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 = 3Â
Output: { { 1, 2, 3 }, { 4, 5, 6 } }Â
Explanation:Â
Since the size of the matrix { { 1, 2, 3 }, { 4, 5, 6 } } is A * B(2 * 3).Â
Therefore, the required output is { { 1, 2, 3 }, { 4, 5, 6 } }.Input: mat[][] = { {1, 2}, { 3, 4 }, { 5, 6 } }, A = 1, B = 6Â
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:Â
1 2 3 4 5 6
Â
Time Complexity: O(N * M)Â
Auxiliary Space: O(N * M)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
