Sort the given matrix

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 ‘i’, 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

Brute Force Using Extra O(n*m) Space:

    The Approach:

here in this approach we store the all the elment in a vector then sort it we need to fill that matrix again.

C++

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int main() {
    vector<vector<int>>v{{5, 4, 7},{1, 3, 8},{2, 9, 6}};
    int n=v.size();
    vector<int>x;
    for(int i=0;i<n;i++){
     for(int j=0;j<n;j++){
      x.push_back(v[i][j]);
      }
    }
    sort(x.begin(),x.end());
    int k=0;
    for(int i=0;i<n;i++){
     for(int j=0;j<n;j++){
       v[i][j]=x[k++];
      }
    }
  cout<<"Sorted Matrix Will be:"<<endl;
    for(auto it:v){
     for(auto vt:it){
       cout<<vt<<" ";
      }cout<<endl;
    }
    return 0;
}

Javascript

// JavaScript program for the above approach
let v = [[5,4,7], [1,3,8], [2,9,6]];
let n = v.length;
let x = [];

for(let i = 0; i<n; i++){
    for(let j = 0; j<n; j++){
        x.push(v[i][j]);
    }
}

x.sort();
let k = 0;
for(let i = 0; i<n; i++){
    for(let j = 0; j<n; j++){
        v[i][j] = x[k++];
    }
}
document.write("Sorted Matrix will be : <br>");
for(let i = 0; i<n; i++){
    for(let j = 0; j<n; j++){
        document.write(v[i][j] + " ");
    }
    document.write("<br>");
}
// this code is contributed by Yash Agarwal(yashagarwal2852002)

Output

Sorted Matrix Will be:
1 2 3 
4 5 6 
7 8 9 

Time Complexity: O(n*n),For traversing
Auxiliary Space: O(n*n),For vector.

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.