int [,]a, int n, int m) { int [,]cell = new int[k,2]; bool [,]visited = new bool[n,m]; int count = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i, j] != 0 && a[i, j] != -1) { cell[count, 0] = i; cell[count, 1] = j; count++; } visited[i, j] = false; } } // Arrays to make grid traversals easier int []dx = { 0, 0, 1, -1 }; int []dy = { 1, -1, 0, 0 }; // Store number of components int component = 0; // Perform BFS and maark every cell // by the component in which it belongs for (int i = 0; i < k; i++) { int x = cell[i, 0], y = cell[i, 1]; if (visited[x, y]) continue; component++; List<pair> cells = new List<pair>(); cells.Add(new pair(x, y)); visited[x, y] = true; while (cells.Count != 0) { pair z = cells[0]; cells.RemoveAt(0); a[z.first,z.second] = component; for (int j = 0; j < 4; j++) { int new_x = z.first + dx[j]; int new_y = z.second + dy[j]; if (new_x < 0 || new_x >= n || new_y < 0 || new_y >= m) continue; if (visited[new_x,new_y] || a[new_x, new_y] == -1) continue; cells.Add(new pair(new_x, new_y)); visited[new_x, new_y] = true; } } } int maximum = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i, j] == -1) { HashSet<int> set = new HashSet<int>(); for (int kk = 0; kk < 4; kk++) { int xx = i + dx[kk]; int yy = j + dy[kk]; if (xx < 0 || xx >= n || yy < 0 || yy >= m) continue; // if the cell doesn't // belong to any component if (a[xx, yy] <= 0) continue; set.Add(a[xx, yy]); } int s = set.Count; maximum = Math.Max(s, maximum); } } } if (maximum == component) { Console.Write("Yes\n"); } else { Console.Write("No\n"); } } public static void Main(String[] args) { int k = 6; int n = 4, m = 4; int [,]a = { { 0, 5, 6, 0 }, { 3, -1, -1, 4 }, { -1, 2, 1, -1 }, { -1, -1, -1, -1 } }; check(k, a, n, m); } } // This code is contributed by 29AjayKumar |
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Output:
Yes
Performance Analysis:
- Time Complexity: Performing BFS on the matrix takes O(N*M) time and O(N*M) time for checking every blocked cell. Hence the overall Time Complexity will be O(N * M).
- Auxiliary Space Complexity: O(N * M)
