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> |
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> |
Yes
Time Complexity: O(N)
Auxiliary Space: O(N)