Given an array arr[] of N non-negative integers, the task is to sort these integers according to their reverse.
Examples:
Input: arr[] = {12, 10, 102, 31, 15}
Output: 10 31 12 15 102
Reversing the numbers:
12 -> 21
10 -> 01
102 -> 201
31 -> 13
15 -> 51
Sorting the reversed numbers: 01 13 21 51 201
Original sorted array: 10 13 12 15 102Input: arr[] = {12, 10}
Output: 10 12
Approach: The idea is to store each element with its reverse in a vector pair and then sort all the elements of the vector according to the reverse stored. Finally, print the elements in order.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the// reverse of nint reverseDigits(int num){ int rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num;}// Function to sort the array according to// the reverse of elementsvoid sortArr(int arr[], int n){ // Vector to store the reverse // with respective elements vector<pair<int, int> > vp; // Inserting reverse with elements // in the vector pair for (int i = 0; i < n; i++) { vp.push_back( make_pair(reverseDigits(arr[i]), arr[i])); } // Sort the vector, this will sort the pair // according to the reverse of elements sort(vp.begin(), vp.end()); // Print the sorted vector content for (int i = 0; i < vp.size(); i++) cout << vp[i].second << " ";}// Driver codeint main(){ int arr[] = { 12, 10, 102, 31, 15 }; int n = sizeof(arr) / sizeof(arr[0]); sortArr(arr, n); return 0;} |
Java
// Java implementation of the approach import java.util.*;import java.lang.*;import java.io.*;class GFG{ // Function to return the // reverse of n static int reverseDigits(int num) { int rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num; } // Function to sort the array according // to the reverse of elements static void sortArr(int arr[], int n) { // Vector to store the reverse // with respective elements ArrayList<int[]> vp = new ArrayList<>(); // Inserting reverse with elements // in the vector pair for(int i = 0; i < n; i++) { vp.add(new int[]{reversDigits(arr[i]), arr[i]}); } // Sort the vector, this will sort the pair // according to the reverse of elements Collections.sort(vp, (a, b) -> a[0] - b[0]); // Print the sorted vector content for(int i = 0; i < vp.size(); i++) System.out.print(vp.get(i)[1] + " "); } // Driver codepublic static void main(String[] args) { int arr[] = { 12, 10, 102, 31, 15 }; int n = arr.length; sortArr(arr, n);}}// This code is contributed by offbeat |
Python3
# Python3 implementation of the approach# Function to return the# reverse of ndef reverseDigits(num) : rev_num = 0; while (num > 0) : rev_num = rev_num * 10 + num % 10; num = num // 10; return rev_num;# Function to sort the array according to# the reverse of elementsdef sortArr(arr, n) : # Vector to store the reverse # with respective elements vp = []; # Inserting reverse with elements # in the vector pair for i in range(n) : vp.append((reversDigits(arr[i]),arr[i])); # Sort the vector, this will sort the pair # according to the reverse of elements vp.sort() # Print the sorted vector content for i in range(len(vp)) : print(vp[i][1],end= " ");# Driver codeif __name__ == "__main__" : arr = [ 12, 10, 102, 31, 15 ]; n = len(arr); sortArr(arr, n);# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG { // Function to return the // reverse of n static int reverseDigits(int num) { int rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num; } // Function to sort the array according to // the reverse of elements static void sortArr(int[] arr, int n) { // Vector to store the reverse // with respective elements List<Tuple<int, int>> vp = new List<Tuple<int, int>>(); // Inserting reverse with elements // in the vector pair for (int i = 0; i < n; i++) { vp.Add(new Tuple<int, int>(reversDigits(arr[i]),arr[i])); } // Sort the vector, this will sort the pair // according to the reverse of elements vp.Sort(); // Print the sorted vector content for (int i = 0; i < vp.Count; i++) Console.Write(vp[i].Item2 + " "); } // Driver code static void Main() { int[] arr = { 12, 10, 102, 31, 15 }; int n = arr.Length; sortArr(arr, n); }}// This code is contributed by divyesh072019 |
Javascript
<script>// Javascript implementation of the// above approach// Function to return the// reverse of nfunction reverseDigits(num){ var rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = Math.floor(num / 10); } return rev_num;} // Function to sort the array according to// the reverse of elementsfunction sortArr(arr, n){ // Vector to store the reverse // with respective elements var vp = new Array(n); for (var i = 0; i < n; i++) { vp[i] = []; } // Inserting reverse with elements // in the vector pair for (var i = 0; i < n; i++) { var pair = []; pair.push(reversDigits(arr[i])); pair.push(arr[i]); vp[i] = pair; } // Sort the vector, this will sort the pair // according to the reverse of elements vp = vp.sort(function(a,b) { return a[0] - b[0]; }); // Print the sorted vector content for (var i = 0; i < n; i++){ document.write(vp[i][1] + " "); }}// Driver codevar arr = [ 12, 10, 102, 31, 15 ];var n = arr.length;sortArr(arr, n);// This code is contributed by Shivanisingh</script> |
Output:
10 31 12 15 102
Time Complexity:
where N is the size of the array
Auxiliary Space: O(N)
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