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Rearrange an array so that arr[i] becomes arr[arr[i]] with O(1) extra space

  • Difficulty Level : Hard
  • Last Updated : 07 Jul, 2021

Given an array arr[] of size n where every element is in range from 0 to n-1. Rearrange the given array so that arr[i] becomes arr[arr[i]]. This should be done with O(1) extra space.

Examples: 

Input: arr[]  = {3, 2, 0, 1}
Output: arr[] = {1, 0, 3, 2}
Explanation: 
In the given array 
arr[arr[0]] is 1 so arr[0] in output array is 1
arr[arr[1]] is 0 so arr[1] in output array is 0
arr[arr[2]] is 3 so arr[2] in output array is 3
arr[arr[3]] is 2 so arr[3] in output array is 2

Input: arr[] = {4, 0, 2, 1, 3}
Output: arr[] = {3, 4, 2, 0, 1}
Explanation:
arr[arr[0]] is 3 so arr[0] in output array is 3
arr[arr[1]] is 4 so arr[1] in output array is 4
arr[arr[2]] is 2 so arr[2] in output array is 2
arr[arr[3]] is 0 so arr[3] in output array is 0
arr[arr[4]] is 1 so arr[4] in output array is 1

Input: arr[] = {0, 1, 2, 3}
Output: arr[] = {0, 1, 2, 3}
Explanation:
arr[arr[0]] is 0 so arr[0] in output array is 0
arr[arr[1]] is 1 so arr[1] in output array is 1
arr[arr[2]] is 2 so arr[2] in output array is 2
arr[arr[3]] is 3 so arr[3] in output array is 3

If the extra space condition is removed, the question becomes very easy. The main part of the question is to do it without extra space.  

We strongly recommend that you click here and practice it, before moving on to the solution.

The credit for following solution goes to Ganesh Ram Sundaram

Approach: The array elements of the given array lies from 0 to n-1. Now an array element is needed that can store two different values at the same time. To achieve this increment every element at ith index is incremented by (arr[arr[i]] % n)*n.After the increment operation of first step, every element holds both old values and new values. Old value can be obtained by arr[i]%n and a new value can be obtained by arr[i]/n.



How this can be achieved? 
Let’s assume an element is a and another element is b, both the elements are less than n. So if an element a is incremented by b*n. So the element becomes a + b*n so when a + b*n is divided by n then the value is b and a + b*n % n is a.

Algorithm:  

  1. Traverse the array from start to end.
  2. For every index increment the element by array[array[index] ] % n. To get the ith element find the modulo with n, i.e array[index]%n.
  3. Again Travsese the array from start to end
  4. Print the ith element after dividing the ith element by n, i.e. array[i]/n.

Implementation: 
 

C++




#include <iostream>
using namespace std;
 
// The function to rearrange an array
// in-place so that arr[i] becomes arr[arr[i]].
void rearrange(int arr[], int n)
{
    // First step: Increase all values by (arr[arr[i]]%n)*n
    for (int i=0; i < n; i++)
        arr[i] += (arr[arr[i]]%n)*n;
 
    // Second Step: Divide all values by n
    for (int i=0; i<n; i++)
        arr[i] /= n;
}
 
// A utility function to print an array of size n
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    cout << endl;
}
 
 
/* Driver program to test above functions*/
int main()
{
    int arr[] = {3, 2, 0, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
 
    cout << "Given array is \n";
    printArr(arr, n);
 
    rearrange(arr, n);
 
    cout << "Modified array is \n";
    printArr(arr, n);
    return 0;
}

Java




class Rearrange
{
    // The function to rearrange an array in-place so that arr[i]
    // becomes arr[arr[i]].
    void rearrange(int arr[], int n)
    {
        // First step: Increase all values by (arr[arr[i]]%n)*n
        for (int i = 0; i < n; i++)
            arr[i] += (arr[arr[i]] % n) * n;
 
        // Second Step: Divide all values by n
        for (int i = 0; i < n; i++)
            arr[i] /= n;
    }
 
    // A utility function to print an array of size n
    void printArr(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
        System.out.println("");
    }
 
    /* Driver program to test above functions */
    public static void main(String[] args)
    {
        Rearrange rearrange = new Rearrange();
        int arr[] = {3, 2, 0, 1};
        int n = arr.length;
 
        System.out.println("Given Array is :");
        rearrange.printArr(arr, n);
 
        rearrange.rearrange(arr, n);
 
        System.out.println("Modified Array is :");
        rearrange.printArr(arr, n);
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python3




# Python3 program to Rearrange
# an array so that arr[i] becomes
# arr[arr[i]]
 
# The function to rearrange an
# array in-place so that arr[i]
# becomes arr[arr[i]].
def rearrange(arr, n):
 
    # First step: Increase all values
    # by (arr[arr[i]] % n) * n
    for i in range(0, n):
        arr[i] += (arr[arr[i]] % n) * n
 
    # Second Step: Divide all values
    # by n
    for i in range(0, n):
        arr[i] = int(arr[i] / n)
 
# A utility function to print
# an array of size n
def printArr(arr, n):
 
    for i in range(0, n):
        print (arr[i], end =" ")
    print ("")
 
# Driver program
arr = [3, 2, 0, 1]
n = len(arr)
 
print ("Given array is")
printArr(arr, n)
 
rearrange(arr, n);
print ("Modified array is")
printArr(arr, n)
 
# This code is contributed by shreyanshi_arun

C#




// C# Program to rearrange an array
// so that arr[i] becomes arr[arr[i]]
// with O(1) extra space
using System;
 
class Rearrange
{
     
    // Function to rearrange an
    // array in-place so that arr[i]
    // becomes arr[arr[i]].
    void rearrange(int []arr, int n)
    {
         
        // First step: Increase all values
        // by (arr[arr[i]] % n) * n
        for (int i = 0; i < n; i++)
            arr[i] += (arr[arr[i]] % n) * n;
 
        // Second Step: Divide all values by n
        for (int i = 0; i < n; i++)
            arr[i] /= n;
    }
 
    // A utility function to
    // print an array of size n
    void printArr(int []arr, int n)
    {
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine("");
    }
 
    //  Driver Code
    public static void Main()
    {
        Rearrange rearrange = new Rearrange();
        int []arr = {3, 2, 0, 1};
        int n = arr.Length;
 
        Console.Write("Given Array is :");
        rearrange.printArr(arr, n);
 
        rearrange.rearrange(arr, n);
 
        Console.Write("Modified Array is :");
        rearrange.printArr(arr, n);
    }
}
 
// This code has been contributed by Nitin Mittal.

PHP




<?php
// The function to rearrange an array
// in-place so that arr[i] becomes arr[arr[i]].
function rearrange(&$arr, $n)
{
    // First step: Increase all values
    // by (arr[arr[i]]%n)*n
    for ($i = 0; $i < $n; $i++)
        $arr[$i] += ($arr[$arr[$i]] % $n) * $n;
 
    // Second Step: Divide all values by n
    for ($i = 0; $i < $n; $i++)
        $arr[$i] = intval($arr[$i] / $n);
}
 
// A utility function to print
// an array of size n
function printArr(&$arr, $n)
{
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i] ." ";
    echo "\n";
}
 
 
// Driver Code
$arr = array(3, 2, 0, 1);
$n = sizeof($arr);
 
echo "Given array is \n";
printArr($arr, $n);
 
rearrange($arr, $n);
 
echo "Modified array is \n";
printArr($arr, $n);
 
// This code is contributed
// by ChitraNayal
?>

Javascript




<script>
// The function to rearrange an array
// in-place so that arr[i] becomes arr[arr[i]].
function rearrange(arr, n)
{
 
    // First step: Increase all values by (arr[arr[i]]%n)*n
    for (let i = 0; i < n; i++)
        arr[i] += (arr[arr[i]] % n) * n;
 
    // Second Step: Divide all values by n
    for (let i = 0; i < n; i++)
        arr[i] = Math.floor(arr[i] / n);
}
 
// A utility function to print an array of size n
function printArr(arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
    document.write("<br>");
}
 
/* Driver program to test above functions*/
    let arr = [3, 2, 0, 1];
    let n = arr.length;
 
    document.write("Given array is <br>");
    printArr(arr, n);
    rearrange(arr, n);
 
    document.write("Modified array is <br>");
    printArr(arr, n);
 
// This code is contributed by Surbhi Tyagi.
</script>

Output:  

Given array is
3 2 0 1
Modified array is
1 0 3 2

Complexity Analysis:  

  • Time Complexity: O(n), Only one traversal of the array is needed. So time complexity is O(n).
  • Auxiliary Space: O(1), No extra space is required.

The problem with the above solution is, it may cause an overflow.

Here is a better solution: 
Rearrange an array such that ‘arr[j]’ becomes ‘i’ if ‘arr[i]’ is ‘j’

This article is contributed by Himanshu Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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