Check if all rows of a Binary Matrix have all ones placed adjacently or not

Given a binary matrix mat[][] of dimension N*M, the task is to check if all 1s in each row are placed adjacently on the given matrix. If all 1s in each row are adjacent, then print “Yes”. Otherwise, print “No”.

Examples:

Input: mat[][] = {{0, 1, 1, 0}, {1, 1, 0, 0}, {0, 0, 0, 1}, {1, 1, 1, 0}
Output: Yes
Explanation:
Elements in the first row are {0, 1, 1, 0}.
Elements in the 2nd row are {1, 1, 0, 0}.
Elements in the 3rd row are {0, 0, 0, 1}.
Elements in the 4th row are {1, 1, 1, 0}.
Therefore, all the rows have all 1s grouped together. Therefore, print Yes.

Input: mat[][] = {{1, 0, 1}, {0, 0, 1}, {0, 0, 0}}
Output: No

Approach: The idea is to perform row-wise traversal on the matrix and check if all the 1s in a row are placed adjacently or not by using the property of Bitwise XOR. The given problem can be solved based on the following observations:

• Calculate the sum of Bitwise XOR of every pair of adjacent elements of i^th row, say X. All 1s will be not together in the i^th row if any of the following conditions are satisfied:
  • If X > 2 and mat[i][0] + mat[i][M – 1] = 0.
  • If X > 1 and mat[i][0] + mat[i][M – 1] = 1.
  • If X > 0 and mat[i][0] + mat[i][M – 1] = 0.

Follow the steps below to solve this problem:

• Traverse the given matrix mat[][] and perform the following operations:
  • For each row, check if the value of M is less than 3, then print “Yes”.
  • Otherwise, find the sum of Bitwise XOR of adjacent array elements and store it in a variable, say X.
  • For every value of X, if any of the above-mentioned conditions holds true, then print “No”.
• After completing the above steps, if any of the above conditions does not hold true for any value of X, then print “No”.

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 if all 1s are
// placed adjacently in an array or not
bool checkGroup(vector<int> arr)
{

    // Base Case
    if (arr.size() <= 2)
        return true;
    int corner = arr[0] + arr[(int)arr.size()-1];

    // Stores the sum of XOR of all
    // pair of adjacent elements
    int xorSum = 0;

    // Calculate sum of XOR of all
    // pair of adjacent elements
    for (int i = 0; i < arr.size() - 1; i++)
        xorSum += (arr[i] ^ arr[i + 1]);

    // Check for corner cases
    if (!corner)
        if (xorSum > 2)
            return false;
    else if (corner == 1)
        if (xorSum > 1)
            return false;
    else
        if (xorSum > 0)
            return false;

    // Return true
    return true;
}

// Function to check if all the rows
// have all 1s grouped together or not
bool isInGroupUtil(vector<vector<int>> mat)
{

    // Traverse each row
    for (auto i:mat)
    {

        // Check if all 1s are placed
        // together in the ith row or not
        if (!checkGroup(i))
            return false;
         }
    return true;
}


// Function to check if all 1s in a row
// are grouped together in a matrix or not
void isInGroup(vector<vector<int>> mat)
{

    bool ans = isInGroupUtil(mat);

    //Print the result
    if (ans)
        printf("Yes");
    else
        printf("No");
}

// Driver Code
int main()
{

  // Given matrix
  vector<vector<int>> mat = {{0, 1, 1, 0},
                            {1, 1, 0, 0},
                            {0, 0, 0, 1},
                            {1, 1, 1, 0}};

  // Function Call
  isInGroup(mat);
}

// This code is contributed by mohit kumar 29.

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // Function to check if all 1s are
    // placed adjacently in an array or not
    static Boolean checkGroup(Vector<Integer> arr)
    {

        // Base Case
        if (arr.size() <= 2)
            return true;
        int corner = arr.get(0) + arr.get(arr.size()-1);

        // Stores the sum of XOR of all
        // pair of adjacent elements
        int xorSum = 0;

        // Calculate sum of XOR of all
        // pair of adjacent elements
        for (int i = 0; i < arr.size() - 1; i++)
            xorSum += (arr.get(i) ^ arr.get(i + 1));

        // Check for corner cases
        if (corner == 0)
            if (xorSum > 2)
                return false;
        else if (corner == 1)
            if (xorSum > 1)
                return false;
        else
            if (xorSum > 0)
                return false;

        // Return true
        return true;
    }

    // Function to check if all the rows
    // have all 1s grouped together or not
    static Boolean isInGroupUtil(int[][] mat)
    {

        // Traverse each row
        for (int i = 0; i < mat.length; i++)
        {
            Vector<Integer> arr = new Vector<Integer>();
            for(int j = 0; j < mat[i].length; j++)
            {
                arr.add(mat[i][j]);
            }
            // Check if all 1s are placed
            // together in the ith row or not
            if (!checkGroup(arr))
                return false;
             }
        return true;
    }


    // Function to check if all 1s in a row
    // are grouped together in a matrix or not
    static void isInGroup(int[][] mat)
    {

        Boolean ans = isInGroupUtil(mat);

        //Print the result
        if (ans)
            System.out.print("Yes");
        else
            System.out.print("No");
    }

    public static void main(String[] args) {
        // Given matrix
        int[][] mat = {{0, 1, 1, 0},
                   {1, 1, 0, 0},
                   {0, 0, 0, 1},
                   {1, 1, 1, 0}};

        // Function Call
        isInGroup(mat);
    }
}

// This code is contributed by decode2207.

Python3

# Python3 program for the above approach

# Function to check if all 1s are
# placed adjacently in an array or not
def checkGroup(arr):

    # Base Case
    if len(arr) <= 2:
        return True

    corner = arr[0] + arr[-1]

    # Stores the sum of XOR of all
    # pair of adjacent elements
    xorSum = 0

    # Calculate sum of XOR of all
    # pair of adjacent elements
    for i in range(len(arr)-1):
        xorSum += (arr[i] ^ arr[i + 1])

    # Check for corner cases
    if not corner:
        if xorSum > 2:
            return False
    elif corner == 1:
        if xorSum > 1:
            return False
    else:
        if xorSum > 0:
            return False

    # Return true
    return True

# Function to check if all the rows
# have all 1s grouped together or not
def isInGroupUtil(mat):

    # Traverse each row
    for i in mat:

        # Check if all 1s are placed
        # together in the ith row or not
        if not checkGroup(i):
            return False

    return True

# Function to check if all 1s in a row
# are grouped together in a matrix or not
def isInGroup(mat):

    ans = isInGroupUtil(mat)

    # Print the result
    if ans:
        print("Yes")
    else:
        print("No")


# Given matrix
mat = [[0, 1, 1, 0], [1, 1, 0, 0],
       [0, 0, 0, 1], [1, 1, 1, 0]]

# Function Call
isInGroup(mat)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {

    // Function to check if all 1s are
    // placed adjacently in an array or not
    static bool checkGroup(List<int> arr)
    {

        // Base Case
        if (arr.Count <= 2)
            return true;
        int corner = arr[0] + arr[arr.Count-1];

        // Stores the sum of XOR of all
        // pair of adjacent elements
        int xorSum = 0;

        // Calculate sum of XOR of all
        // pair of adjacent elements
        for (int i = 0; i < arr.Count - 1; i++)
            xorSum += (arr[i] ^ arr[i + 1]);

        // Check for corner cases
        if (corner == 0)
            if (xorSum > 2)
                return false;
        else if (corner == 1)
            if (xorSum > 1)
                return false;
        else
            if (xorSum > 0)
                return false;

        // Return true
        return true;
    }

    // Function to check if all the rows
    // have all 1s grouped together or not
    static bool isInGroupUtil(int[,] mat)
    {

        // Traverse each row
        for (int i = 0; i < mat.GetLength(1); i++)
        {
            List<int> arr = new List<int>();
            for(int j = 0; j < mat.GetLength(0); j++)
            {
                arr.Add(mat[i,j]);
            }
            // Check if all 1s are placed
            // together in the ith row or not
            if (!checkGroup(arr))
                return false;
            }
        return true;
    }


    // Function to check if all 1s in a row
    // are grouped together in a matrix or not
    static void isInGroup(int[,] mat)
    {

        bool ans = isInGroupUtil(mat);

        //Print the result
        if (ans)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }

  // Driver code
  static void Main()
  {

    // Given matrix
    int[,] mat = {{0, 1, 1, 0},
               {1, 1, 0, 0},
               {0, 0, 0, 1},
               {1, 1, 1, 0}};

    // Function Call
    isInGroup(mat);
  }
}

// This code is contributed by divyeshrabadiya07.

Javascript

<script>
    // Javascript program for the above approach

    // Function to check if all 1s are
    // placed adjacently in an array or not
    function checkGroup(arr)
    {

        // Base Case
        if (arr.length <= 2)
            return true;
        let corner = arr[0] + arr[arr.length-1];

        // Stores the sum of XOR of all
        // pair of adjacent elements
        let xorSum = 0;

        // Calculate sum of XOR of all
        // pair of adjacent elements
        for (let i = 0; i < arr.length - 1; i++)
            xorSum += (arr[i] ^ arr[i + 1]);

        // Check for corner cases
        if (corner == 0)
            if (xorSum > 2)
                return false;
        else if (corner == 1)
            if (xorSum > 1)
                return false;
        else
            if (xorSum > 0)
                return false;

        // Return true
        return true;
    }

    // Function to check if all the rows
    // have all 1s grouped together or not
    function isInGroupUtil(mat)
    {

        // Traverse each row
        for (let i = 0; i < mat.length; i++)
        {
            let arr = []
            for(let j = 0; j < mat[i].length; j++)
            {
                arr.push(mat[i][j]);
            }
            // Check if all 1s are placed
            // together in the ith row or not
            if (!checkGroup(arr))
                return false;
             }
        return true;
    }


    // Function to check if all 1s in a row
    // are grouped together in a matrix or not
    function isInGroup(mat)
    {

        let ans = isInGroupUtil(mat);

        //Print the result
        if (ans)
            document.write("Yes");
        else
            document.write("No");
    }

    // Given matrix
    let mat = [[0, 1, 1, 0],
               [1, 1, 0, 0],
               [0, 0, 0, 1],
               [1, 1, 1, 0]];

    // Function Call
    isInGroup(mat);

// This code is contributed by mukesh07.
</script>

Output:

Yes

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