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Rearrange an array so that arr[i] becomes arr[arr[i]] with O(1) extra space using C++
We are given a positive integer type array, let's say, arr[] of any given size such that elements in an array should value greater than 0 but less than the size of an array. The task is to rearrange an array in such a manner that arr[i] becomes arr[arr[i]] within the given O(1) space only and print the final result.
Let us see various input output scenarios for this −
Input − int arr[] = {0 3 2 1 5 4 }
Output − Array before Arrangement: 0 3 2 1 5 4 Rearrangement of an array so that arr[i] becomes arr[arr[i]] with O(1) extra space is: 0 1 2 3 4 5
Explanation − we are given an integer array of size 6 and all the elements in an array value less than 6. Now, we will rearrange the array i.e. arr[arr[0] is 0, arr[arr[1]] is 1, arr[arr[2]] is 2, arr[arr[3]] is 3, arr[arr[4]] is 4 and arr[arr[5]] is 5. So, the final array after rearrangement is 0 1 2 3 4 5.
Input − int arr[] = {1, 0}
Output − Array before Arrangement: 1 0 Rearrangement of an array so that arr[i] becomes arr[arr[i]] with O(1) extra space is: 0 1
Explanation − we are given an integer array of size 2 and all the elements in an array value less than 2. Now, we will rearrange the array i.e. arr[arr[0] is 1 and arr[arr[1]] is 0. So, the final array after rearrangement is 0 1.
Input − int arr[] = {1, 0, 2, 3}
Output − Array before Arrangement: 1 0 2 3 Rearrangement of an array so that arr[i] becomes arr[arr[i]] with O(1) extra space is: 0 1 2 3
Explanation − we are given an integer array of size 4 and all the elements in an array value less than 4. Now, we will rearrange the array i.e. arr[arr[0] is 0, arr[arr[1]] is 1, arr[arr[2]] is 2 and arr[arr[3]] is 3. So, the final array after rearrangement is 0 1 2 3.
Approach used in the below program is as follows
Input an array of integer type elements and calculate the size of an array
Print the array before arrangement and call the function Rearrangement(arr, size)
Inside the function Rearrangement(arr, size)
Start loop FOR from i to 0 till i less than size. Inside the loop, set temp as arr[arr[i]] % size and arr[i] += temp * size.
Start loop FOR from i to 0 till i less than size. Inside the loop, set arr[i] = arr[i] / size
Print the result.
Example
#include <bits/stdc++.h>
using namespace std;
void Rearrangement(int arr[], int size){
for(int i=0; i < size; i++){
int temp = arr[arr[i]] % size;
arr[i] += temp * size;
}
for(int i = 0; i < size; i++){
arr[i] = arr[i] / size;
}
}
int main(){
//input an array
int arr[] = {0, 3, 2, 1, 5, 4};
int size = sizeof(arr) / sizeof(arr[0]);
//print the original Array
cout<<"Array before Arrangement: ";
for (int i = 0; i < size; i++){
cout << arr[i] << " ";
}
//calling the function to rearrange the array
Rearrangement(arr, size);
//print the array after rearranging the values
cout<<"\nRearrangement of an array so that arr[i] becomes arr[arr[i]] with O(1) extra space is: ";
for(int i = 0; i < size; i++){
cout<< arr[i] << " ";
}
return 0;
}Output
If we run the above code it will generate the following Output
Array before Arrangement: 0 3 2 1 5 4 Rearrangement of an array so that arr[i] becomes arr[arr[i]] with O(1) extra space is: 0 1 2 3 4 5
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