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

Try it on GfG Practice

Given an array arr[] of size n where every element is in the 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: 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

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.

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

Note: This Approach may cause an overflow.

Algorithm:

• Traverse the array from start to end.
• 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.
• Again Traverse the array from start to end
• Print the ith element after dividing the ith element by n, i.e. array[i]/n.

#include <iostream>
#include <vector>
using namespace std;

void rearrange(vector<int>arr)
{
    int n=arr.size();
    
    // 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;
}

void printArr(vector<int>arr)
{
     int n=arr.size();
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    cout << endl;
}

int main()
{
    vector<int>arr = { 3, 2, 0, 1 };
    
    rearrange(arr);
    
    printArr(arr);
    return 0;
}

import java.util.*;

class GfG{
    static void rearrange(int[] arr) {
        int n = arr.length;
        
        // 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;
    }

    static void printArr(int[] arr) {
        int n = arr.length;
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    public static void main(String[] args) {
        int[] arr = { 3, 2, 0, 1 };
        rearrange(arr);
        printArr(arr);
    }
}

def rearrange(arr):
    n = len(arr)
    
    # First step: Increase all values 
    # by (arr[arr[i]]%n)*n
    for i in range(n):
        arr[i] += (arr[arr[i]] % n) * n

    # Second Step: Divide all values by n
    for i in range(n):
        arr[i] //= n


def printArr(arr):
    n = len(arr)
    for i in range(n):
        print(arr[i], end=' ')
    print()

if __name__ == '__main__':
    arr = [3, 2, 0, 1]
    rearrange(arr)
    printArr(arr)

using System;
using System.Collections.Generic;

class GfG{
    static void Rearrange(int[] arr) {
        int n = arr.Length;
        
        // 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;
    }

    static void PrintArr(int[] arr) {
        int n = arr.Length;
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine();
    }

    static void Main() {
        int[] arr = { 3, 2, 0, 1 };
        Rearrange(arr);
        PrintArr(arr);
    }
}

function rearrange(arr) {
    const n = arr.length;
    
    // 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);
}

function printArr(arr) {
    const n = arr.length;
    for (let i = 0; i < n; i++)
        process.stdout.write(arr[i] + ' ');
    console.log();
}

const arr = [3, 2, 0, 1];
rearrange(arr);
printArr(arr);

3 2 0 1 

Related Article: 
Rearrange an array such that ‘arr[j]’ becomes ‘i’ if ‘arr[i]’ is ‘j’

kartik

Follow

Improve

Article Tags :

• Mathematical
• DSA
• Arrays

Practice Tags :

• Arrays
• Mathematical