Given an array of integers arr[] and an integer x, the task is to sort all the elements of the array which are multiples of x in decreasing order in their relative positions i.e. positions of the other elements must not be affected.
Examples:
Input: arr[] = {10, 5, 8, 2, 15}, x = 5
Output: 15 10 8 2 5
We rearrange all multiples of 5 (i.e. 10, 5 and 15) in decreasing order in their relative positions, keeping other elements same.Input: arr[] = {100, 12, 25, 50, 5}, x = 5
Output: 100 12 50 25 5
Approach:
- Traverse the array and check if the number is multiple of x. If it is, store it in a vector.
- Then, sort the vector in decreasing order.
- Again traverse the array and replace the elements which are multiples of 5 with the vector elements one by one.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to sort all the// multiples of x from the// array in decreasing ordervoid sortMultiples(int arr[], int n, int x){ vector<int> v; // Insert all multiples of x to a vector for (int i = 0; i < n; i++) if (arr[i] % x == 0) v.push_back(arr[i]); // Sort the vector in descending sort(v.begin(), v.end(), std::greater<int>()); int j = 0; // update the array elements for (int i = 0; i < n; i++) { if (arr[i] % x == 0) arr[i] = v[j++]; }}// Utility function to print the arrayvoid printArray(int arr[], int N){ // Print the array for (int i = 0; i < N; i++) { cout << arr[i] << " "; }}// Driver codeint main(){ int arr[] = { 125, 3, 15, 6, 100, 5 }; int x = 5; int n = sizeof(arr) / sizeof(arr[0]); sortMultiples(arr, n, x); printArray(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{ // Function to sort all the // multiples of x from the // array in decreasing order static void sortMultiples(int arr[], int n, int x) { Vector<Integer> v = new Vector<Integer>(); // Insert all multiples of x to a vector for (int i = 0; i < n; i++) { if (arr[i] % x == 0) { v.add(arr[i]); } } // Sort the vector in descending Collections.sort(v, Collections.reverseOrder()); int j = 0; // update the array elements for (int i = 0; i < n; i++) { if (arr[i] % x == 0) { arr[i] = v.get(j++); } } } // Utility function to print the array static void printArray(int arr[], int N) { // Print the array for (int i = 0; i < N; i++) { System.out.print(arr[i] + " "); } } // Driver code public static void main(String[] args) { int arr[] = {125, 3, 15, 6, 100, 5}; int x = 5; int n = arr.length; sortMultiples(arr, n, x); printArray(arr, n); }}// This code has been contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to sort all the # multiples of x from the # array in decreasing order def sortMultiples(arr, n, x) : v = [] # Insert all multiples of x # to a vector for i in range(n) : if (arr[i] % x == 0) : v.append(arr[i]) # Sort the vector in descending v.sort(reverse = True) j = 0 # update the array elements for i in range(n) : if (arr[i] % x == 0) : arr[i] = v[j] j += 1 # Utility function to print the array def printArray(arr, N) : # Print the array for i in range(N) : print(arr[i], end = " ")# Driver code if __name__ == "__main__" : arr= [ 125, 3, 15, 6, 100, 5 ] x = 5 n = len(arr) sortMultiples(arr, n, x) printArray(arr, n) # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG{ // Function to sort all the // multiples of x from the // array in decreasing order static void sortMultiples(int []arr, int n, int x) { List<int> v = new List<int>(); // Insert all multiples of x to a vector for (int i = 0; i < n; i++) { if (arr[i] % x == 0) { v.Add(arr[i]); } } // Sort the vector in descending v.Sort(); v.Reverse(); int j = 0; // update the array elements for (int i = 0; i < n; i++) { if (arr[i] % x == 0) { arr[i] = v[j++]; } } } // Utility function to print the array static void printArray(int []arr, int N) { // Print the array for (int i = 0; i < N; i++) { Console.Write(arr[i] + " "); } } // Driver code public static void Main() { int []arr = {125, 3, 15, 6, 100, 5}; int x = 5; int n = arr.Length; sortMultiples(arr, n, x); printArray(arr, n); }}/* This code contributed by PrinciRaj1992 */ |
PHP
<?php // PHP implementation of the approach// Function to sort all the multiples // of x from the array in decreasing orderfunction sortMultiples(&$arr, $n, $x){ $v = array(); // Insert all multiples of x to a vector for ($i = 0; $i < $n; $i++) if ($arr[$i] % $x == 0) array_push($v, $arr[$i]); // Sort the vector in descending rsort($v); $j = 0; // update the array elements for ($i = 0; $i < $n; $i++) { if ($arr[$i] % $x == 0) $arr[$i] = $v[$j++]; }}// Utility function to print the arrayfunction printArray($arr, $N){ // Print the array for ($i = 0; $i < $N; $i++) { echo $arr[$i] . " "; }}// Driver code$arr = array(125, 3, 15, 6, 100, 5 );$x = 5;$n = sizeof($arr);sortMultiples($arr, $n, $x);printArray($arr, $n);// This code is contributed by ita_c?> |
Javascript
<script>// JavaScript implementation of the approach// Function to sort all the// multiples of x from the// array in decreasing orderfunction sortMultiples(arr, n, x) { var v = []; // Insert all multiples of x to a vector for(var i = 0; i < n; i++) if (arr[i] % x == 0) v.push(arr[i]); // Sort the vector in descending v.sort((a, b) => b - a); var j = 0; // update the array elements for(var i = 0; i < n; i++) { if (arr[i] % x == 0) arr[i] = v[j++]; }}// Utility function to print the arrayfunction printArray(arr, N){ // Print the array for(var i = 0; i < N; i++) { document.write(arr[i] + " "); }}// Driver codevar arr = [ 125, 3, 15, 6, 100, 5 ];var x = 5;var n = arr.length;sortMultiples(arr, n, x);printArray(arr, n);// This code is contributed by rdtank</script> |
Output
125 3 100 6 15 5
Complexity Analysis:
- Time Complexity: O(nlogn)
- Auxiliary Space: O(n)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
