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)
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