Sort the Matrix based on the given column number
Last Updated :
14 Apr, 2023
Given a Matrix of size M * N (0-indexed) and an integer K, the matrix contains distinct integers only, the task is to sort the Matrix (i.e., the rows of the matrix) by their Values in the Kth column, from Maximum to Minimum. Return the matrix after sorting it.
Examples:
Input: Matrix = [[10, 6, 9, 1], [7, 5, 11, 2], [4, 8, 3, 15]], K = 2
Output: [[7, 5, 11, 2], [10, 6, 9, 1], [4, 8, 3, 15]]
Explanation: In the above Example:
- The index Row 1 has 11 in Column 2, which is the highest, so put it first place.
- The index Row 0 has 9 in Column 2, which is the second highest, so put it in second place.
- The index Row 2 has 3 in Column 2, which is the lowest, so put it in third place.
Input: Matrix = [[3, 4], [5, 6]], K = 0
Output: [[5, 6], [3, 4]]
Explanation: In the above example:
- The index Row 1 has 5 in Column 0, which is the highest, so put it first place.
- The index Row 0 has 3 in Column 0, which is the lowest, so put it in last place.
Approach 1: Using Bubble Sort
This problem can simply be solved by using Bubble sort or some other basic sorting techniques.
- Declare a 2D vector matrix of type integer and initialize it with some values. This matrix represents the input data that needs to be sorted.
- Declare an integer variable k and initialize it with some value. This variable is used to specify the column on which the sorting should be performed.
- Declare two integer variables i and j for use in the following loops.
- Determine the number of rows in the matrix by calling the size method on the matrix vector.
- Use a nested for loop to iterate over the rows and columns of the matrix, starting with the first row and the first column, and comparing the adjacent kth element in each row. If the kth element in the current row is less than the kth element in the next row, swap the two rows.
- Continue this process until all rows have been compared and sorted.
- Print the sorted matrix to the console using nested for loops, iterating over each row and column of the matrix.
Below is the implementation of the above approach.
C++
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<vector<int> > matrix{ { 4, 8, 3, 5 },
{ 7, 5, 11, 2 },
{ 10, 6, 9, 1 } };
int k = 2;
int i, j;
int rows = matrix.size();
// Bubble sort logic
for (i = 0; i < rows - 1; i++) {
for (j = 0; j < rows - i - 1; j++) {
// Comparing the adjacent kth element
if (matrix[j][k] < matrix[j + 1][k]) {
// Storing whole row in temp
vector<int> temp = matrix[j];
matrix[j] = matrix[j + 1];
matrix[j + 1] = temp;
}
}
}
// Printing the matrix
cout << "Sorted matrix is : " << endl;
for (int i = 0; i < matrix.size(); i++) {
for (int j = 0; j < matrix[i].size(); j++)
cout << matrix[i][j] << " ";
cout << endl;
}
return 0;
}
// This code is contributed by aeroabrar_31
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static void main(String[] args)
{
int[][] matrix = { { 4, 8, 3, 5 },
{ 7, 5, 11, 2 },
{ 10, 6, 9, 1 } };
int k = 2;
int rows = matrix.length;
// Bubble sort logic
for (int i = 0; i < rows - 1; i++) {
for (int j = 0; j < rows - i - 1; j++) {
// comparing the adjacent kth element
if (matrix[j][k] < matrix[j + 1][k]) {
// swapping adjacent rows
int[] temp = matrix[j]; // storing whole
// row in temp
matrix[j] = matrix[j + 1];
matrix[j + 1] = temp;
}
}
}
System.out.println("Sorted matrix : ");
// printing the matrix after sorting
for (int[] temp : matrix) {
for (int it : temp) {
System.out.print(it + " ");
}
System.out.println();
}
}
}
// This code is contributed by aeroabrar_31
def sort_the_matrix(matrix, k):
n = len(matrix)
# bubble sort logic
for i in range(0, n-1):
for j in range(0, n-i-1):
# comparing the adjacent kth element
if matrix[j][k] < matrix[j+1][k]:
matrix[j], matrix[j+1] = matrix[j +
1], matrix[j] # swapping rows
# printing the sorted matrix
print("Sorted matrix : ")
N = len(matrix)
M = len(matrix[0])
for i in range(N):
for j in range(M):
print(matrix[i][j], end=" ")
print()
# Driver's code
matrix = [[4, 8, 3, 5],
[7, 5, 11, 2],
[10, 6, 9, 1]]
K = 2
sort_the_matrix(matrix, K)
let matrix = [[4, 8, 3, 5], [7, 5, 11, 2], [10, 6, 9, 1]];
let k = 2;
// Bubble sort logic
for (let i = 0; i < matrix.length - 1; i++)
{
for (let j = 0; j < matrix.length - i - 1; j++)
{
// Comparing the adjacent kth element
if (matrix[j][k] < matrix[j + 1][k])
{
// Storing whole row in temp
let temp = matrix[j];
matrix[j] = matrix[j + 1];
matrix[j + 1] = temp;
}
}
}
// Printing the matrix
console.log("Sorted matrix is:");
for (let i = 0; i < matrix.length; i++) {
console.log(matrix[i].join(" "));
}
// This code is contributed by rutikbhosale
using System;
using System.Collections.Generic;
class Program
{
static void Main(string[] args)
{
List<List<int>> matrix = new List<List<int>>()
{
new List<int> { 4, 8, 3, 5 },
new List<int> { 7, 5, 11, 2 },
new List<int> { 10, 6, 9, 1 }
};
int k = 2;
int i, j;
int rows = matrix.Count;
// Bubble sort logic
for (i = 0; i < rows - 1; i++)
{
for (j = 0; j < rows - i - 1; j++)
{
// Comparing the adjacent kth element
if (matrix[j][k] < matrix[j + 1][k])
{
// Storing whole row in temp
List<int> temp = matrix[j];
matrix[j] = matrix[j + 1];
matrix[j + 1] = temp;
}
}
}
// Printing the matrix
Console.WriteLine("Sorted matrix is : ");
for (int m = 0; m < matrix.Count; m++)
{
for (int n = 0; n < matrix[m].Count; n++)
{
Console.Write(matrix[m][n] + " ");
}
Console.WriteLine();
}
Console.ReadLine();
}
}
OutputSorted matrix is :
7 5 11 2
10 6 9 1
4 8 3 5
Time Complexity: O ( M2 ) M rows
Auxiliary Space: O( N ) N is for storing temp row with N elements while swapping
Approach 2: Using sort( ) function
To solve the problem follow the below steps:
- Create a vector that consists of a pair of elements vector<pair<int, int> v.
- Push the Kth Column Elements In that Vector Along with each row.
- Sort that Column by using the sort function.
- Reverse the vector.
- Now create a 2D vector and push each element from the vector to the corresponding row along with the same column number of the vector resultant.
- Return new 2D vector.
Below is the implementation of the above approach:
C++
// C++ Program to sort the matrix based
// on the column values.
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> >
sortTheMatrix(vector<vector<int> >& matrix, int k)
{
// Rows
int n = matrix.size();
vector<pair<int, int> > v;
for (int i = 0; i < n; i++) {
// Kth col in ith row and
// their ith row value
v.push_back({ matrix[i][k], i });
}
// Based on increasing manner
// of kth col
sort(v.begin(), v.end());
// Based on decreasing manner
// of kth col
reverse(v.begin(), v.end());
vector<vector<int> > a;
for (auto& it : v) {
a.push_back(
// Now put each row one by
// one into our ans in
// decresing manner(based
// on Kth col)
matrix[it.second]);
}
return a;
}
// Print function
void print(vector<vector<int> > matrix, int k)
{
vector<vector<int> > ans = sortTheMatrix(matrix, k);
cout << "Sorted Matrix Based on column 2 Values :"
<< endl;
int N = ans.size();
int M = ans[0].size();
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
cout << ans[i][j] << " ";
}
cout << '\n';
}
}
// Drivers code
int main()
{
vector<vector<int> > matrix = { { 10, 6, 9, 1 },
{ 7, 5, 11, 2 },
{ 4, 8, 3, 15 } };
int K = 2;
// Function Call
print(matrix, K);
return 0;
}
import java.util.*;
public class Main {
public static List<List<Integer>> sortTheMatrix(List<List<Integer>> matrix, int k) {
// Rows
int n = matrix.size();
// Create a list of pairs containing the value at
// the k-th column and the index of the row
List<Tuple<Integer, Integer>> v = new ArrayList<>();
for (int i = 0; i < n; i++) {
v.add(new Tuple<>(matrix.get(i).get(k), i));
}
// Sort the list based on increasing order of the
// k-th column
Collections.sort(v, new Comparator<Tuple<Integer, Integer>>() {
@Override
public int compare(Tuple<Integer, Integer> x, Tuple<Integer, Integer> y) {
return x.item1.compareTo(y.item1);
}
});
// Reverse the list to get decreasing order of the
// k-th column
Collections.reverse(v);
// Create a new matrix and add the rows to it in
// decreasing order of the k-th column
List<List<Integer>> a = new ArrayList<>();
for (Tuple<Integer, Integer> item : v) {
a.add(matrix.get(item.item2));
}
return a;
}
// print function
public static void print(List<List<Integer>> matrix, int k) {
List<List<Integer>> ans = sortTheMatrix(matrix, k);
System.out.println("Sorted Matrix Based on column " + (k ) + " Values:");
for (List<Integer> row : ans) {
for (int value : row) {
System.out.print(value + " ");
}
System.out.println();
}
}
// Driver's code
public static void main(String[] args) {
List<List<Integer>> matrix = Arrays.asList(
Arrays.asList(10, 6, 9, 1),
Arrays.asList(7, 5, 11, 2),
Arrays.asList(4, 8, 3, 15)
);
int K = 2;
// function call
print(matrix, K);
}
}
class Tuple<X, Y> {
public final X item1;
public final Y item2;
public Tuple(X x, Y y) {
this.item1 = x;
this.item2 = y;
}
}
# Python Program to sort the matrix based
# on the column values.
def sortTheMatrix(matrix, k):
# Rows
n = len(matrix)
v = []
for i in range(n):
# Kth col in ith row and
# their ith row value
v.append((matrix[i][k], i))
# Based on increasing manner
# of kth col
v.sort()
# Based on decreasing manner of kth col
v = v[::-1]
a = []
for it in v:
# Now put each row one by
# one into our ans in
# decresing manner(based
# on Kth col)
a.append(matrix[it[1]])
return a
# Print function
def print_matrix(matrix, k):
ans = sortTheMatrix(matrix, k)
print("Sorted Matrix Based on column {} Values :".format(k))
N = len(ans)
M = len(ans[0])
for i in range(N):
for j in range(M):
print(ans[i][j], end=" ")
print()
# Drivers code
matrix = [[10, 6, 9, 1],
[7, 5, 11, 2],
[4, 8, 3, 15]]
K = 2
# Function Call
print_matrix(matrix, K)
# This code is contributed by prasad264
// C# program to sort the matrix based
// on the column values
using System;
using System.Collections.Generic;
using System.Linq;
class Program {
static List<List<int> >
sortTheMatrix(List<List<int> > matrix, int k)
{
// Rows
int n = matrix.Count;
// Create a list of pairs containing the value at
// the k-th column and the index of the row
List<Tuple<int, int> > v
= new List<Tuple<int, int> >();
for (int i = 0; i < n; i++) {
v.Add(Tuple.Create(matrix[i][k], i));
}
// Sort the list based on increasing order of the
// k-th column
v.Sort((x, y) = > x.Item1.CompareTo(y.Item1));
// Reverse the list to get decreasing order of the
// k-th column
v.Reverse();
// Create a new matrix and add the rows to it in
// decreasing order of the k-th column
List<List<int> > a = new List<List<int> >();
foreach(Tuple<int, int> item in v)
{
a.Add(matrix[item.Item2]);
}
return a;
}
// print function
static void print(List<List<int> > matrix, int k)
{
List<List<int> > ans = sortTheMatrix(matrix, k);
Console.WriteLine(
"Sorted Matrix Based on column 2 Values :");
foreach(List<int> row in ans)
{
foreach(int value in row)
{
Console.Write(value + " ");
}
Console.WriteLine();
}
}
// Driver's code
static void Main()
{
List<List<int> > matrix = new List<List<int> >{
new List<int>{ 10, 6, 9, 1 },
new List<int>{ 7, 5, 11, 2 },
new List<int>{ 4, 8, 3, 15 }
};
int K = 2;
// function call
print(matrix, K);
}
}
// JavaScript Program to sort the matrix based
// on the column values.
// Function to sort the matrix based on column values
function sortTheMatrix(matrix, k) {
// Rows
const n = matrix.length;
// Create a list of pairs containing the value at
// the k-th column and the index of the row
const v = [];
for (let i = 0; i < n; i++) {
v.push([(matrix[i][k]), i]);
}
// Sort the list based on increasing order of the
// k-th column
v.sort((a, b) => a[0] - b[0]);
// Reverse the list to get decreasing order of the
// k-th column
v.reverse();
// Create a new matrix and add the rows to it in
// decreasing order of the k-th column
const a = [];
for (const [value, index] of v) {
a.push(matrix[index]);
}
return a;
}
// Print function
function print(matrix, k) {
const ans = sortTheMatrix(matrix, k);
console.log(`Sorted Matrix Based on column ${k + 1} Values:`);
for (const row of ans) {
console.log(row.join(' '));
}
}
// Driver's code
const matrix = [
[10, 6, 9, 1],
[7, 5, 11, 2],
[4, 8, 3, 15],
];
const K = 2;
// Function call
print(matrix, K);
OutputSorted Matrix Based on column 2 Values :
7 5 11 2
10 6 9 1
4 8 3 15
Time Complexity: O(N*logN) i., e O(N) running for loop and LogN for STL.
Auxiliary Space: O(N)
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