Given a matrix of N rows and M columns, the task is to sort the matrix in the strict order that is every row is sorted in increasing order and the first element of every row is greater than the first element of the previous row.
Examples:
Input: M[][] = { {5, 4, 7},
{1, 3, 8},
{2, 9, 6} }
Output: 1 2 3
4 5 6
7 8 9
Explanation:
Please refer above image
Input: M[][] = { {5, 4, 7},
{1, 3, 8} }
Output: 1 3 4
5 7 8
Approach: 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 / cols ][ i % cols ]
Algorithm:
- Find the number of rows(say rows) and columns(say cols) in the matrix by finding the length of the number of rows in the 2D-Array and the elements in each row in the Array.
- Iterate over each element of the matrix from 0 to the number of elements (rows * cols).
- Find the appropriate position of the element in the matrix using the above formulae for each element.
- Compare each element with the next element (For the last element in the row, the next element will be the next row first element) in the matrix, and if the next element is, less then swap these elements.
Illustration with Example:
| I |
J |
Comparison Elements |
Matrix |
Comments |
| 0 |
0 |
(0, 0) & (0, 1) |
5 6 7
1 4 8 |
No Swap |
| 0 |
1 |
(0, 1) & (0, 2) |
5 6 7
1 4 8 |
No Swap |
| 0 |
2 |
(0, 2) & (1, 0) |
5 6 1
7 4 8 |
Swapped |
| 0 |
3 |
(1, 0) & (1, 1) |
5 6 1
4 7 8 |
Swapped |
| 0 |
4 |
(1, 1) & (1, 2) |
5 6 1
4 7 8 |
No Swap |
| 1 |
0 |
(0, 0) & (0, 1) |
5 6 1
4 7 8 |
No Swap |
| 1 |
1 |
(0, 1) & (0, 2) |
5 1 6
4 7 8 |
Swapped |
| 1 |
2 |
(0, 2) & (1, 0) |
5 1 4
6 7 8 |
Swapped |
| 1 |
3 |
(1, 0) & (1, 1) |
5 1 4
6 7 8 |
No Swap |
| 1 |
4 |
(1, 1) & (1, 2) |
5 1 4
4 7 8 |
No Swap |
| 2 |
0 |
(0, 0) & (0, 1) |
1 5 4
6 7 8 |
Swapped |
| 2 |
1 |
(0, 1) & (0, 2) |
1 4 5
6 7 8 |
Swapped |
| 2 |
2 |
(0, 2) & (1, 0) |
1 4 5
6 7 8 |
No Swap |
| 2 |
3 |
(1, 0) & (1, 1) |
5 1 4
6 7 8 |
No Swap |
| 2 |
4 |
(1, 1) & (1, 2) |
5 1 4
4 7 8 |
No Swap |
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
#define N 3
#define M 3
void sortMat(int data[N][M], int row, int col)
{
int size = row * col;
for (int i = 0; i < size; i++)
{
for (int j = 0; j < size - 1; j++)
{
if (data[j / col][j % col] > data[(j + 1)
/ col][(j + 1) % col])
{
int temp = data[j / col][j % col];
data[j / col][j % col] = data[(j + 1)
/ col][(j + 1) % col];
data[(j + 1) / col][(j + 1) % col] = temp;
}
}
}
}
void printMat(int mat[N][M], int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
{
cout << mat[i][j] << " ";
}
cout << endl;
}
}
int main()
{
int mat[N][M] = { { 5, 4, 7 },
{ 1, 3, 8 },
{ 2, 9, 6 } };
int row = N;
int col = M;
sortMat(mat, row, col);
printMat(mat, row, col);
return 0;
}
|
Java
class GFG
{
static void sortMat(int[][] data, int row, int col)
{
int size = row * col;
for (int i = 0; i < size; i++)
{
for (int j = 0; j < size - 1; j++)
{
if (data[j / col][j % col] > data[(j + 1)
/ col][(j + 1) % col])
{
int temp = data[j / col][j % col];
data[j / col][j % col] = data[(j + 1)
/ col][(j + 1) % col];
data[(j + 1) / col][(j + 1) % col] = temp;
}
}
}
}
static void printMat(int[][] mat, int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
{
System.out.print(mat[i][j] + " ");
}
System.out.println();
}
}
public static void main(String[] args)
{
int[][] mat = { { 5, 4, 7 },
{ 1, 3, 8 },
{ 2, 9, 6 } };
int row = mat.length;
int col = mat[0].length;
sortMat(mat, row, col);
printMat(mat, row, col);
}
}
|
Python3
def sortMat(data, row, col):
size = row * col
for i in range(0, size):
for j in range(0, size-1):
if ( data[j//col][j % col] >\
data[(j + 1)//col][(j + 1)% col] ):
temp = data[j//col][j % col]
data[j//col][j % col] =\
data[(j + 1)//col][(j + 1)% col]
data[(j + 1)//col][(j + 1)% col] =\
temp
def printMat(mat, row, col):
for i in range(row):
for j in range(col):
print(mat[i][j], end =" ")
print()
if __name__ == "__main__":
mat = [ [5, 4, 7],
[1, 3, 8],
[2, 9, 6] ]
row = len(mat)
col = len(mat[0])
sortMat(mat, row, col)
printMat(mat, row, col)
|
C#
using System;
class GFG
{
static void sortMat(int[,] data, int row, int col)
{
int size = row * col;
for (int i = 0; i < size; i++)
{
for (int j = 0; j < size - 1; j++)
{
if (data[j / col,j % col] > data[(j + 1)
/ col,(j + 1) % col])
{
int temp = data[j / col,j % col];
data[j / col,j % col] = data[(j + 1)
/ col,(j + 1) % col];
data[(j + 1) / col,(j + 1) % col] = temp;
}
}
}
}
static void printMat(int[,] mat, int row, int col)
{
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
{
Console.Write(mat[i,j] + " ");
}
Console.WriteLine();
}
}
public static void Main(String[] args)
{
int[,] mat = { { 5, 4, 7 },
{ 1, 3, 8 },
{ 2, 9, 6 } };
int row = mat.GetLength(0);
int col = mat.GetLength(1);
sortMat(mat, row, col);
printMat(mat, row, col);
}
}
|
Javascript
<script>
let N = 3;
let M = 3;
function sortMat(data, row, col)
{
let size = row * col;
for (let i = 0; i < size; i++)
{
for (let j = 0; j < size - 1; j++)
{
if (data[(Math.floor(j / col))][j % col] >
data[(Math.floor((j + 1) / col))][(j + 1) % col])
{
let temp = data[(Math.floor(j / col))][j % col];
data[(Math.floor(j / col))][j % col] =
data[(Math.floor((j + 1) / col))][(j + 1) % col];
data[(Math.floor((j + 1) / col))][(j + 1) % col] = temp;
}
}
}
}
function printMat(mat, row, col)
{
for (let i = 0; i < row; i++)
{
for (let j = 0; j < col; j++)
{
document.write(mat[i][j] + " ");
}
document.write("<br>");
}
}
let mat = [ [ 5, 4, 7 ],
[ 1, 3, 8 ],
[ 2, 9, 6 ] ];
let row = N;
let col = M;
sortMat(mat, row, col);
printMat(mat, row, col);
</script>
|
Output:
1 2 3
4 5 6
7 8 9
Performance Analysis:
- Time Complexity: In the given approach, we are sorting the elements in the matrix by considering the elements in the 1D-Array using Bubble sort, so the overall complexity will be O(N * M>)
- Space Complexity: In the given approach, no extra space is used, so the overall space complexity will be O(1)