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Daily/Number_Of_Enclaves.cpp

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+/*
+You are given an n x m binary matrix grid, where 0 represents a sea cell and 1 represents a land cell.
+
+A move consists of walking from one land cell to another adjacent (4-directionally) land cell or walking off the boundary of the grid.
+
+Find the number of land cells in grid for which we cannot walk off the boundary of the grid in any number of moves.
+
+Example 1:
+
+Input:
+grid[][] = {{0, 0, 0, 0},
+            {1, 0, 1, 0},
+            {0, 1, 1, 0},
+            {0, 0, 0, 0}}
+Output:
+3
+Explanation:
+0 0 0 0
+1 0 1 0
+0 1 1 0
+0 0 0 0
+The highlighted cells represents the land cells.
+
+Example 2:
+
+Input:
+grid[][] = {{0, 0, 0, 1},
+            {0, 1, 1, 0},
+            {0, 1, 1, 0},
+            {0, 0, 0, 1},
+            {0, 1, 1, 0}}
+Output:
+4
+Explanation:
+0 0 0 1
+0 1 1 0
+0 1 1 0
+0 0 0 1
+0 1 1 0
+The highlighted cells represents the land cells.*/
+
+//{ Driver Code Starts
+// Initial Template for C++
+
+#include <bits/stdc++.h>
+using namespace std;
+
+
+// } Driver Code Ends
+// User function Template for C++
+
+class Solution {
+  private:
+    void dfs(int i,int j,vector<vector<int>> &visit,vector<vector<int>> &grid){
+        visit[i][j] = 1;
+        int n = grid.size();
+        int m = grid[0].size();
+
+        int row[] = {1,0,-1,0};
+        int col[] = {0,1,0,-1};
+
+        for(int k=0;k<4;k++){
+            int nr = i+row[k];
+            int nc = j+col[k];
+
+            if(nr<0||nr>=n||nc<0||nc>=m) continue;
+
+            if(grid[nr][nc]==1 && !visit[nr][nc]){
+                visit[nr][nc] = 1;
+                dfs(nr,nc,visit,grid);
+            }
+        }
+    }
+  public:
+    int numberOfEnclaves(vector<vector<int>> &grid) {
+        // Code here
+
+        int n = grid.size();
+        int m = grid[0].size();
+
+        vector<vector<int>> visit(n,vector<int>(m,0));
+
+        for(int i=0;i<n;i++){
+            for(int j=0;j<m;j++){
+                if(i==0||j==0||j==m-1||i==n-1){
+                    if(grid[i][j]==1 && !visit[i][j]){
+                        dfs(i,j,visit,grid);
+                    }
+                }
+            }
+        }
+        int count = 0;
+        for(int i=0;i<n;i++){
+            for(int j=0;j<m;j++){
+                if(grid[i][j]==1 && !visit[i][j])
+                    count++;
+            }
+        }
+        return count;
+    }
+};
+
+
+//{ Driver Code Starts.
+
+int main() {
+    int t;
+    cin >> t;
+    while (t--) {
+        int n, m;
+        cin >> n >> m;
+        vector<vector<int>> grid(n, vector<int>(m));
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                cin >> grid[i][j];
+            }
+        }
+        Solution obj;
+        cout << obj.numberOfEnclaves(grid) << endl;
+    }
+}
+// } Driver Code Ends