Rearrange all elements of array which are multiples of x in increasing order

Last Updated : 07 Sep, 2022

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 order
void 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 code
int 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 order
def 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 code
if __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# implementation of the approach
using 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 order
function 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 code
var arr = [ 125, 3, 15, 6, 100, 5 ];
var x = 5;
var n = arr.length;

sortMultiples(arr, n, x);

// Print the result
for(var i = 0; i < n; i++) 
{
    document.write(arr[i] + "&nbsp;&nbsp;");
}

// 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.

Analysis of Algorithms

Shashank_Sharma

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Rearrange all elements of array which are multiples of x in increasing order

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