Given an array of integers ‘arr’ and a number x, the task is to sort all the elements which are multiples of x of the array in ascending order in their relative positions i.e. other positions of the other elements must not be affected.
Examples:
Input: arr[] = {10, 5, 8, 2, 15}, x = 5
Output: 5 10 8 2 15
We rearrange all multiples of 5 in increasing order, keeping other elements same.Input: arr[] = {100, 12, 25, 50, 5}, x = 5
Output: 5 12 25 50 100
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 ascending 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 ascending ordervoid sortMultiples(int arr[], int n, int x){ vector<int> v; // Insert all multiples of 5 to a vector for (int i = 0; i < n; i++) if (arr[i] % x == 0) v.push_back(arr[i]); // Sort the vector sort(v.begin(), v.end()); int j = 0; // update the array elements for (int i = 0; i < n; i++) { if (arr[i] % x == 0) arr[i] = v[j++]; }}// 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); // Print the result for (int i = 0; i < n; i++) { cout << arr[i] << " "; } return 0;} |
Java
import java.util.Collections;import java.util.Vector;// Java implementation of the approach class GFG {// Function to sort all the // multiples of x from the // array in ascending order static void sortMultiples(int arr[], int n, int x) { Vector<Integer> v = new Vector<Integer>(); // Insert all multiples of 5 to a vector for (int i = 0; i < n; i++) { if (arr[i] % x == 0) { v.add(arr[i]); } } // Sort the vector Collections.sort(v); //sort(v.begin(), v.end()); int j = 0; // update the array elements for (int i = 0; i < n; i++) { if (arr[i] % x == 0) { arr[i] = v.get(j++); } } }// 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); // Print the result for (int i = 0; i < n; i++) { System.out.print(arr[i]+" "); } }}// This code is contributed by Rajput-Ji |
Python3
# Python 3 implementation of the approach# Function to sort all the multiples of x # from the array in ascending orderdef sortMultiples(arr, n, x): v = [] # Insert all multiples of 5 to a vector for i in range(0, n, 1): if (arr[i] % x == 0): v.append(arr[i]) # Sort the vector v.sort(reverse=False) j = 0 # update the array elements for i in range(0, n, 1): if (arr[i] % x == 0): arr[i] = v[j] j += 1# Driver codeif __name__ == '__main__': arr = [ 125, 3, 15, 6, 100, 5] x = 5 n = len(arr) sortMultiples(arr, n, x) # Print the result for i in range(0, n, 1): print(arr[i], end = " ")# This code is contributed by # Surendra _Gangwar |
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 ascending order static void sortMultiples(int []arr, int n, int x) { List<int> v = new List<int>(); int i; // Insert all multiples of 5 to a vector for (i = 0; i < n; i++) if (arr[i] % x == 0) v.Add(arr[i]); // Sort the vector v.Sort(); int j = 0; // update the array elements for (i = 0; i < n; i++) { if (arr[i] % x == 0) arr[i] = v[j++]; } } // 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); // Print the result for (int i = 0; i < n; i++) { Console.Write(arr[i] + " "); } }}// This code is contributed by// Shivi_Aggarwal |
Javascript
<script>// JavaScript implementation of the approach// Function to sort all the// multiples of x from the// array in ascending orderfunction sortMultiples(arr, n, x){ var v = []; // Insert all multiples of 5 to a vector for(var i = 0; i < n; i++) { if (arr[i] % x == 0) { v.push(arr[i]); } } // Sort the vector v.sort((a, b) => a - b); var j = 0; // update the array elements for(var i = 0; i < n; i++) { if (arr[i] % x == 0) arr[i] = v[j++]; }}// Driver codevar arr = [ 125, 3, 15, 6, 100, 5 ];var x = 5;var n = arr.length;sortMultiples(arr, n, x);// Print the resultfor(var i = 0; i < n; i++) { document.write(arr[i] + " ");}// This code is contributed by rdtank</script> |
Output
5 3 15 6 100 125
Complexity Analysis:
- Time Complexity: O(N*logN), as we are using inbuilt sort function which cost the afore mentioned time.
- Auxiliary Space: O(N), as we are using extra space for array/vector v.
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