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Count all 0s which are blocked by 1s in binary matrix
  • Difficulty Level : Medium
  • Last Updated : 18 Mar, 2020

Given Binary matrix. Task is count all zeros which are surrounded by one (may not be immediate neighbor).
Note: here we are only taking four direction up, left, down, right.
Examples:

Input :  Int M[][] = {{ 0, 1, 1, 0},
                      { 1, 0, 0, 1},
                      { 0, 1, 0, 1},
                      { 1, 0, 1, 1}}
Output : 3 
Explanation : All zeros which are surrounded 
by 1 are (1, 1), (1, 2) and (2, 2)    

Idea is based on the DFS.
First we remove all the zero value cells in the matrix which are reachable from boundary of Matrix using DFS. Note that any cell which is reachable from a boundary 0 cell is not surrounded by ones.

  Int M[][] =  {{  0, 1, 1,  0},
                { 1, 0, 0, 1},
                { 0, 1, 1, 1},
                { 1,  0, 1, 1}}
All zero cell which are reachable from boundary are in read color.

After that we count all zeros which are left in binary matrix.
Below is the implementation of above idea.

C++




// C++ program to count number of zeros
// surrounded by 1
#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;
}
  
// DFS function to mark all reachable cells from
// (x, y)
void DFS(int x, int y, int M[][Col])
{
    // make it's 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);
}
  
// function return count of 0's which are surrounded by 1
int CountAllZero(int M[][Col])
{
    // first we remove all zeros which are not
    // surrounded by 1 that means we only remove 
    // those zeros which are reachable from
    // any boundary of given matrix.
    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);
  
    // count 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;
}
  
// driver program to test above function
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;
}


Java




// Java program to count number of zeros
// surrounded by 1
class GFG {
  
    final static int Row = 4;
    final static int Col = 5;
    static int r[] = {0, 0, 1, -1};
    static int c[] = {1, -1, 0, 0};
  
    static boolean isSafe(int x, int y, int M[][]) {
        if (x >= 0 && x <Row && y >= 0
                && y < Col && M[x][y] == 0) {
            return true;
        }
        return false;
    }
  
// DFS function to mark all reachable cells from
// (x, y)
    static void DFS(int x, int y, int M[][]) {
        // make it's 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);
            }
        }
    }
  
// function return count of 0's which are surrounded by 1
    static int CountAllZero(int M[][]) {
        // first we remove all zeros which are not
        // surrounded by 1 that means we only remove 
        // those zeros which are reachable from
        // any boundary of given matrix.
        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);
            }
        }
  
        // count 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;
    }
  
// driver program to test above function
    public static void main(String[] args) {
        int M[][] = {{1, 1, 1, 0, 1},
        {1, 0, 0, 1, 0},
        {1, 0, 1, 0, 1},
        {0, 1, 1, 1, 1}};
        System.out.print(CountAllZero(M));
    }
}
// This code is contributed by Rajput-Ji


Python3




# Python3 program to count number of zeros 
# surrounded by 1 
Row = 4
Col = 5
r = [0, 0, 1, -1 ]
c = [1, -1, 0, 0 ]
  
def isSafe(x, y, M):
      
    if (x >= 0 and x < Row and y >= 0 and y < Col):
        if M[x][y] == 0:
            return True
      
    return False
  
# DFS function to mark all reachable cells from 
# (x, y) 
def DFS(x, y, M):
      
    # make it's visited 
    M[x][y] = 1
    for k in range(4):
        if (isSafe(x + r[k], y + c[k], M)):
            DFS(x + r[k], y + c[k], M) 
  
# function return count of 0's which are surrounded by 1 
def CountAllZero(M):
  
    # first we remove all zeros which are not 
    # surrounded by 1 that means we only remove 
    # those zeros which are reachable from 
    # any boundary of given matrix. 
    for i in range(Col):
        if (M[0][i] == 0):
            DFS(0, i, M) 
    for i in range(Col):
        if (M[Row - 1][i] == 0):
            DFS(Row - 1, i, M) 
    for i in range(Row): 
        if (M[i][0] == 0):
            DFS(i, 0, M) 
    for i in range(Row): 
        if (M[i][Col - 1] == 0):
            DFS(i, Col - 1, M) 
      
    # count all zeros which are surrounded by 1 
    result = 0
    for i in range(Row): 
        for j in range(Col):
            if (M[i][j] == 0):
                result += 1
      
    return result 
  
# Driver code 
M = [[1, 1, 1, 0, 1], [1, 0, 0, 1, 0 ], 
    [1, 0, 1, 0, 1] , [ 0, 1, 1, 1, 1 ]] 
print(CountAllZero(M))
  
# This code is conributed by shubhamsingh10


C#




      
// C# program to count number of zeros
// surrounded by 1
using System;
public class GFG {
   
    readonly static int Row = 4;
    readonly static int Col = 5;
    static int []r = {0, 0, 1, -1};
    static int []c = {1, -1, 0, 0};
   
    static bool isSafe(int x, int y, int [,]M) {
        if (x >= 0 && x <Row && y >= 0
                && y < Col && M[x,y] == 0) {
            return true;
        }
        return false;
    }
   
// DFS function to mark all reachable cells from
// (x, y)
    static void DFS(int x, int y, int [,]M) {
        // make it's 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);
            }
        }
    }
   
// function return count of 0's which are surrounded by 1
    static int CountAllZero(int [,]M) {
        // first we remove all zeros which are not
        // surrounded by 1 that means we only remove 
        // those zeros which are reachable from
        // any boundary of given matrix.
        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);
            }
        }
   
        // count 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;
    }
   
// driver program to test above function
    public static void Main() {
        int [,]M = {{1, 1, 1, 0, 1},
        {1, 0, 0, 1, 0},
        {1, 0, 1, 0, 1},
        {0, 1, 1, 1, 1}};
        Console.Write(CountAllZero(M));
    }
// This code is contributed by 29AjayKumar



Output:



4

Time Complexity: O(Row*Col)

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