Sort Matrix in alternating ascending and descending order rowwise
Given an N x N matrix, our task is to print the row of the matrix in ascending and descending order alternatively.
Examples:
Input:
5 7 3 4
9 5 8 2
6 3 8 1
5 8 9 3
Output:
3 4 5 7
9 8 5 2
1 3 6 8
9 8 5 3
Explanation:
Here the first row is sorted in ascending order, second row sorted in descending order, third row sorted in ascending order and so on.
Input:
7 3 4
3 8 2
3 6 1
Output:
3 4 7
8 3 2
1 3 6
Explanation:
Here the first row is sorted in ascending order, second row sorted in descending order, third row sorted in ascending order.
Approach to solve
To solve the problem mentioned above we iterate 0 to N and check if the ith row is even or odd, if it is even then we sort the row in ascending order otherwise sort the ith row in descending order. Return the matrix after N iterations.
Below is the implementation of the above approach:
C++
// C++ implementation to print row of // matrix in ascending or descending // order alternatively #include <stdio.h> #define N 4 void func( int a[][N]) { // Iterate matrix rowwise for ( int i = 0; i < N; i++) { // Sort even rows in ascending order if (i % 2 == 0) { for ( int j = 0; j < N; j++) { for ( int k = j + 1; k < N; ++k) { // compare adjacent elements if (a[i][j] > a[i][k]) { // swap adjacent element int temp = a[i][j]; a[i][j] = a[i][k]; a[i][k] = temp; } } } } // Sort even rows in descending order else { for ( int j = 0; j < N; j++) { for ( int k = j + 1; k < N; ++k) { // compare adjacent elements if (a[i][j] < a[i][k]) { // swap adjacent element int temp = a[i][j]; a[i][j] = a[i][k]; a[i][k] = temp; } } } } } // Printing the final Output for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) { printf ( "%d " , a[i][j]); } printf ( "\n" ); } } // Driver code int main() { int a[N][N] = { { 5, 7, 3, 4 }, { 9, 5, 8, 2 }, { 6, 3, 8, 1 }, { 5, 8, 9, 3 } }; func(a); return 0; } |
Java
// Java implementation to print row of // matrix in ascending or descending // order alternatively import java.io.*; class GFG{ static int N = 4 ; static void func( int a[][]) { int i, j, k; // Iterate matrix rowwise for (i = 0 ; i < N; i++) { // Sort even rows in ascending order if (i % 2 == 0 ) { for (j = 0 ; j < N; j++) { for (k = j + 1 ; k < N; ++k) { // Compare adjacent elements if (a[i][j] > a[i][k]) { // Swap adjacent element int temp = a[i][j]; a[i][j] = a[i][k]; a[i][k] = temp; } } } } // Sort even rows in descending order else { for (j = 0 ; j < N; j++) { for (k = j + 1 ; k < N; ++k) { // Compare adjacent elements if (a[i][j] < a[i][k]) { // Swap adjacent element int temp = a[i][j]; a[i][j] = a[i][k]; a[i][k] = temp; } } } } } // Printing the final Output for (i = 0 ; i < N; i++) { for (j = 0 ; j < N; j++) { System.out.print(a[i][j] + " " ); } System.out.print( "\n" ); } } // Driver code public static void main (String []args) { int a[][] = { { 5 , 7 , 3 , 4 }, { 9 , 5 , 8 , 2 }, { 6 , 3 , 8 , 1 }, { 5 , 8 , 9 , 3 } }; func(a); } } // This code is contributed by chitranayal |
Python3
# Python3 implementation to print row of # matrix in ascending or descending # order alternatively N = 4 def func(a): # Iterate matrix rowwise for i in range (N): # Sort even rows in ascending order if i % 2 = = 0 : for j in range (N): for k in range (j + 1 , N): # Compare adjacent elements if a[i][j] > a[i][k]: # Swap adjacent element temp = a[i][j] a[i][j] = a[i][k] a[i][k] = temp # Sort even rows in descending order else : for j in range (N): for k in range (j + 1 , N): # Compare adjacent elements if a[i][j] < a[i][k]: # Swap adjacent element temp = a[i][j] a[i][j] = a[i][k] a[i][k] = temp # Printing the final output for i in range (N): for j in range (N): print (a[i][j], end = " " ) print () # Driver code if __name__ = = '__main__' : a = [ [ 5 , 7 , 3 , 4 ], [ 9 , 5 , 8 , 2 ], [ 6 , 3 , 8 , 1 ], [ 5 , 8 , 9 , 3 ] ] func(a) # This code is contributed by mohit kumar 29 |
C#
// C# implementation to print row of // matrix in ascending or descending // order alternatively using System; class GFG{ static int N = 4; static void func( int [,]a) { int i, j, k; // Iterate matrix rowwise for (i = 0; i < N; i++) { // Sort even rows in ascending order if (i % 2 == 0) { for (j = 0; j < N; j++) { for (k = j + 1; k < N; ++k) { // Compare adjacent elements if (a[i, j] > a[i, k]) { // Swap adjacent element int temp = a[i, j]; a[i, j] = a[i, k]; a[i, k] = temp; } } } } // Sort even rows in descending order else { for (j = 0; j < N; j++) { for (k = j + 1; k < N; ++k) { // Compare adjacent elements if (a[i, j] < a[i, k]) { // Swap adjacent element int temp = a[i, j]; a[i, j] = a[i, k]; a[i, k] = temp; } } } } } // Printing the readonly Output for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { Console.Write(a[i, j] + " " ); } Console.Write( "\n" ); } } // Driver code public static void Main(String []args) { int [,]a = { { 5, 7, 3, 4 }, { 9, 5, 8, 2 }, { 6, 3, 8, 1 }, { 5, 8, 9, 3 } }; func(a); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript implementation to print row of // matrix in ascending or descending // order alternatively let N = 4; function func(a) { // Iterate matrix rowwise for (let i = 0; i < N; i++) { // Sort even rows in ascending order if (i % 2 == 0) { for (let j = 0; j < N; j++) { for (let k = j + 1; k < N; ++k) { // compare adjacent elements if (a[i][j] > a[i][k]) { // swap adjacent element let temp = a[i][j]; a[i][j] = a[i][k]; a[i][k] = temp; } } } } // Sort even rows in descending order else { for (let j = 0; j < N; j++) { for (let k = j + 1; k < N; ++k) { // compare adjacent elements if (a[i][j] < a[i][k]) { // swap adjacent element let temp = a[i][j]; a[i][j] = a[i][k]; a[i][k] = temp; } } } } } // Printing the final Output for (let i = 0; i < N; i++) { for (let j = 0; j < N; j++) { document.write( " " + a[i][j]); } document.write( "<br>" ); } } // Driver code let a = [ [ 5, 7, 3, 4 ], [ 9, 5, 8, 2 ], [ 6, 3, 8, 1 ], [ 5, 8, 9, 3 ] ]; func(a); // This code is contributed by subham348. </script> |
3 4 5 7 9 8 5 2 1 3 6 8 9 8 5 3
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient approach: We can optimize the above approach by sorting any given row by using the sort() function. It will sort any given row in O(n*log(n)) time complexity.
Steps involved in the approach:
- Traverse through each row and check if the row is at an even number or odd.
- If even row is there then sort it in ascending order by directly using the sort function.
- If odd row then sort it in descending order by passing the “greater()” function in sort.
Below is the code for the above approach:
C++
// C++ implementation to print row of // matrix in ascending or descending // order alternatively #include <bits/stdc++.h> using namespace std; #define N 4 void func( int a[][N]) { // Iterate matrix rowwise for ( int i = 0; i < N; i++) { // Sort even rows in ascending order if (i % 2 == 0) { sort(a[i],a[i]+N); } // Sort even rows in descending order else { sort(a[i],a[i]+N,greater< int >()); } } // Printing the final Output for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) { printf ( "%d " , a[i][j]); } printf ( "\n" ); } } // Driver code int main() { int a[N][N] = { { 5, 7, 3, 4 }, { 9, 5, 8, 2 }, { 6, 3, 8, 1 }, { 5, 8, 9, 3 } }; func(a); return 0; } // This code is contributed by Pushpesh Raj |
Java
import java.util.Arrays; import java.util.Collections; public class Main { public static void func(Integer[][] a) { for ( int i = 0 ; i < a.length; i++) { if (i % 2 == 0 ) { Arrays.sort(a[i]); } else { Arrays.sort(a[i], Collections.reverseOrder()); } } for ( int i = 0 ; i < a.length; i++) { for ( int j = 0 ; j < a[i].length; j++) { System.out.print(a[i][j] + " " ); } System.out.println(); } } public static void main(String[] args) { Integer[][] a = {{ 5 , 7 , 3 , 4 }, { 9 , 5 , 8 , 2 }, { 6 , 3 , 8 , 1 }, { 5 , 8 , 9 , 3 }}; func(a); } } |
Python
import numpy as np def func(a): for i in range ( len (a)): if i % 2 = = 0 : a[i] = np.sort(a[i]) else : a[i] = np.sort(a[i])[:: - 1 ] for i in range ( len (a)): for j in range ( len (a[i])): print (a[i][j]) print () a = np.array([[ 5 , 7 , 3 , 4 ], [ 9 , 5 , 8 , 2 ], [ 6 , 3 , 8 , 1 ], [ 5 , 8 , 9 , 3 ]]) func(a) |
Javascript
function func(a) { for (let i = 0; i < a.length; i++) { if (i % 2 === 0) { a[i].sort((a, b) => a - b); } else { a[i].sort((a, b) => b - a); } } for (let i = 0; i < a.length; i++) { for (let j = 0; j < a[i].length; j++) { console.log(a[i][j]); } console.log( "\n" ); } } let a = [[5, 7, 3, 4], [9, 5, 8, 2], [6, 3, 8, 1], [5, 8, 9, 3]]; func(a); |
C#
using System; using System.Collections.Generic; using System.Linq; class MainClass { public static void func( int [][] a) { for ( int i = 0; i < a.Length; i++) { if (i % 2 == 0) { Array.Sort(a[i]); } else { Array.Sort(a[i], new Comparison< int >((x, y) => y.CompareTo(x))); // Using Comparison delegate to sort in reverse order } } for ( int i = 0; i < a.Length; i++) { for ( int j = 0; j < a[i].Length; j++) { Console.Write(a[i][j] + " " ); } Console.WriteLine(); } } public static void Main ( string [] args) { int [][] a = new int [][] { new int [] {5, 7, 3, 4}, new int [] {9, 5, 8, 2}, new int [] {6, 3, 8, 1}, new int [] {5, 8, 9, 3} }; func(a); } } |
3 4 5 7 9 8 5 2 1 3 6 8 9 8 5 3
Time Complexity: O(N2log(N))
Auxiliary Space: O(1)
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