Count all 0s which are blocked by 1s in binary matrix in C++

In this tutorial, we will be discussing a program to find the count of 0s which are blocked by 1s in a binary matrix.

For this we will be provided with a binary matrix. Our task is to find and count all the 0s in the matrix that are blocked by 1s.

Example

 Live Demo

#include <iostream>
using namespace std;
#define Row 4
#define Col 5
int r[4] = { 0, 0, 1, -1 };
int c[4] = { 1, -1, 0, 0 };
bool isSafe(int x, int y, int M[][Col]) {
   if (x >= 0 && x <= Row && y >= 0 &&
      y <= Col && M[x][y] == 0)
      return true;
   return false;
}
//performing DFS in the matrix
void DFS(int x, int y, int M[][Col]) {
   //marking the node as visited
   M[x][y] = 1;
   for (int k = 0; k < 4; k++)
      if (isSafe(x + r[k], y + c[k], M))
         DFS(x + r[k], y + c[k], M);
}
//returning count of blocked 0s
int CountAllZero(int M[][Col]){
   for (int i = 0; i < Col; i++)
      if (M[0][i] == 0)
         DFS(0, i, M);
   for (int i = 0; i < Col; i++)
      if (M[Row - 1][i] == 0)
         DFS(Row - 1, i, M);
   for (int i = 0; i < Row; i++)
      if (M[i][0] == 0)
         DFS(i, 0, M);
   for (int i = 0; i < Row; i++)
      if (M[i][Col - 1] == 0)
         DFS(i, Col - 1, M);
//counting all zeros which are surrounded by 1
int result = 0;
for (int i = 0; i < Row; i++)
   for (int j = 0; j < Col; j++)
      if (M[i][j] == 0)
         result++;
   return result;
}
int main(){
   int M[][Col] = { { 1, 1, 1, 0, 1 },{ 1, 0, 0, 1, 0 },{ 1, 0, 1, 0, 1 },{ 0, 1, 1, 1, 1 } };
   cout << CountAllZero(M) << endl;
   return 0;
}

Output

4

Ayush Gupta

Published on 05-Feb-2020 07:37:05

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