Rearrange an array such that ‘arr[j]’ becomes ‘i’ if ‘arr[i]’ is ‘j’ | Set 1

Given an array of size n where all elements are distinct and in the range from 0 to n-1, change the contents of arr[] so that arr[i] = j is changed to arr[j] = i. 

Examples: 

Example 1:
Input: arr[]  = {1, 3, 0, 2};
Output: arr[] = {2, 0, 3, 1};
Explanation for the above output.
Since arr[0] is 1, arr[1] is changed to 0
Since arr[1] is 3, arr[3] is changed to 1
Since arr[2] is 0, arr[0] is changed to 2
Since arr[3] is 2, arr[2] is changed to 3

Example 2:
Input: arr[]  = {2, 0, 1, 4, 5, 3};
Output: arr[] = {1, 2, 0, 5, 3, 4};

Example 3:
Input: arr[]  = {0, 1, 2, 3};
Output: arr[] = {0, 1, 2, 3};

Example 4:
Input: arr[]  = {3, 2, 1, 0};
Output: arr[] = {3, 2, 1, 0};

A Simple Solution is to create a temporary array and one by one copy ‘i’ to ‘temp[arr[i]]’ where i varies from 0 to n-1. 

Below is the implementation of the above idea. 

Given array is 
1302
Modified array is 
2031

Time complexity: O(n) 
Auxiliary space :O(n)

Can we solve this in O(n) time and O(1) auxiliary space? 
The idea is based on the fact that the modified array is basically a permutation of the input array. We can find the target permutation by storing the next item before updating it. 
Let us consider array ‘{1, 3, 0, 2}’ for example. We start with i = 0, arr[i] is 1. So we go to arr[1] and change it to 0 (because i is 0). Before we make the change, we store the old value of arr[1] as the old value is going to be our new index i. In the next iteration, we have i = 3, arr[3] is 2, so we change arr[2] to 3. Before making the change we store next i as old value of arr[2]. 

The below code gives idea about this approach. 

// This function works only when output is a permutation
// with one cycle.
void rearrangeUtil(int arr[], int n)
{
    // 'val' is the value to be stored at 'arr[i]'
    int val = 0;   // The next value is determined
                  // using current index
    int i = arr[0];  // The next index is determined
                     // using current value

    // While all elements in cycle are not processed
    while (i != 0)
    {
        // Store value at index as it is going to be
        // used as next index
        int new_i = arr[i];

        // Update arr[]
        arr[i] = val;

        // Update value and index for next iteration
        val = i;
        i = new_i;
    }

    arr[0] = val;  // Update the value at arr[0]
}

The above function doesn’t work for inputs like {2, 0, 1, 4, 5, 3}; as there are two cycles. One cycle is (2, 0, 1) and other cycle is (4, 5, 3). 
How to handle multiple cycles with the O(1) space constraint? 
The idea is to process all cycles one by one. To check whether an element is processed or not, we change the value of processed items arr[i] as -arr[i]. Since 0 can not be made negative, we first change all arr[i] to arr[i] + 1. In the end, we make all values positive and subtract 1 to get old values back. 

Given array is 
2 0 1 4 5 3 
Modified array is 
1 2 0 5 3 4 

The time complexity of this method seems to be more than O(n) at first look. If we take a closer look, we can notice that no element is processed more than a constant number of times.

Another Method: The idea is to store each element’s new and old value as quotient and remainder of n, respectively (n being the size of the array). 
For example, Suppose an element’s new value is 2, the old value is 1 and n is 3, then the element’s value is stored as 1 + 2*3 = 7. We can retrieve its old value by 7%3 = 1 and its new value by 7/3 = 2. 
Thanks, Prateek Oraon for suggesting this method. 

Given array is : 
2 0 1 4 5 3 
Modified array is :
1 2 0 5 3 4 

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

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