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Check if Bitwise AND of concatenation of diagonals exceeds that of middle row/column elements of a Binary Matrix

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Given a binary matrix mat[][] of dimensions N * N, the task is to check if Bitwise AND of the decimal numbers obtained by concatenating primary and secondary diagonals elements is greater than the Bitwise AND of the decimal numbers obtained by the elements present in the middle row and column. If found to be true, then print “Yes”. Otherwise, print “No”.

Note: Concatenate matrix elements from left to right and top to bottom only. If N is even, then take the first middle row/column out of the two.

Examples:

Input: M[][] = {{1, 0, 1}, {0, 0, 1}, {0, 1, 1}}
Output: No
Explanation: 
The number formed by concatenating principal diagonal elements is “101”.
The number formed by concatenating cross diagonal elements is “001”.
The number formed by concatenating elements in the middle row is “001”.
The number formed by concatenating elements in the middle column is “001”.
Therefore, the Bitwise AND of “101” and “001” is the same as the Bitwise AND of “001” and “001”. 

Input: M[][] = {{0, 1, 1}, {0, 0, 0}, {0, 1, 1}}
Output: Yes

 

Naive Approach: The simplest approach to solve the problem is to traverse the given matrix and append the corresponding number to a variable, say P, if the current row is equal to the current column, to a variable, say S, if the row is N-column, to a variable,  say MR, if the row is equal to N/2, and to a variable, say MC, if the column is N/2. After completing the above steps, if Bitwise AND of P and S is greater than Bitwise AND of MR and MC, print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to convert obtained binary
// representation to decimal value
int convert(vector<int> p)
{
 
    // Stores the resultant number
    int ans = 0;
 
    // Traverse string arr
    for(int i : p)
    {
        ans = (ans << 1) | i;
    }
 
    // Return the number formed
    return ans;
}
 
// Function to count the number of
// set bits in the number num
int count(int num)
{
 
    // Stores the count of set bits
    int ans = 0;
 
    // Iterate until num > 0
    while (num > 0)
    {
        ans += num & 1;
        num >>= 1;
    }
    return ans;
}
 
// Function to check if the given matrix
// satisfies the given condition or not
void checkGoodMatrix(vector<vector<int> > mat)
{
    vector<int> P;
    vector<int> S;
    vector<int> MR;
    vector<int> MC;
 
    // To get P, S, MR, and MC
    for(int i = 0; i < mat.size(); i++)
    {
        for(int j = 0; j < mat[0].size(); j++)
        {
            if (i == j)
                P.push_back(mat[i][j]);
 
            if (i + j == mat.size() - 1)
                S.push_back(mat[i][j]);
 
            if (i == floor((mat.size() - 1) / 2))
                MR.push_back(mat[i][j]);
 
            if (j == floor((mat.size() - 1) / 2))
                MC.push_back(mat[i][j]);
        }
    }
    reverse(S.begin(), S.end());
 
    // Stores decimal equivalents
    // of binary representations
    int P0 = convert(P);
    int S0 = convert(S);
    int MR0 = convert(MR);
    int MC0 = convert(MC);
 
    // Get the number of set bits
    int setBitsPS = count((P0 & S0));
    int setBitsMM = count((MR0 & MC0));
 
    // Print the answer
    if (setBitsPS > setBitsMM)
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver code
int main()
{
    vector<vector<int>> mat = { { 1, 0, 1 },
                                { 0, 0, 1 },
                                { 0, 1, 1 } };
 
    checkGoodMatrix(mat);
}
 
// This code is contributed by nirajgusain5


Java




// Java program for the above approach
import java.util.ArrayList;
import java.util.Collections;
 
class GFG{
     
// Function to convert obtained binary
// representation to decimal value
static int convert(ArrayList<Integer> p)
{
     
    // Stores the resultant number
    int ans = 0;
     
    // Traverse string arr
    for(int i: p)
    {
        ans = (ans << 1) | i;
    }
     
    // Return the number formed
    return ans;
}
 
// Function to count the number of
// set bits in the number num
static int count(int num)
{
     
    // Stores the count of set bits
    int ans = 0;
     
    // Iterate until num > 0
    while (num > 0)
    {
        ans += num & 1;
        num >>= 1;
    }
    return ans;
}
 
// Function to check if the given matrix
// satisfies the given condition or not
static void checkGoodMatrix(int mat[][])
{
    ArrayList<Integer> P = new ArrayList<Integer>();
    ArrayList<Integer> S = new ArrayList<Integer>();
    ArrayList<Integer> MR = new ArrayList<Integer>();
    ArrayList<Integer> MC = new ArrayList<Integer>();
     
    // To get P, S, MR, and MC
    for(int i = 0; i < mat.length; i++)
    {
        for(int j = 0; j < mat[0].length; j++)
        {
            if (i == j)
                P.add(mat[i][j]);
 
            if (i + j == mat.length - 1)
                S.add(mat[i][j]);
 
            if (i == Math.floor((mat.length - 1) / 2))
                MR.add(mat[i][j]);
 
            if (j == Math.floor((mat.length - 1) / 2))
                MC.add(mat[i][j]);
        }
    }
    Collections.reverse(S);
 
    // Stores decimal equivalents
    // of binary representations
    int P0 = convert(P);
    int S0 = convert(S);
    int MR0 = convert(MR);
    int MC0 = convert(MC);
 
    // Get the number of set bits
    int setBitsPS = count((P0 & S0));
    int setBitsMM = count((MR0 & MC0));
     
    // Print the answer
    if (setBitsPS > setBitsMM)
       System.out.print("Yes");
    else
        System.out.print("No");
}
 
// Driver code
public static void main(String[] args)
{
    int mat[][] = { { 1, 0, 1 },
                    { 0, 0, 1 },
                    { 0, 1, 1 } };
    checkGoodMatrix(mat);
}
}
 
// This code is contributed by abhinavjain194


Python3




# Python3 program for the above approach
 
# Functio to convert obtained binary
# representation to decimal value
def convert(arr):
   
      # Stores the resultant number
    ans = 0
     
    # Traverse string arr
    for i in arr:
        ans = (ans << 1) | i
         
    # Return the number formed
    return ans
 
# Function to count the number of
# set bits in the number num
def count(num):
   
    # Stores the count of set bits
    ans = 0
     
    # Iterate until num > 0
    while num:
        ans += num & 1
        num >>= 1
    return ans
 
# Function to check if the given matrix
# satisfies the given condition or not
def checkGoodMatrix(mat):
    P = []
    S = []
    MR = []
    MC = []
 
    # To get P, S, MR, and MC
    for i in range(len(mat)):
        for j in range(len(mat[0])):
 
            if i == j:
                P.append(mat[i][j])
 
            if i + j == len(mat)-1:
                S.append(mat[i][j])
 
            if i == (len(mat)-1)//2:
                MR.append(mat[i][j])
 
            if j == (len(mat)-1)//2:
                MC.append(mat[i][j])
 
    S.reverse()
 
    # Stores decimal equivalents
    # of binary representations
    P = convert(P)
    S = convert(S)
    MR = convert(MR)
    MC = convert(MC)
 
    # Get the number of set bits
    setBitsPS = count(P & S)
    setBitsMM = count(MR & MC)
 
    # Print the answer
    if setBitsPS > setBitsMM:
        print("Yes")
    else:
        print("No")
 
# Driver Code
 
# Given Matrix
mat = [[1, 0, 1], [0, 0, 1], [0, 1, 1]]
 
checkGoodMatrix(mat)


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to convert obtained binary
// representation to decimal value
static int convert(List<int> p)
{
     
    // Stores the resultant number
    int ans = 0;
 
    // Traverse string arr
    foreach(int i in p)
    {
        ans = (ans << 1) | i;
    }
 
    // Return the number formed
    return ans;
}
 
// Function to count the number of
// set bits in the number num
static int count(int num)
{
     
    // Stores the count of set bits
    int ans = 0;
 
    // Iterate until num > 0
    while (num > 0)
    {
        ans += num & 1;
        num >>= 1;
    }
    return ans;
}
 
// Function to check if the given matrix
// satisfies the given condition or not
static void checkGoodMatrix(int[, ] mat)
{
    List<int> P = new List<int>();
    List<int> S = new List<int>();
    List<int> MR = new List<int>();
    List<int> MC = new List<int>();
 
    // To get P, S, MR, and MC
    for(int i = 0; i < mat.GetLength(0); i++)
    {
        for(int j = 0; j < mat.GetLength(1); j++)
        {
            if (i == j)
                P.Add(mat[i, j]);
 
            if (i + j == mat.GetLength(0) - 1)
                S.Add(mat[i, j]);
 
            if (i == Math.Floor(
                (mat.GetLength(0) - 1) / 2.0))
                MR.Add(mat[i, j]);
 
            if (j == Math.Floor(
                (mat.GetLength(0) - 1) / 2.0))
                MC.Add(mat[i, j]);
        }
    }
    S.Reverse();
 
    // Stores decimal equivalents
    // of binary representations
    int P0 = convert(P);
    int S0 = convert(S);
    int MR0 = convert(MR);
    int MC0 = convert(MC);
 
    // Get the number of set bits
    int setBitsPS = count((P0 & S0));
    int setBitsMM = count((MR0 & MC0));
 
    // Print the answer
    if (setBitsPS > setBitsMM)
        Console.Write("Yes");
    else
        Console.Write("No");
}
 
// Driver code
public static void Main(string[] args)
{
    int[,] mat = { { 1, 0, 1 },
                   { 0, 0, 1 },
                   { 0, 1, 1 } };
                    
    checkGoodMatrix(mat);
}
}
 
// This code is contributed by ukasp


Javascript




<script>
// Javascript program for above approach
 
// Function to convert obtained binary
// representation to decimal value
function convert(p) {
 
    // Stores the resultant number
    let ans = 0;
 
    // Traverse string arr
    for (let i of p) {
        ans = (ans << 1) | i;
    }
 
    // Return the number formed
    return ans;
}
 
// Function to count the number of
// set bits in the number num
function count(num) {
 
    // Stores the count of set bits
    let ans = 0;
 
    // Iterate until num > 0
    while (num > 0) {
        ans += num & 1;
        num >>= 1;
    }
    return ans;
}
 
// Function to check if the given matrix
// satisfies the given condition or not
function checkGoodMatrix(mat) {
    let P = [], S = [], MR = [], MC = [];
 
    // To get P, S, MR, and MC
    for (let i = 0; i < mat.length; i++) {
        for (let j = 0; j < mat[0].length; j++) {
            if (i == j)
                P.push(mat[i][j]);
 
            if (i + j == mat.length - 1)
                S.push(mat[i][j]);
 
            if (i == Math.floor((mat.length - 1) / 2))
                MR.push(mat[i][j]);
 
            if (j == Math.floor((mat.length - 1) / 2))
                MC.push(mat[i][j]);
        }
    }
    S.reverse();
 
    // Stores decimal equivalents
    // of binary representations
    let P0 = convert(P);
    let S0 = convert(S);
    let MR0 = convert(MR);
    let MC0 = convert(MC);
 
    // Get the number of set bits
    let setBitsPS = count((P0 & S0));
    let setBitsMM = count((MR0 & MC0));
 
    // Print the answer
    if (setBitsPS > setBitsMM)
        document.write("Yes");
    else
        document.write("No");
}
 
// Driver code
 
let mat = [[1, 0, 1],
           [0, 0, 1],
           [0, 1, 1]];
 
checkGoodMatrix(mat);
 
 
// This code is contributed by _saurabh_jaiswal
</script>


Output: 

No

 

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient approach: To optimize the above approach, the above approach can be optimized by traversing on each element’s diagonals, middle row, and middle column only. Follow the steps below to solve the problem:

  • Initialize auxiliary vectors, say P, S, MR, and MC to store the connected elements of the main diagonal, cross diagonal, mid-row, and mid-column respectively.
  • Iterate over the range [0, N – 1]:
    • Append the element from (i, i) to P, i.e., main diagonal.
    • Append the element from (N – 1 – i, i) to S, i.e., cross diagonal.
    • Append the element from ((N-1)/2, i) to MR, i.e., mid-row.
    • Append the element from ((N-1)/2, i) to MC, i.e., mid-column.
  • Iterate over the range [0, N – 1]:
    • Check if P[i] & S[i] > MR[i] & MC[i], then print “Yes” and return.
    • Otherwise, check if p[i] & s[i] < MR[i] & MC[i], then print “No” and return.
  • If none of the above conditions satisfy, 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 the matrix
// satisfy the given condition or not
void checkGoodMatrix(
    vector<vector<int> > M, int N)
{
    // Stores the binary representation
    vector<int> p, s, MR, MC;
 
    // Iterate over the range [0, N]
    for (int i = 0; i < N; i++) {
 
        // Push element of main diagonal
        p.push_back(M[i][i]);
 
        // Push element of cross diagona
        s.push_back(M[N - 1 - i][i]);
 
        // Push element of Mid row
        MR.push_back(M[(N - 1) / 2][i]);
 
        // Push element of Mid column
        MC.push_back(M[i][(N - 1) / 2]);
    }
 
    // Check if S & P > MR & MC
    for (int i = 0; i < N; i++) {
 
        if (p[i] & s[i] > MR[i] & MC[i]) {
            cout << "Yes";
            return;
        }
        else if (p[i] & s[i] < MR[i] & MC[i]) {
            cout << "No";
            return;
        }
    }
 
    cout << "No";
}
 
// Driver Code
int main()
{
    // Given matrix
    vector<vector<int> > M{ { 0, 1, 1 },
                            { 0, 0, 0 },
                            { 0, 1, 1 } };
 
    // Size of the matrix
    int N = M.size();
 
    checkGoodMatrix(M, N);
 
    return 0;
}


Java




// Java program for the above approach
import java.util.Vector;
 
class GFG{
 
static void checkGoodMatrix(int[][] M, int N)
{
     
    // Stores the binary representation
    Vector<Integer> p = new Vector<Integer>();
    Vector<Integer> s = new Vector<Integer>();
    Vector<Integer> MR = new Vector<Integer>();
    Vector<Integer> MC = new Vector<Integer>();
 
    // Iterate over the range [0, N]
    for(int i = 0; i < N; i++)
    {
         
        // Push element of main diagonal
        p.add(M[i][i]);
 
        // Push element of cross diagona
        s.add(M[N - 1 - i][i]);
 
        // Push element of Mid row
        MR.add(M[(N - 1) / 2][i]);
 
        // Push element of Mid column
        MC.add(M[i][(N - 1) / 2]);
    }
 
    // Check if S & P > MR & MC
    for(int i = 0; i < N; i++)
    {
        int P = p.get(i);
        int S = s.get(i);
        int Mr = MR.get(i);
        int Mc = MC.get(i);
 
        if ((P & S) > (Mr & Mc))
        {
            System.out.print("Yes");
            return;
        }
        else if ((P & S) < (Mr & Mc))
        {
            System.out.print("No");
            return;
        }
    }
    System.out.print("No");
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given matrix
    int[][] M = { { 0, 1, 1 },
                  { 0, 0, 0 },
                  { 0, 1, 1 } };
 
    // Size of the matrix
    int N = M.length;
 
    checkGoodMatrix(M, N);
}
}
 
// This code is contributed by abhinavjain194


Python3




# Python3 program for the above approach
 
# Function to check if the matrix
# satisfy the given condition or not
def checkGoodMatrix(M, N):
     
    # Stores the binary representation
    p = []
    s = []
    MR = []
    MC = []
     
    # Iterate over the range [0, N]
    for i in range(N):
         
        # Push element of main diagonal
        p.append(M[i][i])
         
        # Push element of cross diagona
        s.append(M[N - 1 - i][i])
         
        # Push element of Mid row
        MR.append(M[(N - 1) // 2][i])
         
        # Push element of Mid column
        MC.append(M[i][(N - 1) // 2])
         
    # Check if S & P > MR & MC
    for i in range(N):
        if (p[i] & s[i] > MR[i] & MC[i]):
            print("Yes")
            return
        elif (p[i] & s[i] < MR[i] & MC[i]):
            print("No")
            return
     
    print("No")
 
# Driver Code
 
# Given matrix
M = [ [ 0, 1, 1 ],
      [ 0, 0, 0 ],
      [ 0, 1, 1 ] ] 
 
# Size of the matrix
N = len(M)
 
checkGoodMatrix(M, N)
 
# This code is contributed by SHUBHAMSINGH10


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
     
    static void checkGoodMatrix(int[,] M, int N)
    {
          
        // Stores the binary representation
        List<int> p = new List<int>();
        List<int> s = new List<int>();
        List<int> MR = new List<int>();
        List<int> MC = new List<int>();
      
        // Iterate over the range [0, N]
        for(int i = 0; i < N; i++)
        {
              
            // Push element of main diagonal
            p.Add(M[i,i]);
      
            // Push element of cross diagona
            s.Add(M[N - 1 - i,i]);
      
            // Push element of Mid row
            MR.Add(M[(N - 1) / 2,i]);
      
            // Push element of Mid column
            MC.Add(M[i,(N - 1) / 2]);
        }
      
        // Check if S & P > MR & MC
        for(int i = 0; i < N; i++)
        {
            int P = p[i];
            int S = s[i];
            int Mr = MR[i];
            int Mc = MC[i];
      
            if ((P & S) > (Mr & Mc))
            {
                Console.WriteLine("Yes");
                return;
            }
            else if ((P & S) < (Mr & Mc))
            {
                Console.WriteLine("No");
                return;
            }
        }
        Console.WriteLine("No");
    }
 
  static void Main() {
    // Given matrix
    int[,] M = { { 0, 1, 1 },
                  { 0, 0, 0 },
                  { 0, 1, 1 } };
  
    // Size of the matrix
    int N = 3;
  
    checkGoodMatrix(M, N);
  }
}
 
// This code is contributed by mukesh07.


Javascript




<script>
 
    // JavaScript program for the above approach
     
    function checkGoodMatrix(M, N)
    {
 
        // Stores the binary representation
        let p = [];
        let s = [];
        let MR = [];
        let MC = [];
 
        // Iterate over the range [0, N]
        for(let i = 0; i < N; i++)
        {
 
            // Push element of main diagonal
            p.push(M[i][i]);
 
            // Push element of cross diagona
            s.push(M[N - 1 - i][i]);
 
            // Push element of Mid row
            MR.push(M[parseInt((N - 1) / 2, 10)][i]);
  
            // Push element of Mid column
            MC.push(M[i][parseInt((N - 1) / 2, 10)]);
        }
 
        // Check if S & P > MR & MC
        for(let i = 0; i < N; i++)
        {
            let P = p[i];
            let S = s[i];
            let Mr = MR[i];
            let Mc = MC[i];
 
            if ((P & S) > (Mr & Mc))
            {
                document.write("Yes");
                return;
            }
            else if ((P & S) < (Mr & Mc))
            {
                document.write("No");
                return;
            }
        }
        document.write("No");
    }
     
    // Given matrix
    let M = [ [ 0, 1, 1 ],
               [ 0, 0, 0 ],
               [ 0, 1, 1 ] ];
  
    // Size of the matrix
    let N = M.length;
  
    checkGoodMatrix(M, N);
     
</script>


Output: 

Yes

 

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

 



Last Updated : 27 Feb, 2023
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