Check if elements of a Binary Matrix can be made alternating

Last Updated : 30 Oct, 2023

Given a 2D array grid[][] of size N * M, consisting of the characters “1”, “0”, and “*”, where “*” denotes an empty space and can be replaced by either a “1” or a “0”. The task is to fill the grid such that “0” and “1” occur alternatively and no two consecutive characters occur together, i.e. (101010) is valid and (101101) is not because two “1” occurs simultaneously. If it is possible to do so, print Yes and the possible 2-D array. Otherwise, print No.

Examples:

Input: N = 4, M = 4 grid[][] = { {**10}, {****}, {****}, {**01}}
Output: Yes
1010
0101
1010
0101
Explanation: Create a grid with alternating “1” and “0” characters so the answer is Yes, followed by the filled characters.

Input: N = 4, M = 4, grid[][] = {{*1*0}, {****}, {**10}, {****}}
Output: No
Explanation: In the first row, 1 and 0 have one cell blank which can neither be filled with 1 nor 0.

Approach: There are only 2 possibilities for a possible 2-D array one, with starting 1 and one with starting 0. Generate both of them and check if any one of them matches with the given 2-D array grid[][]. Follow the steps below to solve the problem:

Define a function createGrid(char grid[][1001], bool is1, int N, int M) and perform the following tasks:
- Iterate over the range [0, N] using the variable i and performing the following tasks:
  - Iterate over the range [0, M] using the variable j and performing the following tasks:
    - If is1 is true, then set grid[i][j] to ‘0’ and set is1 to false.
    - Else, then set grid[i][j] to ‘1’ and set is1 to true.
  - If M%2 is equal to 0, then set the value of is1 to not of is1.
Define a function testGrid(char testGrid[][1001], char Grid[][1001], int N, int M) and perform the following tasks:
- Iterate over the range [0, N] using the variable i and performing the following tasks:
  - Iterate over the range [0, M] using the variable j and performing the following tasks:
    - If Grid[i][j] is not equal to ‘*’ and testGrid[i][j] is not equal to Grid[i][j], then return false.
- After performing the above steps, return the value of true as the answer.
Define a function printGrid(char grid[][1001], int N, int M) and perform the following tasks:
- Iterate over the range [0, N] using the variable i and performing the following tasks:
  - Iterate over the range [0, M] using the variable j and performing the following tasks:
    - Print the value of grid[i][j].
Initialize two 2-D arrays gridTest1[N][1001] and gridTest2[N][1001] to store the possible alternating grids.
Call the function createGrid(gridTest1, true, N, M) and createGrid(gridTest2, false, N, M) to form the possible alternating grids.
If the function testGrid(gridTest1, grid, N, M) returns true, then call the function printGrid(gridTest1, N, M) to print the grid as the answer.
Else If, the function testGrid(gridTest2, grid, N, M) returns true, then call the function printGrid(gridTest2, N, M) to print the grid as the answer.
Else, print No as the answer.

Below is the implementation of the above approach.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to create the possible grids
void createGrid(char grid[][1001], bool is1,
                int N, int M)
{
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            if (is1) {
                grid[i][j] = '0';
                is1 = false;
            }
            else {
                grid[i][j] = '1';
                is1 = true;
            }
        }
        if (M % 2 == 0)
            is1 = !is1;
    }
}
 
// Function to test if any one of them
// matches with the given 2-D array
bool testGrid(char testGrid[][1001],
              char Grid[][1001],
              int N, int M)
{
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
 
            if (Grid[i][j] != '*') {
 
                if (Grid[i][j] != testGrid[i][j]) {
                    return false;
                }
            }
        }
    }
    return true;
}
 
// Function to print the grid, if possible
void printGrid(char grid[][1001], int N, int M)
{
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
            cout << grid[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Function to check if the grid
// can be made alternating or not
void findPossibleGrid(int N, int M,
                      char grid[][1001])
{
    // Grids to store the possible grids
    char gridTest1[N][1001], gridTest2[N][1001];
 
    createGrid(gridTest1, true, N, M);
 
    createGrid(gridTest2, false, N, M);
 
    if (testGrid(gridTest1, grid, N, M)) {
 
        cout << "Yes\n";
        printGrid(gridTest1, N, M);
    }
    else if (testGrid(gridTest2, grid, N, M)) {
 
        cout << "Yes\n";
        printGrid(gridTest2, N, M);
    }
    else {
        cout << "No\n";
    }
}
 
// Driver Code
int main()
{
    int N = 4, M = 4;
    char grid[][1001] = { { '*', '*', '1', '0' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '0', '1' } };
 
    findPossibleGrid(N, M, grid);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to create the possible grids
public static void createGrid(char[][] grid,
                              boolean is1, 
                              int N, int M)
{
    for(int i = 0; i < N; i++) 
    {
        for(int j = 0; j < M; j++) 
        {
            if (is1)
            {
                grid[i][j] = '0';
                is1 = false;
            }
            else
            {
                grid[i][j] = '1';
                is1 = true;
            }
        }
        if (M % 2 == 0)
            is1 = !is1;
    }
}
 
// Function to test if any one of them
// matches with the given 2-D array
public static boolean testGrid(char[][] testGrid,
                               char[][] Grid, int N,
                               int M)
{
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < M; j++) 
        {
            if (Grid[i][j] != '*')
            {
                if (Grid[i][j] != testGrid[i][j]) 
                {
                    return false;
                }
            }
        }
    }
    return true;
}
 
// Function to print the grid, if possible
public static void printGrid(char[][] grid, int N,
                             int M)
{
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < M; j++) 
        {
            System.out.print(grid[i][j] + " ");
        }
        System.out.println();
    }
}
 
// Function to check if the grid
// can be made alternating or not
public static void findPossibleGrid(int N, int M,
                                    char[][] grid)
{
     
    // Grids to store the possible grids
    char[][] gridTest1 = new char[N][1001];
    char[][] gridTest2 = new char[N][1001];
 
    createGrid(gridTest1, true, N, M);
    createGrid(gridTest2, false, N, M);
 
    if (testGrid(gridTest1, grid, N, M))
    {
        System.out.println("Yes");
        printGrid(gridTest1, N, M);
    }
    else if (testGrid(gridTest2, grid, N, M)) 
    {
        System.out.println("Yes");
        printGrid(gridTest2, N, M);
    }
    else
    {
        System.out.println("No");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4, M = 4;
    char[][] grid = { { '*', '*', '1', '0' },
                      { '*', '*', '*', '*' },
                      { '*', '*', '*', '*' },
                      { '*', '*', '0', '1' } };
 
    findPossibleGrid(N, M, grid);
}
}
 
// This code is contributed by maddler

Python3

# python 3 program for the above approach
 
# Function to create the possible grids
def createGrid(grid, is1, N, M):
    for i in range(N):
        for j in range(M):
            if (is1):
                grid[i][j] = '0'
                is1 = False
 
            else:
                grid[i][j] = '1'
                is1 = True
 
        if (M % 2 == 0):
            is1 = True if is1 == False else False
 
# Function to test if any one of them
# matches with the given 2-D array
def testGrid(testGrid, Grid, N, M):
    for i in range(N):
        for j in range(M):
            if (Grid[i][j] != '*'):
                if (Grid[i][j] != testGrid[i][j]):
                    return False
 
    return True
 
# Function to print the grid, if possible
def printGrid(grid, N, M):
    for i in range(N):
        for j in range(M):
            print(grid[i][j],end = " ")
        print("\n",end = "")
 
# Function to check if the grid
# can be made alternating or not
def findPossibleGrid(N, M, grid):
    # Grids to store the possible grids
    gridTest1 = [['' for i in range(1001)] for j in range(N)]
    gridTest2 = [['' for i in range(1001)] for j in range(N)]
 
    createGrid(gridTest1, True, N, M)
 
    createGrid(gridTest2, False, N, M)
 
    if(testGrid(gridTest1, grid, N, M)):
        print("Yes")
        printGrid(gridTest1, N, M)
 
    elif(testGrid(gridTest2, grid, N, M)):
        print("Yes")
        printGrid(gridTest2, N, M)
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
    N = 4
    M = 4
    grid  = [['*', '*', '1', '0'],
             ['*', '*', '*', '*'],
             ['*', '*', '*', '*'],
             ['*', '*', '0', '1']]
 
    findPossibleGrid(N, M, grid)
     
    # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to create the possible grids
public static void createGrid(char[,] grid,
                              bool is1, 
                              int N, int M)
{
    for(int i = 0; i < N; i++) 
    {
        for(int j = 0; j < M; j++) 
        {
            if (is1)
            {
                grid[i, j] = '0';
                is1 = false;
            }
            else
            {
                grid[i, j] = '1';
                is1 = true;
            }
        }
        if (M % 2 == 0)
            is1 = !is1;
    }
}
 
// Function to test if any one of them
// matches with the given 2-D array
public static bool testGrid(char[,] testGrid,
                            char[,] Grid, int N,
                            int M)
{
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < M; j++) 
        {
            if (Grid[i, j] != '*')
            {
                if (Grid[i, j] != testGrid[i, j]) 
                {
                    return false;
                }
            }
        }
    }
    return true;
}
 
// Function to print the grid, if possible
public static void printGrid(char[,] grid, int N,
                             int M)
{
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < M; j++) 
        {
            Console.Write(grid[i, j] + " ");
        }
        Console.WriteLine();
    }
}
 
// Function to check if the grid
// can be made alternating or not
public static void findPossibleGrid(int N, int M,
                                    char[,] grid)
{
     
    // Grids to store the possible grids
    char[,] gridTest1 = new char[N, 1001];
    char[,] gridTest2 = new char[N, 1001];
 
    createGrid(gridTest1, true, N, M);
    createGrid(gridTest2, false, N, M);
 
    if (testGrid(gridTest1, grid, N, M))
    {
        Console.WriteLine("Yes");
        printGrid(gridTest1, N, M);
    }
    else if (testGrid(gridTest2, grid, N, M)) 
    {
        Console.WriteLine("Yes");
        printGrid(gridTest2, N, M);
    }
    else
    {
        Console.WriteLine("No");
    }
}
 
// Driver code
public static void Main()
{
    int N = 4, M = 4;
    char[,] grid = { { '*', '*', '1', '0' },
                     { '*', '*', '*', '*' },
                     { '*', '*', '*', '*' },
                     { '*', '*', '0', '1' } };
     
    findPossibleGrid(N, M, grid);
}
}
 
// This code is contributed by sanjoy_62

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function to create the possible grids
function createGrid(grid, is1, N, M) {
  for (let i = 0; i < N; i++) {
    for (let j = 0; j < M; j++) {
      if (is1) {
        grid[i][j] = "0";
        is1 = false;
      } else {
        grid[i][j] = "1";
        is1 = true;
      }
    }
    if (M % 2 == 0) is1 = !is1;
  }
}
 
// Function to test if any one of them
// matches with the given 2-D array
function testGrid(testGrid, Grid, N, M) {
  for (let i = 0; i < N; i++) {
    for (let j = 0; j < M; j++) {
      if (Grid[i][j] != "*") {
        if (Grid[i][j] != testGrid[i][j]) {
          return false;
        }
      }
    }
  }
  return true;
}
 
// Function to print the grid, if possible
function printGrid(grid, N, M) {
  for (let i = 0; i < N; i++) {
    for (let j = 0; j < M; j++) {
      document.write(grid[i][j] + " ");
    }
    document.write("<br>");
  }
}
 
// Function to check if the grid
// can be made alternating or not
function findPossibleGrid(N, M, grid) {
  // Grids to store the possible grids
  let gridTest1 = new Array(N).fill(0).map(() => new Array(1001));
  let gridTest2 = new Array(N).fill(0).map(() => new Array(1001));
 
  createGrid(gridTest1, true, N, M);
 
  createGrid(gridTest2, false, N, M);
 
  if (testGrid(gridTest1, grid, N, M)) {
    document.write("Yes<br>");
    printGrid(gridTest1, N, M);
  } else if (testGrid(gridTest2, grid, N, M)) {
    document.write("Yes<br>");
    printGrid(gridTest2, N, M);
  } else {
    document.write("No<br>");
  }
}
 
// Driver Code
 
let N = 4,
  M = 4;
let grid = [
  ["*", "*", "1", "0"],
  ["*", "*", "*", "*"],
  ["*", "*", "*", "*"],
  ["*", "*", "0", "1"],
];
 
findPossibleGrid(N, M, grid);
 
</script>

Output

Time Complexity: O(N*M)
Auxiliary Space: O(N*M)

Efficient Approach: The idea is to use two variables instead of creating the 2-D arrays to maintain the possible alternating arrays. Follow the steps below to solve the problem:

Define a function posCheck(char grid[][1001], int N, int M, char check) and perform the following tasks:
- Iterate over the range [0, N] using the variable i and performing the following tasks:
  - Iterate over the range [0, M] using the variable j and performing the following tasks:
    - If grid[i][j] is equal to ‘*’, then continue.
    - Else, if (i+j)%2 is equal to 1 and grid[i][j] is not equal to check, then return false.
    - Else, if (i+j)%2 is equal to 0 and grid[i][j] is equal to check, then return false.
- After performing the above steps, return the value of true as the answer.
Define a function posCheck(char grid[][1001], int N, int M, char odd, char even) and perform the following tasks:
- Iterate over the range [0, N] using the variable i and performing the following tasks:
  - Iterate over the range [0, M] using the variable j and performing the following tasks:
    - If (i+j)%2 is equal to 1, then set the value of grid[i][j] to odd else even.
Initialize the bool variable flag as true.
Initialize the variables k and o as -1 to store the value of the first cell which doesn’t have the character ‘*’.
Iterate over the range [0, N] using the variable i and performing the following tasks:
- Iterate over the range [0, M] using the variable j and performing the following tasks:
  - If Grid[i][j] is not equal to ‘*’, then set the value of k as i and o as j and break.
- If k is not equal to -1, then break.
If k is not equal to -1, then call the function PosCheck(grid, n, m, ‘1’) and store the value returned by the function in the variable flag and if flag is true, then call the function fill(grid, n, m, ‘1’, ‘0’) to fill the grid in one of the possible ways.
If flag is false, then call the function PosCheck(grid, n, m, ‘0’) and store the value returned by the function in the variable flag and if flag is true, then call the function fill(grid, n, m, ‘0’, ‘1’) to fill the grid in one of the possible ways.
If k is equal to -1, then, initialize the char variable h as ‘0’.
Iterate over the range [0, M] using the variable i and performing the following tasks:
- Set the value of grid[0][i] as h and if h is ‘0‘, then set it to ‘1‘, otherwise ‘0’.
Iterate over the range [0, N] using the variable i and performing the following tasks:
- Iterate over the range [0, M] using the variable j and performing the following tasks:
  - If i-1 is less than 0, then continue.
  - If grid[i-1][j] is ‘1’, then set it to ‘0’, else ‘1’.
Set the value of flag as true as the grid is made alternating.
If flag is false, then print “NO”.
Else, print “YES” and print the elements of the grid[][].

Below is the implementation of the above approach.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check for the grid in
// one of the alternating ways
bool PosCheck(char a[][1001], int n,
              int m, char check)
{
 
    // Iterate over the range
    for (int i = 0; i < n; i++) {
 
        // Iterate over the range
        for (int j = 0; j < m; j++) {
 
            if (a[i][j] == '*') {
                continue;
            }
            else {
 
                // (i+j)%2==1 cells should be with
                // the character check and the rest
                // should be with the other character
                if (((i + j) & 1) && a[i][j] != check) {
                    return false;
                }
                if (!((i + j) & 1) && a[i][j] == check) {
                    return false;
                }
            }
        }
    }
    return true;
}
 
// Function to fill the grid in a possible way
void fill(char a[][1001], int n, int m,
          char odd, char even)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if ((i + j) & 1) {
                a[i][j] = odd;
            }
            else {
                a[i][j] = even;
            }
        }
    }
}
 
// Function to find if the grid can be made alternating
void findPossibleGrid(int n, int m, char a[][1001])
{
    bool flag = true;
    int k = -1, o = -1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
 
            if (a[i][j] != '*') {
 
                // If grid contains atleast
                // one 1 or 0
                k = i;
                o = j;
                break;
            }
        }
        if (k != -1) {
            break;
        }
    }
    if (k != -1) {
        flag = PosCheck(a, n, m, '1');
        if (flag) {
            fill(a, n, m, '1', '0');
        }
        else {
            flag = PosCheck(a, n, m, '0');
            if (flag) {
                fill(a, n, m, '0', '1');
            }
        }
    }
    else {
        // Fill the grid in any possible way
        char h = '1';
        for (int i = 0; i < m; i++) {
            a[0][i] = h;
            if (h == '1') {
                h = '0';
            }
            else {
                h = '1';
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i - 1 < 0) {
                    continue;
                }
                if (a[i - 1][j] == '1') {
                    a[i][j] = '0';
                }
                else {
                    a[i][j] = '1';
                }
            }
        }
        flag = true;
    }
 
    if (!flag) {
        cout << "NO\n";
    }
    else {
        cout << "YES\n";
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cout << a[i][j];
            }
            cout << endl;
        }
    }
}
 
// Driver Code
int main()
{
    int n = 4, m = 4;
    char grid[][1001] = { { '*', '*', '1', '0' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '0', '1' } };
 
    findPossibleGrid(n, m, grid);
 
    return 0;
}

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    public static boolean PosCheck(char[][] a, int n, int m,
                                   char check)
    {
 
        // Iterate over the range
        for (int i = 0; i < n; i++) {
 
            // Iterate over the range
            for (int j = 0; j < m; j++) {
 
                if (a[i][j] == '*') {
                    continue;
                }
                else {
 
                    // (i+j)%2==1 cells should be with
                    // the character check and the rest
                    // should be with the other character
                    if ((((i + j) & 1) == 1)
                        && a[i][j] != check) {
                        return false;
                    }
                    if (!(((i + j) & 1) == 1)
                        && a[i][j] == check) {
                        return false;
                    }
                }
            }
        }
        return true;
    }
 
    // Function to fill the grid in a possible way
    public static void fill(char[][] a, int n, int m,
                            char odd, char even)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (((i + j) & 1) == 1) {
                    a[i][j] = odd;
                }
                else {
                    a[i][j] = even;
                }
            }
        }
    }
 
    // Function to find if the grid can be made alternating
    public static void findPossibleGrid(int n, int m,
                                        char[][] a)
    {
        boolean flag = true;
        int k = -1, o = -1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
 
                if (a[i][j] != '*') {
 
                    // If grid contains atleast
                    // one 1 or 0
                    k = i;
                    o = j;
                    break;
                }
            }
            if (k != -1) {
                break;
            }
        }
        if (k != -1) {
            flag = PosCheck(a, n, m, '1');
            if (flag) {
                fill(a, n, m, '1', '0');
            }
            else {
                flag = PosCheck(a, n, m, '0');
                if (flag) {
                    fill(a, n, m, '0', '1');
                }
            }
        }
        else {
            // Fill the grid in any possible way
            char h = '1';
            for (int i = 0; i < m; i++) {
                a[0][i] = h;
                if (h == '1') {
                    h = '0';
                }
                else {
                    h = '1';
                }
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (i - 1 < 0) {
                        continue;
                    }
                    if (a[i - 1][j] == '1') {
                        a[i][j] = '0';
                    }
                    else {
                        a[i][j] = '1';
                    }
                }
            }
            flag = true;
        }
 
        if (!flag) {
            System.out.println("No");
        }
        else {
            System.out.println("Yes");
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    System.out.print(a[i][j]);
                }
                System.out.println();
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 4, m = 4;
        char[][] grid = { { '*', '*', '1', '0' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '0', '1' } };
 
        findPossibleGrid(n, m, grid);
    }
}
 
// This code is contributed by maddler.

Python3

# Python program for the above approach
 
# Function to check for the grid in
# one of the alternating ways
def PosCheck(a, n, m, check):
 
  # Iterate over the range
  for i in range(n):
   
    # Iterate over the range
    for j in range(m):
 
        if (a[i][j] == "*"):
            continue
        else:
       
            # (i+j)%2==1 cells should be with
            # the character check and the rest
            # should be with the other character
            if (((i + j) & 1) and a[i][j] != check):
                return False
            if (((i + j) & 1) == 0 and a[i][j] == check):
                return False
    return True
 
# Function to fill the grid in a possible way
def fill(a, n, m, odd, even):
     
    for i in range(n):
        for j in range(m):
            if ((i + j) & 1):
                a[i][j] = odd
            else:
                a[i][j] = even
 
# Function to find if the grid can be made alternating
def findPossibleGrid(n, m, a):
    flag,k,O = True,-1,-1
    for i in range(n):
        for j in range(m):
            if (a[i][j] != "*"):
       
                # If grid contains atleast
                # one 1 or 0
                k = i
                O = j
                break
 
        if (k != -1):
            break
    if (k != -1):
        flag = PosCheck(a, n, m, "1")
        if (flag):
            fill(a, n, m, "1", "0")
        else:
            flag = PosCheck(a, n, m, "0")
            if (flag):
                fill(a, n, m, "0", "1")
    else:
   
        # Fill the grid in any possible way
        h = "1"
        for i in range(m):
            a[0][i] = h
            if (h == "1"):
                h = "0"
            else:
                h = "1"
        for i in range(n):
            for j in range(m):
                if (i - 1 < 0):
                    continue
                if (a[i - 1][j] == "1"):
                    a[i][j] = "0"
                else:
                    a[i][j] = "1"
        flag = True
 
    if (flag == 0):
        print("NO")
    else:
        print("YES")
         
        for i in range(n):
            for j in range(m): 
                print(a[i][j],end="")
            print()
 
# Driver Code
n,m = 4,4
grid = [
  ["*", "*", "1", "0"],
  ["*", "*", "*", "*"],
  ["*", "*", "*", "*"],
  ["*", "*", "0", "1"],
]
 
findPossibleGrid(n, m, grid)
 
# This code is contributed by shinjanpatra

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
    public static bool PosCheck(char[,] a, int n, int m,
                                   char check)
    {
 
        // Iterate over the range
        for (int i = 0; i < n; i++) {
 
            // Iterate over the range
            for (int j = 0; j < m; j++) {
 
                if (a[i, j] == '*') {
                    continue;
                }
                else {
 
                    // (i+j)%2==1 cells should be with
                    // the character check and the rest
                    // should be with the other character
                    if ((((i + j) & 1) == 1)
                        && a[i, j] != check) {
                        return false;
                    }
                    if (!(((i + j) & 1) == 1)
                        && a[i, j] == check) {
                        return false;
                    }
                }
            }
        }
        return true;
    }
 
    // Function to fill the grid in a possible way
    public static void fill(char[,] a, int n, int m,
                            char odd, char even)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (((i + j) & 1) == 1) {
                    a[i, j] = odd;
                }
                else {
                    a[i, j] = even;
                }
            }
        }
    }
 
    // Function to find if the grid can be made alternating
    public static void findPossibleGrid(int n, int m,
                                        char[,] a)
    {
        bool flag = true;
        int k = -1, o = -1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
 
                if (a[i, j] != '*') {
 
                    // If grid contains atleast
                    // one 1 or 0
                    k = i;
                    o = j;
                    break;
                }
            }
            if (k != -1) {
                break;
            }
        }
        if (k != -1) {
            flag = PosCheck(a, n, m, '1');
            if (flag) {
                fill(a, n, m, '1', '0');
            }
            else {
                flag = PosCheck(a, n, m, '0');
                if (flag) {
                    fill(a, n, m, '0', '1');
                }
            }
        }
        else {
            // Fill the grid in any possible way
            char h = '1';
            for (int i = 0; i < m; i++) {
                a[0, i] = h;
                if (h == '1') {
                    h = '0';
                }
                else {
                    h = '1';
                }
            }
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (i - 1 < 0) {
                        continue;
                    }
                    if (a[i - 1, j] == '1') {
                        a[i, j] = '0';
                    }
                    else {
                        a[i, j] = '1';
                    }
                }
            }
            flag = true;
        }
 
        if (!flag) {
            Console.WriteLine("No");
        }
        else {
            Console.WriteLine("Yes");
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    Console.Write(a[i, j]);
                }
                Console.WriteLine();
            }
        }
    }
 
// Driver Code
public static void Main()
{
        int n = 4, m = 4;
        char[,] grid = { { '*', '*', '1', '0' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '*', '*' },
                          { '*', '*', '0', '1' } };
 
        findPossibleGrid(n, m, grid);
}
}
 
// This code is contributed by splevel62.

Javascript

<script>
// Javascript program for the above approach
 
// Function to check for the grid in
// one of the alternating ways
function PosCheck(a, n, m, check) 
{
 
  // Iterate over the range
  for (let i = 0; i < n; i++)
  {
   
    // Iterate over the range
    for (let j = 0; j < m; j++)
    {
      if (a[i][j] == "*")
      {
        continue;
      } 
      else
      {
       
        // (i+j)%2==1 cells should be with
        // the character check and the rest
        // should be with the other character
        if ((i + j) & 1 && a[i][j] != check) {
          return false;
        }
        if (!((i + j) & 1) && a[i][j] == check) {
          return false;
        }
      }
    }
  }
  return true;
}
 
// Function to fill the grid in a possible way
function fill(a, n, m, odd, even) {
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
      if ((i + j) & 1) {
        a[i][j] = odd;
      } else {
        a[i][j] = even;
      }
    }
  }
}
 
// Function to find if the grid can be made alternating
function findPossibleGrid(n, m, a) {
  let flag = true;
  let k = -1,
    o = -1;
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
      if (a[i][j] != "*") 
      {
       
        // If grid contains atleast
        // one 1 or 0
        k = i;
        o = j;
        break;
      }
    }
    if (k != -1) {
      break;
    }
  }
  if (k != -1) {
    flag = PosCheck(a, n, m, "1");
    if (flag) {
      fill(a, n, m, "1", "0");
    } else {
      flag = PosCheck(a, n, m, "0");
      if (flag) {
        fill(a, n, m, "0", "1");
      }
    }
  } 
  else
  {
   
    // Fill the grid in any possible way
    let h = "1";
    for (let i = 0; i < m; i++) {
      a[0][i] = h;
      if (h == "1") {
        h = "0";
      } else {
        h = "1";
      }
    }
    for (let i = 0; i < n; i++) {
      for (let j = 0; j < m; j++) {
        if (i - 1 < 0) {
          continue;
        }
        if (a[i - 1][j] == "1") {
          a[i][j] = "0";
        } else {
          a[i][j] = "1";
        }
      }
    }
    flag = true;
  }
 
  if (!flag) {
    document.write("NO<br>");
  } else {
    document.write("YES<br>");
    for (let i = 0; i < n; i++)
    {
      for (let j = 0; j < m; j++) 
      {
        document.write(a[i][j]);
      }
      document.write("<br>");
    }
  }
}
 
// Driver Code
 
let n = 4,
  m = 4;
let grid = [
  ["*", "*", "1", "0"],
  ["*", "*", "*", "*"],
  ["*", "*", "*", "*"],
  ["*", "*", "0", "1"],
];
 
findPossibleGrid(n, m, grid);
 
// This code is contributed by saurabh_jaiswal.
</script>

Output

Time Complexity: O(N*M)
Auxiliary Space: O(1)

Suggest improvement

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Count N-digit numbers having sum of digits equal to a Prime Number

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Check if elements of a Binary Matrix can be made alternating

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