Given a matrix, mat[][] of size N * M and two positive integers A and B, the task is to construct a matrix of size A * B from the given matrix elements. If multiple solutions exist, then print any one of them. Otherwise, print -1.
Input: mat[][] = { { 1, 2, 3, 4, 5, 6 } }, A = 2, B = 3
Output: { { 1, 2, 3 }, { 4, 5, 6 } }
Explanation:
Since the size of the matrix { { 1, 2, 3 }, { 4, 5, 6 } } is A * B(2 * 3).
Therefore, the required output is { { 1, 2, 3 }, { 4, 5, 6 } }.Input: mat[][] = { {1, 2}, { 3, 4 }, { 5, 6 } }, A = 1, B = 6
Output: { { 1, 2, 3, 4, 5, 6 } }
Approach: Follow the steps below to solve the problem:
- Initialize a matrix of size A * B say, res[][].
- Traverse the matrix, mat[][] and insert each element of the matrix into the matrix, res[][].
- Finally, print the matrix res[][].
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to construct a matrix of size // A * B from the given matrix elements void ConstMatrix( int * mat, int N, int M,
int A, int B)
{ if (N * M != A * B)
return ;
int idx = 0;
// Initialize a new matrix
int res[A][B];
// Traverse the matrix, mat[][]
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < M; j++)
{
res[idx / B][idx % B] = *((mat + i * M) + j);
// Update idx
idx++;
}
}
// Print the resultant matrix
for ( int i = 0; i < A; i++)
{
for ( int j = 0; j < B; j++)
cout << res[i][j] << " " ;
cout << "\n" ;
}
} // Driver Code int main()
{ int mat[][6] = { { 1, 2, 3, 4, 5, 6 } };
int A = 2;
int B = 3;
int N = sizeof (mat) / sizeof (mat[0]);
int M = sizeof (mat[0]) / sizeof ( int );
ConstMatrix(( int *)mat, N, M, A, B);
return 0;
} // This code is contributed by subhammahato348 |
// Java program to implement // the above approach import java.util.*;
class GFG
{ // Function to construct a matrix of size // A * B from the given matrix elements static void ConstMatrix( int [][] mat, int N, int M,
int A, int B)
{ if (N * M != A * B)
return ;
int idx = 0 ;
// Initialize a new matrix
int [][]res = new int [A][B];
// Traverse the matrix, mat[][]
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < M; j++)
{
res[idx / B][idx % B] = mat[i][j];
// Update idx
idx++;
}
}
// Print the resultant matrix
for ( int i = 0 ; i < A; i++)
{
for ( int j = 0 ; j < B; j++)
System.out.print(res[i][j] + " " );
System.out.print( "\n" );
}
} // Driver Code public static void main(String[] args)
{ int mat[][] = { { 1 , 2 , 3 , 4 , 5 , 6 } };
int A = 2 ;
int B = 3 ;
int N = mat.length;
int M = mat[ 0 ].length;
ConstMatrix(mat, N, M, A, B);
} } // This code is contributed by 29AjayKumar |
# Python3 program to implement # the above approach # Function to construct a matrix of size A * B # from the given matrix elements def ConstMatrix(mat, N, M, A, B):
if (N * M ! = A * B):
return - 1
idx = 0 ;
# Initialize a new matrix
res = [[ 0 for i in range (B)] for i in range (A)]
# Traverse the matrix, mat[][]
for i in range (N):
for j in range (M):
res[idx / / B][idx % B] = mat[i][j];
# Update idx
idx + = 1
# Print the resultant matrix
for i in range (A):
for j in range (B):
print (res[i][j], end = " " )
print ()
# Driver Code if __name__ = = '__main__' :
mat = [ [ 1 , 2 , 3 , 4 , 5 , 6 ] ]
A = 2
B = 3
N = len (mat)
M = len (mat[ 0 ])
ConstMatrix(mat, N, M, A, B)
|
// C# program to implement // the above approach using System;
class GFG
{ // Function to construct a matrix of size // A * B from the given matrix elements static void ConstMatrix( int [,] mat, int N, int M,
int A, int B)
{ if (N * M != A * B)
return ;
int idx = 0;
// Initialize a new matrix
int [,]res = new int [A,B];
// Traverse the matrix, [,]mat
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < M; j++)
{
res[idx / B, idx % B] = mat[i, j];
// Update idx
idx++;
}
}
// Print the resultant matrix
for ( int i = 0; i < A; i++)
{
for ( int j = 0; j < B; j++)
Console.Write(res[i, j] + " " );
Console.Write( "\n" );
}
} // Driver Code public static void Main(String[] args)
{ int [,]mat = {{ 1, 2, 3, 4, 5, 6 }};
int A = 2;
int B = 3;
int N = mat.GetLength(0);
int M = mat.GetLength(1);
ConstMatrix(mat, N, M, A, B);
} } // This code is contributed by 29AjayKumar |
<script> // javascript program of the above approach // Function to construct a matrix of size // A * B from the given matrix elements function ConstMatrix(mat, N, M,
A, B)
{ if (N * M != A * B)
return ;
let idx = 0;
// Initialize a new matrix
let res = new Array(A);
// Loop to create 2D array using 1D array
for ( var i = 0; i < res.length; i++) {
res[i] = new Array(2);
}
// Traverse the matrix, mat[][]
for (let i = 0; i < N; i++)
{
for (let j = 0; j < M; j++)
{
res[(Math.floor(idx / B))][idx % B] = mat[i][j];
// Update idx
idx++;
}
}
// Print the resultant matrix
for (let i = 0; i < A; i++)
{
for (let j = 0; j < B; j++)
document.write(res[i][j] + " " );
document.write( "<br/>" );
}
} // Driver Code
let mat = [[ 1, 2, 3, 4, 5, 6 ]];
let A = 2;
let B = 3;
let N = mat.length;
let M = mat[0].length;
ConstMatrix(mat, N, M, A, B);
// This code is contributed by chinmoy1997pal.
</script> |
1 2 3 4 5 6
Time Complexity: O(N * M)
Auxiliary Space: O(N * M)