Given an array **arr[] **of size **N **that contains elements from the range [1, N], the task is to calculate the number of right shifts required for each element to reach its respective position if the array was sorted.
Examples:
_Input: __arr[] = {1, 4, 3, 2, 5}, N = 5 _
_Output: __0 2 0 3 0 _
Explanation:
_The sorted array is {1, 2, 3, 4, 5}. _
_4 is at index 1, so right shift 2 times to reach index 3. (1->2->3) _
_2 is at index 3, so right shift 3 times to reach index 1. (3->4->0->1) _
All the other elements are at their respective positions in the sorted array.
_Input: __arr[]={2, 4, 3, 1, 5}, N = 5 _
_Output: __2 1 0 2 0 _
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
**Approach: **
The idea is to calculate the difference between the actual position and the sorted position of each element of the array. As the elements are from 1 to**N, **the sorted position of each element can be determined without sorting the array. The sorted position of each element is given by **(arr[i]-1). **Therefore, the number of right shifts is given by **(arr[i] – 1 – i + N) % N.**
Below is the implementation of the above approach:
- C++
// C++ Program to implement
// the approach
#include <bits/stdc++.h>
**using** **namespace** std;
// Function to find the right
// shifts required for each
// element to reach its sorted
// array position in A[]
**void** findShifts(``**int** A[], **int** N)
{
// Stores required number of
// shifts for each element
**int** shift[N];
**for** (``**int** i = 0; i < N; i++) {
// If the element is
// at sorted position
**if** (i == A[i] - 1)
shift[i] = 0;
// Otherwise
**else**
// Calculate right shift
shift[i]
= (A[i] - 1 - i + N)
% N;
}
// Print the respective shifts
**for** (``**int** i = 0; i < N; i++)
cout << shift[i] << " "``;
}
// Driver Code
**int** main()
{
**int** arr[] = { 1, 4, 3, 2, 5 };
**int** N = **sizeof**``(arr) / **sizeof**``(arr[0]);
findShifts(arr, N);
**return** 0;
}
Output:
0 2 0 3 0
_Time Complexity: __O(N) _
_Auxiliary Space: _O(N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
#arrays #greedy #mathematical #array-rearrange
