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Check if Bitwise AND of concatenation of diagonals exceeds that of middle row/column elements of a Binary Matrix
  • Last Updated : 15 Apr, 2021

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: 
Number formed by concatenating principal diagonal elements is “101”.
Number formed by concatenating cross diagonal elements is “001”.
Number formed by concatenating elements in the middle row is “001”.
Number formed by concatenating elements in the middle column is “001”.
Therefore, the Bitwise AND of “101” and “001” is same as 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:

Java




// Java program for the above approach
import java.util.ArrayList;
import java.util.Collections;
 
// Java Program for above approach
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);
 
    // Gett 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)
 
    # Gett 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);
 
    // Gett 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
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 at (i, i) to P, i.e., main diagonal.
    • Append the element at (N – 1 – i, i) to S, i.e., cross diagonal.
    • Append the element at ((N-1)/2, i) to MR, i.e., mid-row.
    • Append the element at ((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
Output: 
Yes

 

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

 

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