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Sort the given matrix

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  • Difficulty Level : Basic
  • Last Updated : 02 Aug, 2022
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Given a n x n matrix. The problem is to sort the given matrix in strict order. Here strict order means that the matrix is sorted in a way such that all elements in a row are sorted in increasing order and for row ā€˜iā€™, where 1 <= i <= n-1, the first element of row ‘i’ is greater than or equal to the last element of row ‘i-1’.

Examples: 

Input : mat[][] = { {5, 4, 7},
                    {1, 3, 8},
                    {2, 9, 6} }
Output : 1 2 3
         4 5 6
         7 8 9
Recommended Practice

Approach: Create a temp[] array of size n^2. Starting with the first row one by one copy the elements of the given matrix into temp[]. Sort temp[]. Now one by one copy the elements of temp[] back to the given matrix.

Below is the implementation of the above approach:

C++




// C++ implementation to sort the given matrix
#include <bits/stdc++.h>
using namespace std;
 
#define SIZE 10
 
// function to sort the given matrix
void sortMat(int mat[SIZE][SIZE], int n)
{
    // temporary matrix of size n^2
    int temp[n * n];
    int k = 0;
 
    // copy the elements of matrix one by one
    // into temp[]
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            temp[k++] = mat[i][j];
 
    // sort temp[]
    sort(temp, temp + k);
     
    // copy the elements of temp[] one by one
    // in mat[][]
    k = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            mat[i][j] = temp[k++];
}
 
// function to print the given matrix
void printMat(int mat[SIZE][SIZE], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cout << mat[i][j] << " ";
        cout << endl;
    }
}
 
// Driver program to test above
int main()
{
    int mat[SIZE][SIZE] = { { 5, 4, 7 },
                            { 1, 3, 8 },
                            { 2, 9, 6 } };
    int n = 3;
 
    cout << "Original Matrix:\n";
    printMat(mat, n);
 
    sortMat(mat, n);
 
    cout << "\nMatrix After Sorting:\n";
    printMat(mat, n);
 
    return 0;
}


Java




// Java implementation to
// sort the given matrix
import java.io.*;
import java.util.*;
 
class GFG {
     
    static int SIZE = 10;
 
    // function to sort the given matrix
    static void sortMat(int mat[][], int n)
    {
        // temporary matrix of size n^2
        int temp[] = new int[n * n];
        int k = 0;
     
        // copy the elements of matrix
        // one by one into temp[]
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                temp[k++] = mat[i][j];
     
        // sort temp[]
        Arrays.sort(temp);
         
        // copy the elements of temp[]
        // one by one in mat[][]
        k = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                mat[i][j] = temp[k++];
    }
     
    // function to print the given matrix
    static void printMat(int mat[][], int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                System.out.print( mat[i][j] + " ");
            System.out.println();
        }
    }
     
    // Driver program to test above
    public static void main(String args[])
    {
        int mat[][] = { { 5, 4, 7 },
                        { 1, 3, 8 },
                        { 2, 9, 6 } };
        int n = 3;
     
        System.out.println("Original Matrix:");
        printMat(mat, n);
     
        sortMat(mat, n);
     
        System.out.println("Matrix After Sorting:");
        printMat(mat, n);
     
    }
}
 
// This code is contributed by Nikita Tiwari.


Python3




# Python3 implementation to sort
# the given matrix
 
SIZE = 10
 
# Function to sort the given matrix
def sortMat(mat, n) :
     
    # Temporary matrix of size n^2
    temp = [0] * (n * n)
    k = 0
 
    # Copy the elements of matrix 
    # one by one into temp[]
    for i in range(0, n) :
         
        for j in range(0, n) :
             
            temp[k] = mat[i][j]
            k += 1
 
    # sort temp[]
    temp.sort()
     
    # copy the elements of temp[]
    # one by one in mat[][]
    k = 0
     
    for i in range(0, n) :
         
        for j in range(0, n) :
            mat[i][j] = temp[k]
            k += 1
 
 
# Function to print the given matrix
def printMat(mat, n) :
     
    for i in range(0, n) :
         
        for j in range( 0, n ) :
             
            print(mat[i][j] , end = " ")
             
        print()
     
     
# Driver program to test above
mat = [ [ 5, 4, 7 ],
        [ 1, 3, 8 ],
        [ 2, 9, 6 ] ]
n = 3
 
print( "Original Matrix:")
printMat(mat, n)
 
sortMat(mat, n)
 
print("\nMatrix After Sorting:")
printMat(mat, n)
 
 
# This code is contributed by Nikita Tiwari.


C#




// C# implementation to
// sort the given matrix
using System;
 
class GFG {
    static int SIZE = 10;
 
    // function to sort the given matrix
    static void sortMat(int[, ] mat, int n)
    {
        // temporary matrix of size n^2
        int[] temp = new int[n * n];
        int k = 0;
 
        // copy the elements of matrix
        // one by one into temp[]
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                temp[k++] = mat[i, j];
 
        // sort temp[]
        Array.Sort(temp);
 
        // copy the elements of temp[]
        // one by one in mat[][]
        k = 0;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
            mat[i, j] = temp[k++];
    }
 
    // function to print the given matrix
    static void printMat(int[, ] mat, int n)
    {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
            Console.Write(mat[i, j] + " ");
            Console.WriteLine();
        }
    }
 
    // Driver code
    public static void Main()
    {
        int[, ] mat = { { 5, 4, 7 },
                        { 1, 3, 8 },
                        { 2, 9, 6 } };
        int n = 3;
 
        Console.WriteLine("Original Matrix:");
        printMat(mat, n);
 
        sortMat(mat, n);
 
        Console.WriteLine("Matrix After Sorting:");
        printMat(mat, n);
    }
}
 
// This code is contributed by Sam007


Javascript




<script>
 
// JavaScript implementation to sort
// the given matrix
 
let SIZE  = 10
 
// function to sort the given matrix
function sortMat(mat, n)
{
    // temporary matrix of size n^2
    let temp = new Array(n * n);
    let k = 0;
 
    // copy the elements of matrix one by one
    // into temp[]
    for (let i = 0; i < n; i++)
        for (let j = 0; j < n; j++)
            temp[k++] = mat[i][j];
 
    // sort temp[]
    temp.sort();
     
    // copy the elements of temp[] one by one
    // in mat[][]
    k = 0;
    for (let i = 0; i < n; i++)
        for (let j = 0; j < n; j++)
            mat[i][j] = temp[k++];
}
 
// function to print the given matrix
function printMat(mat, n)
{
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++)
            document.write( mat[i][j] + " ");
        document.write( "<br>");
    }
}
 
// Driver program to test above
 
    let mat = [ [ 5, 4, 7 ],
                 [ 1, 3, 8 ],
                [ 2, 9, 6 ] ];
    let n = 3;
 
    document.write( "Original Matrix: " + "<br>");
    printMat(mat, n);
 
    sortMat(mat, n);
    document.write( "<br>");
    document.write( "\nMatrix After Sorting: " + "<br>");
    printMat(mat, n);
 
// This code is contributed by Manoj
 
</script>


Output

Original Matrix:
5 4 7 
1 3 8 
2 9 6 

Matrix After Sorting:
1 2 3 
4 5 6 
7 8 9 

Time Complexity: O(n2log2n). 
Auxiliary Space: O(n2), since n * n extra space has been taken.


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