Convert a Matrix into another Matrix of given dimensions

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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++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

1 2 3 
4 5 6

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

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