Sort the given matrix (original) (raw)
Last Updated : 22 Oct, 2024
Given a m 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 <= m-1, the first element 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**Input: mat[][] = { {5, 4, 7},
{1, 3, 8} }
**Output: 1 3 4
5 7 8
Naive Approach - O(mn Log mn) Time and O(mn) Space
We first store the matrix elements in a m x n sized 1D array, then sort the array and finally copy the elements back to the matrix,
C++ `
#include #include #include using namespace std;
void sortMatrix(vector<vector>& mat) { vector temp;
// Collect all elements in a temporary vector
for (auto& row : mat) {
for (int x : row) {
temp.push_back(x);
}
}
// Sort the vector
sort(temp.begin(), temp.end());
// Put sorted values back into the matrix
int k = 0;
for (auto& row : mat) {
for (int& x : row) {
x = temp[k++];
}
}}
int main() { vector<vector> mat{{5, 4, 7}, {1, 3, 8}, {2, 9, 6}};
sortMatrix(mat);
for (auto& row : mat) {
for (int x : row) {
cout << x << " ";
}
cout << endl;
}
return 0;}
Java
import java.util.*;
public class Main { public static void main(String[] args) { // Initialize the 2D vector with some values List<List > v = new ArrayList<>(Arrays.asList( new ArrayList<>(Arrays.asList(5, 4, 7)), new ArrayList<>(Arrays.asList(1, 3, 8)), new ArrayList<>(Arrays.asList(2, 9, 6))));
int n = v.size();
List<Integer> x = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
x.add(v.get(i).get(j));
}
}
Collections.sort(x);
int k = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
v.get(i).set(j, x.get(k++));
}
}
System.out.println("Sorted Matrix Will be:");
for (List<Integer> row : v) {
for (int num : row) {
System.out.print(num + " ");
}
System.out.println();
}
}}
Python
Python program for the above approach
driver code
v = [[5,4,7], [1,3,8], [2,9,6]] n = len(v)
x = [] for i in range(n): for j in range(n): x.append(v[i][j])
x.sort() k = 0 for i in range(n): for j in range(n): v[i][j] = x[k] k += 1
print("Sorted Matrix will be: ") for i in range(n): for j in range(n): print(v[i][j], end=" ") print("")
THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)
C#
// C# implementation to // sort the given matrix using System;
class GFG { // Driver code public static void Main() { int[, ] mat = { { 5, 4, 7 }, { 1, 3, 8 }, { 2, 9, 6 } }; int n = 3; int temp = 0; int[] x = new int[n*n]; for(int i = 0; i<n; i++){ for(int j = 0; j<n; j++){ x[temp++] = mat[i,j]; } }
Array.Sort(x);
temp = 0;
for(int i = 0; i<n; i++){
for(int j = 0; j<n; j++){
mat[i,j] = x[temp++];
}
}
Console.WriteLine("Sorted Matrix will be : ");
for(int i = 0; i<n; i++){
for(int j = 0; j<n; j++){
Console.Write(mat[i,j] + " ");
}
Console.WriteLine("");
}
}}
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((a, b) => a - b);
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 :
");
for(let i = 0; i<n; i++){
for(let j = 0; j<n; j++){
document.write(v[i][j] + " ");
}
document.write("
");
}
// 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(mnlog2mn)
**Auxiliary Space: O(mn) for the auxiliary 1D array.
Treating the Given Array as 1D Array - O(1) Space
The idea is to treat the 2D-Array as a 1D-Array to sort the matrix without using extra space. This can also be explained with the help of the following example.
**For Example:
There is a 2*2 Matrix with 4 elements,
The idea is to treat the elements of the matrix
as 1D Array of 4 elements.
1 2
3 4
As In the given matrix each element can be accessed as -
1st Element - 0th Row, 0th Col
2nd Element - 0th Row, 1st Col
3rd Element - 1st Row, 0th Col
4th Element - 1st Row, 1st Col
So, for Accessing ith element of the matrix, the relation can be defined as:
Ith Element of the Matrix = Mat[ i / n ][ i % n ]
Where n is number of columns
Please refer Sort the given Matrix | Memory Efficient Approach for implementation