Open In App

Reorder an array according to given indexes

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Share
Report issue
Report

Given two integer arrays of the same size, “arr[]” and “index[]”, reorder elements in “arr[]” according to the given index array.

Example: 

Input:  arr[]   = [10, 11, 12];
            index[] = [1, 0, 2];
Output: arr[]   = [11, 10, 12]
           index[] = [0,  1,  2] 

Input:  arr[]   = [50, 40, 70, 60, 90]
          index[] = [3,  0,  4,  1,  2]
Output: arr[]   = [40, 60, 90, 50, 70]
          index[] = [0,  1,  2,  3,   4]

Reorder an array according to given indexes using Sorting:

The idea is to use an array of pairs to associate elements from the arr[] array with their respective indices from the index[] array and then sort these pairs based on the indices.

Follow the steps to solve the problem:

  • Create an array of pairs to and associate elements from the arr[] array with their respective indices from the index[].
  • Then sort the array of pair according based on indices using comparator.
  • Copy the rearranged elements back into the arr[] array, effectively reordering the elements according to the specified indices.

Below is the implementation of above approach:

C++




#include <bits/stdc++.h>
 
using namespace std;
 
// Comparator function to sort pairs based on the second
// element
bool comp(const pair<int, int>& v1,
          const pair<int, int>& v2)
{
    // Sort in ascending order of index values
    return v1.second < v2.second;
}
 
// Function to reorder elements of arr[] according to
// index[]
void reorder(int arr[], int index[], int n)
{
    // Create a vector of pairs, where each pair stores the
    // original element (arr[i]) and its corresponding index
    // (index[i])
    vector<pair<int, int> > a(n);
 
    for (int i = 0; i < n; i++) {
        a[i].first = arr[i];
        a[i].second = index[i];
    }
 
    // Sort the vector of pairs based on the second element
    // (index) using the comp() comparator function
    sort(a.begin(), a.end(), comp);
 
    // Copy the sorted elements back to the original arr[]
    for (int i = 0; i < n; i++) {
        arr[i] = a[i].first;
    }
}
 
// Driver program
int main()
{
    int arr[] = { 50, 40, 70, 60, 90 };
    int index[] = { 3, 0, 4, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Call the reorder function to rearrange elements in
    // arr[] based on index[]
    reorder(arr, index, n);
 
    cout << "Reordered array is: \n";
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
 
    return 0;
}


Java




import java.util.Arrays;
import java.util.Comparator;
 
public class ReorderArray {
 
    // Comparator function to sort pairs based on the second
    // element
    static class PairComparator
        implements Comparator<int[]> {
        @Override
        public int compare(int[] pair1, int[] pair2)
        {
            // Sort in ascending order of index values
            return Integer.compare(pair1[1], pair2[1]);
        }
    }
 
    // Function to reorder elements of arr[] according to
    // index[]
    static void reorder(int[] arr, int[] index, int n)
    {
        // Create a 2D array to store pairs (original
        // element, index)
        int[][] pairs = new int[n][2];
 
        // Populate the pairs array
        for (int i = 0; i < n; i++) {
            pairs[i][0] = arr[i];
            pairs[i][1] = index[i];
        }
 
        // Sort the pairs array based on the second element
        // (index)
        Arrays.sort(pairs, new PairComparator());
 
        // Copy the sorted elements back to the original
        // arr[]
        for (int i = 0; i < n; i++) {
            arr[i] = pairs[i][0];
        }
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int[] arr = { 50, 40, 70, 60, 90 };
        int[] index = { 3, 0, 4, 1, 2 };
        int n = arr.length;
 
        // Call the reorder function to rearrange elements
        // in arr[] based on index[]
        reorder(arr, index, n);
 
        System.out.println("Reordered array is: ");
        for (int i : arr) {
            System.out.print(i + " ");
        }
    }
}


Python




def reorder(arr, index):
    n = len(arr)
    result = [0] * n
 
    for i in range(n):
        result[index[i]] = arr[i]
 
    # Copy the reordered elements back to the original arr[]
    for i in range(n):
        arr[i] = result[i]
 
 
if __name__ == "__main__":
    arr = [50, 40, 70, 60, 90]
    index = [3, 0, 4, 1, 2]
 
    # Call the reorder function to rearrange elements in arr[] based on index[]
    reorder(arr, index)
 
    print("Reordered array is:")
    print(" ".join(map(str, arr)))


C#




using System;
 
class Program {
    // Function to reorder elements of arr[] according to
    // index[]
    static void Reorder(int[] arr, int[] index)
    {
        int n = arr.Length;
        int[] result = new int[n];
 
        for (int i = 0; i < n; i++) {
            result[index[i]] = arr[i];
        }
 
        // Copy the reordered elements back to the original
        // arr[]
        Array.Copy(result, arr, n);
    }
 
    static void Main()
    {
        int[] arr = { 50, 40, 70, 60, 90 };
        int[] index = { 3, 0, 4, 1, 2 };
        int n = arr.Length;
 
        // Call the Reorder function to rearrange elements
        // in arr[] based on index[]
        Reorder(arr, index);
 
        Console.WriteLine("Reordered array is:");
        Console.WriteLine(string.Join(" ", arr));
    }
}


Javascript




// Comparator function to sort pairs based on the second element
function comp(v1, v2) {
    // Sort in ascending order of index values
    return v1[1] - v2[1];
}
 
// Function to reorder elements of arr[] according to index[]
function reorder(arr, index) {
    // Create an array of pairs, where each pair stores the
    // original element (arr[i]) and its corresponding index (index[i])
    const a = arr.map((value, i) => [value, index[i]]);
 
    // Sort the array of pairs based on the second element (index)
    a.sort(comp);
 
    // Copy the sorted elements back to the original arr[]
    for (let i = 0; i < arr.length; i++) {
        arr[i] = a[i][0];
    }
}
 
// Driver program
const arr = [50, 40, 70, 60, 90];
const index = [3, 0, 4, 1, 2];
 
// Call the reorder function to rearrange elements in arr[] based on index[]
reorder(arr, index);
 
console.log("Reordered array is:");
console.log(arr.join(" "));


Output

Reordered array is: 
40 60 90 50 70 






Time Complexity: O(n log n)
Auxiliary Space: O(n)

Reorder an array according to given indexes using Auxiliary Array:

The idea is to use an auxiliary array temp[] of same size as given arrays. Traverse the given array and put all elements at their correct place in temp[] using index[].

Follow the steps to solve the problem:

  • Create an auxiliary array temp[] to store the recorded array.
  • Iterate through the given array arr[] and put all elements at their correct position in temp[] array using the index array.
  • Copy the temp[] array to arr[] array.

Below is the implementation of above approach:

C++




// C++ program to sort an array according to given
// indexes
#include<iostream>
 
using namespace std;
 
// Function to reorder elements of arr[] according
// to index[]
void reorder(int arr[], int index[], int n)
{
    int temp[n];
 
    // arr[i] should be present at index[i] index
    for (int i=0; i<n; i++)
        temp[index[i]] = arr[i];
 
    // Copy temp[] to arr[]
    for (int i=0; i<n; i++)
    {
       arr[i]   = temp[i];
       index[i] = i;
    }
}
 
// Driver program
int main()
{
    int arr[] = {50, 40, 70, 60, 90};
    int index[] = {3,  0,  4,  1,  2};
    int n = sizeof(arr)/sizeof(arr[0]);
 
    reorder(arr, index, n);
 
    cout << "Reordered array is: \n";
    for (int i=0; i<n; i++)
        cout << arr[i] << " ";
 
}


Java




//Java to find positions of zeroes flipping which
// produces maximum number of consecutive 1's
 
import java.util.Arrays;
 
class Test
{
    static int arr[] = new int[]{50, 40, 70, 60, 90};
    static int index[] = new int[]{30412};
     
    // Method to reorder elements of arr[] according
    // to index[]
    static void reorder()
    {
        int temp[] = new int[arr.length];
      
        // arr[i] should be present at index[i] index
        for (int i=0; i<arr.length; i++)
            temp[index[i]] = arr[i];
      
        // Copy temp[] to arr[]
        for (int i=0; i<arr.length; i++)
        {
           arr[i]   = temp[i];
           index[i] = i;
        }
    }
     
    // Driver method to test the above function
    public static void main(String[] args)
    {
         
        reorder();
         
        System.out.println("Reordered array is: ");
        System.out.println(Arrays.toString(arr));
        System.out.println("Modified Index array is:");
        System.out.println(Arrays.toString(index));
         
    }
}


Python3




# Python3 program to sort
# an array according to given
# indexes
 
# Function to reorder
# elements of arr[] according
# to index[]
def reorder(arr,index, n):
 
    temp = [0] * n;
 
    # arr[i] should be
        # present at index[i] index
    for i in range(0,n):
        temp[index[i]] = arr[i]
 
    # Copy temp[] to arr[]
    for i in range(0,n):
        arr[i] = temp[i]
        index[i] = i
     
# Driver program
arr = [50, 40, 70, 60, 90]
index = [3, 0, 4, 1, 2]
n = len(arr)
 
reorder(arr, index, n)
 
print("Reordered array is:")
for i in range(0,n):
    print(arr[i],end = " ")
 
print("\nModified Index array is:")
for i in range(0,n):
    print(index[i],end = " ")
 
# This code is contributed by
# Smitha Dinesh Semwal


C#




// C# to find positions of zeroes flipping which
// produces maximum number of consecutive 1's
using System;
  
public class Test{
     
    static int []arr = new int[]{50, 40, 70, 60, 90};
    static int []index = new int[]{3,  0,  4,  1,  2};
      
    // Method to reorder elements of arr[] according
    // to index[]
    static void reorder()
    {
        int []temp = new int[arr.Length];
       
        // arr[i] should be present at index[i] index
        for (int i=0; i<arr.Length; i++)
            temp[index[i]] = arr[i];
       
        // Copy temp[] to arr[]
        for (int i=0; i<arr.Length; i++)
        {
           arr[i]   = temp[i];
           index[i] = i;
        }
    }
      
    // Driver method to test the above function
    public static void Main()
    {
          
        reorder();
          
        Console.WriteLine("Reordered array is: ");
        Console.WriteLine(string.Join(",", arr));
        Console.WriteLine("Modified Index array is:");
        Console.WriteLine(string.Join(",", index));
          
    }
}
/*This code is contributed by 29AjayKumar*/


Javascript




<script>
      // JavaScript program to sort an array according to given
      // indexes
 
      // Function to reorder elements of arr[] according
      // to index[]
      function reorder(arr, index, n) {
        var temp = [...Array(n)];
 
        // arr[i] should be present at index[i] index
        for (var i = 0; i < n; i++) temp[index[i]] = arr[i];
 
        // Copy temp[] to arr[]
        for (var i = 0; i < n; i++) {
          arr[i] = temp[i];
          index[i] = i;
        }
      }
 
      // Driver program
 
      var arr = [50, 40, 70, 60, 90];
      var index = [3, 0, 4, 1, 2];
      var n = arr.length;
 
      reorder(arr, index, n);
 
      document.write("Reordered array is: ");
      document.write("<br>");
      for (var i = 0; i < n; i++) document.write(arr[i] + " ");
      document.write("<br>");
      document.write("Modified Index array is: ");
      document.write("<br>");
      for (var i = 0; i < n; i++) document.write(index[i] + " ");
       
      // This code is contributed by rdtank.
    </script>


PHP




<?php
// PHP program to sort an array
// according to given indexes
 
// Function to reorder elements
// of arr[] according to index[]
function reorder($arr, $index, $n)
{
    // $temp[$n];
 
    // arr[i] should be present
    // at index[i] index
    for ($i = 0; $i < $n; $i++)
    {
        $temp[$index[$i]] = $arr[$i];
    }
     
     
    // Copy temp[] to arr[]
    for ($i = 0; $i < $n; $i++)
    {
        $arr[$i] = $temp[$i];
        $index[$i] = $i;
    }
     
    echo "Reordered array is: \n";
for ($i = 0; $i < $n; $i++)
{
    echo $arr[$i] . " ";
}
 
echo "\nModified Index array is: \n";
for ($i = 0; $i < $n; $i++)
{
    echo $index[$i] . " ";
}
}
 
// Driver Code
$arr = array(50, 40, 70, 60, 90);
$index = array(3, 0, 4, 1, 2);
$n = sizeof($arr);
 
reorder($arr, $index, $n);
 
// This code is contributed
// by Abby_akku
?>


Output

Reordered array is: 
40 60 90 50 70 






Time Complexity: O(n) 
Auxiliary Space: O(n)

Reorder an array according to given indexes using Cyclic Sort:

The idea is use cyclic sort technique to reorder elements in the arr[] array based on the specified index[]. It iterates through the elements of arr[] and, for each element, continuously swaps it with the element at its correct position according to index[]. The process continues until each element is at its intended position, ensuring the desired order is achieved.

Follow the steps to solve the problem:

  • Iterate through the array and do following for every element arr[i]:
    • While the current element’s index is not equal to its position (i.e., index[i] != i), perform the following steps:
      • Store the value and index of the target (correct) position before placing the current element there.
      • Place the current element at its target position (arr[index[i]] and index[index[i]]), effectively swapping elements.
      • Update the current element’s index and value with the stored target values.
    • Continue this process for all elements in the array until each element is at its intended position according to the index[] array.
  • The arr[] array will be reordered according to the specified indices in index[] after the cyclic sort is complete.

Below is the implementation of the above algorithm. 

C++




// sort an array according to given indexes
#include<iostream>
 
using namespace std;
 
// Function to reorder elements of arr[] according
// to index[]
void reorder(int arr[], int index[], int n)
{
    // Fix all elements one by one
    for (int i=0; i<n; i++)
    {
        // While index[i] and arr[i] are not fixed
        while (index[i] != i)
        {
            // Store values of the target (or correct)
            // position before placing arr[i] there
            int  oldTargetI  = index[index[i]];
            char oldTargetE  = arr[index[i]];
 
            // Place arr[i] at its target (or correct)
            // position. Also copy corrected index for
            // new position
            arr[index[i]] = arr[i];
            index[index[i]] = index[i];
 
            // Copy old target values to arr[i] and
            // index[i]
            index[i] = oldTargetI;
            arr[i]   = oldTargetE;
        }
    }
}
 
// Driver program
int main()
{
    int arr[] = {50, 40, 70, 60, 90};
    int index[] = {3,  0,  4,  1,  2};
    int n = sizeof(arr)/sizeof(arr[0]);
 
    reorder(arr, index, n);
 
    cout << "Reordered array is: \n";
    for (int i=0; i<n; i++)
        cout << arr[i] << " ";
 
    
    return 0;
}


Java




//A O(n) time and O(1) extra space Java program to
//sort an array according to given indexes
 
import java.util.Arrays;
 
class Test
{
    static int arr[] = new int[]{50, 40, 70, 60, 90};
    static int index[] = new int[]{30412};
     
    // Method to reorder elements of arr[] according
    // to index[]
    static void reorder()
    {
        // Fix all elements one by one
        for (int i=0; i<arr.length; i++)
        {
            // While index[i] and arr[i] are not fixed
            while (index[i] != i)
            {
                // Store values of the target (or correct)
                // position before placing arr[i] there
                int  oldTargetI  = index[index[i]];
                char oldTargetE  = (char)arr[index[i]];
      
                // Place arr[i] at its target (or correct)
                // position. Also copy corrected index for
                // new position
                arr[index[i]] = arr[i];
                index[index[i]] = index[i];
      
                // Copy old target values to arr[i] and
                // index[i]
                index[i] = oldTargetI;
                arr[i]   = oldTargetE;
            }
        }
    }
     
    // Driver method to test the above function
    public static void main(String[] args)
    {
         
        reorder();
         
        System.out.println("Reordered array is: ");
        System.out.println(Arrays.toString(arr));
        System.out.println("Modified Index array is:");
        System.out.println(Arrays.toString(index));
         
    }
}


Python3




# A O(n) time and O(1) extra space Python3 program to
# sort an array according to given indexes
 
# Function to reorder elements of arr[] according
# to index[]
def reorder(arr, index, n):
 
    # Fix all elements one by one
    for i in range(0,n):
 
           #  While index[i] and arr[i] are not fixed
           while (index[i] != i):
         
               # Store values of the target (or correct)
               # position before placing arr[i] there
               oldTargetI = index[index[i]]
               oldTargetE = arr[index[i]]
 
               # Place arr[i] at its target (or correct)
               # position. Also copy corrected index for
               # new position
               arr[index[i]] = arr[i]
               index[index[i]] = index[i]
 
               # Copy old target values to arr[i] and
               # index[i]
               index[i] = oldTargetI
               arr[i] = oldTargetE
 
 
# Driver program
arr = [50, 40, 70, 60, 90]
index= [3, 0, 4, 1, 2]
n = len(arr)
 
reorder(arr, index, n)
 
print("Reordered array is:")
for  i in range(0, n):
    print(arr[i],end=" ")
 
print("\nModified Index array is:")
for i in range(0, n):
    print(index[i] ,end=" ")
 
# This code is contributed by
# Smitha Dinesh Semwal


C#




//A O(n) time and O(1) extra space C# program to
//sort an array according to given indexes
using System;
 
public class Test
{
    static int []arr = new int[]{50, 40, 70, 60, 90};
    static int []index = new int[]{3,  0,  4,  1,  2};
      
    // Method to reorder elements of arr[] according
    // to index[]
    static void reorder()
    {
        // Fix all elements one by one
        for (int i=0; i<arr.Length; i++)
        {
            // While index[i] and arr[i] are not fixed
            while (index[i] != i)
            {
                // Store values of the target (or correct)
                // position before placing arr[i] there
                int  oldTargetI  = index[index[i]];
                char oldTargetE  = (char)arr[index[i]];
       
                // Place arr[i] at its target (or correct)
                // position. Also copy corrected index for
                // new position
                arr[index[i]] = arr[i];
                index[index[i]] = index[i];
       
                // Copy old target values to arr[i] and
                // index[i]
                index[i] = oldTargetI;
                arr[i]   = oldTargetE;
            }
        }
    }
      
    // Driver method to test the above function
    public static void Main()
    {
          
        reorder();
          
        Console.WriteLine("Reordered array is: ");
        Console.WriteLine(String.Join(" ",arr));
        Console.WriteLine("Modified Index array is:");
        Console.WriteLine(String.Join(" ",index));
          
    }
}
 
// This code is contributed by PrinciRaj1992


Javascript




<script>
// A O(n) time and O(1) extra space JavaScript program to
// sort an array according to given indexes
 
// Function to reorder elements of arr[] according
// to index[]
function reorder(arr, index, n)
{
    // Fix all elements one by one
    for (let i=0; i<n; i++)
    {
        // While index[i] and arr[i] are not fixed
        while (index[i] != i)
        {
            // Store values of the target (or correct)
            // position before placing arr[i] there
            let oldTargetI = index[index[i]];
            let oldTargetE = arr[index[i]];
 
            // Place arr[i] at its target (or correct)
            // position. Also copy corrected index for
            // new position
            arr[index[i]] = arr[i];
            index[index[i]] = index[i];
 
            // Copy old target values to arr[i] and
            // index[i]
            index[i] = oldTargetI;
            arr[i] = oldTargetE;
        }
    }
}
 
// Driver program
    let arr = [50, 40, 70, 60, 90];
    let index = [3, 0, 4, 1, 2];
    let n = arr.length;
 
    reorder(arr, index, n);
 
    document.write("Reordered array is: <br>");
    for (let i=0; i<n; i++)
        document.write(arr[i] + " ");
 
    document.write("<br>Modified Index array is: <br>");
    for (let i=0; i<n; i++)
        document.write(index[i] + " ");
 
 
 
 
// This code is contributed by Surbhi Tyagi.
</script>


Output

Reordered array is: 
40 60 90 50 70 






Time Complexity: O(n) 
Auxiliary Space: O(1)

Reorder an array according to given indexes using Mathematics Approach:

Follow the steps to solve the problem by the above approach:

  • Calculate the max(array, n) and value = max_value + 1; This is needed to identify the upper range of values present in the array as the modulus operation will always return a value from 0 to N-1  (used in next step for storing two elements together)
  • For each element, place the element at index=i at it’s desired position at index=j such that it’s possible to retrieve both elements when needed. The following formula has been used to store the array[i] at array[j]
    array[index[i]] = (array[index[i]] + array[i]%value*value)
  • Once the elements are placed at each position, traverse the array once and update each element by element value.

Below is the implementation of the above algorithm. 

C++




// C++ code for the above approach
#include <climits>
#include <iostream>
using namespace std;
 
 
int findMax(int arr[], int n)
{
    int Max = INT_MIN;
    for (int i = 0; i < n; i++) {
        if (Max < arr[i])
            Max = arr[i];
    }
    return Max;
}
 
// result = (original + update%z*z)  ..
void rearrange(int arr[], int index[], int n)
{
    int z = findMax(arr, n) + 1;
    for (int i = 0; i < n; i++) {
        arr[index[i]] = arr[index[i]] % z + arr[i] % z * z;
    }
    for (int i = 0; i < n; i++) {
        arr[i] = arr[i] / z;
    }
}
 
int main()
{
    int arr[] = { 23, 12, 20, 10, 23 };
    int index[] = { 4, 0, 1, 2, 3 };
 
    rearrange(arr, index, 5);
    cout << "Reordered array is: \n";
    for (int i = 0; i < 5; i++)
        cout << arr[i] << " ";
 
    return 0;
}
 
// This code is contributed by lokesh


Java




/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
    public static void main(String[] args)
    {
        int arr[] = { 23, 12, 20, 10, 23 };
        int index[] = { 4, 0, 1, 2, 3 };
 
        rearrange(arr, index);
        printArray(arr);
    }
    private static void printArray(int arr[])
    {
        for (int i = 0; i < arr.length; i++)
            System.out.print(arr[i] + " ");
    }
    //    result = (original + update%z*z)  ..
    private static void rearrange(int[] arr, int[] index)
    {
        int z = findMax(arr) + 1;
        for (int i = 0; i < arr.length; i++) {
            arr[index[i]]
                = arr[index[i]] % z + arr[i] % z * z;
        }
        for (int i = 0; i < arr.length; i++) {
            arr[i] = arr[i] / z;
        }
    }
 
    private static int findMax(int[] arr)
    {
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < arr.length; i++) {
            if (max < arr[i])
                max = arr[i];
        }
        return max;
    }
}


Python3




# Python implementation for the above approach
def printArray(arr):
    for i in range(len(arr)):
        print(arr[i], end=" ")
 
# result = (original + update%z*z) ..
def rearrange(arr, index):
    z = findMax(arr) + 1
    for i in range(len(arr)):
        arr[index[i]] = arr[index[i]] % z + arr[i] % z * z
 
    for i in range(len(arr)):
        arr[i] = arr[i]//z
 
 
def findMax(arr):
    Max = float("-inf")
    for i in range(len(arr)):
        if(Max < arr[i]):
            Max = arr[i]
 
    return Max
 
arr = [23, 12, 20, 10, 23]
index = [4, 0, 1, 2, 3]
 
rearrange(arr, index)
printArray(arr)
 
# This code is contributed by lokesh


C#




// C# implementation for the above approach
using System;
 
public class GFG {
 
    static public void Main()
    {
 
        // Code
        int[] arr = { 23, 12, 20, 10, 23 };
        int[] index = { 4, 0, 1, 2, 3 };
 
        rearrange(arr, index);
        printArray(arr);
    }
 
    private static void printArray(int[] arr)
    {
        for (int i = 0; i < arr.Length; i++)
            Console.Write(arr[i] + " ");
    }
    //    result = (original + update%z*z)  ..
    private static void rearrange(int[] arr, int[] index)
    {
        int z = findMax(arr) + 1;
        for (int i = 0; i < arr.Length; i++) {
            arr[index[i]]
                = arr[index[i]] % z + arr[i] % z * z;
        }
        for (int i = 0; i < arr.Length; i++) {
            arr[i] = arr[i] / z;
        }
    }
 
    private static int findMax(int[] arr)
    {
        int max = Int32.MinValue;
        for (int i = 0; i < arr.Length; i++) {
            if (max < arr[i])
                max = arr[i];
        }
        return max;
    }
}
 
// This code is contributed by lokeshmvs21.


Javascript




// JavaScript implementation for the above approach
 
function printArray(){
    for(let i = 0; i < arr.length; i++){
        console.log(Math.trunc(arr[i]) + " ");
    }
}
 
//    result = (original + update%z*z)  ..
function rearrange(arr, index){
    var z = findMax(arr) + 1;
    for(let i = 0; i < arr.length; i++){
        arr[index[i]] = arr[index[i]] % z + arr[i] % z * z;
    }
    for(let i = 0; i < arr.length; i++){
        arr[i] = arr[i]/z;
    }
}
 
function findMax(arr){
    let max = Number.MIN_VALUE;
    for(let i = 0; i < arr.length; i++){
        if(max < arr[i]){
            max = arr[i];
        }
    }
    return max;
}
 
let arr = [ 23, 12, 20, 10, 23 ];
let index = [ 4, 0, 1, 2, 3 ];
 
rearrange(arr, index);
printArray(arr);
 
// This code is contributed by lokeshmvs21.


Output

12 20 10 23 23 






Time Complexity: O(N)
Auxiliary Space: O(1)



Last Updated : 07 Dec, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads