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

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:

1. Traverse the array and check if the number is multiple of x. If it is, store it in a vector.
2. Then, sort the vector in ascending order.
3. 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++ implementation of the approach    #include 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 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; }

 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 v = new Vector();            // 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

 # 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 v = new List();         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 cntributed by // Shivi_Aggarwal

Output:
5 3 15 6 100 125

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