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
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; } |
Java
// Java program to implement // the approach class GFG{ // Function to find the right // shifts required for each // element to reach its sorted // array position in A[] public static void findShifts(int[] A, int N) { // Stores required number of // shifts for each element int[] shift = new int[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++) System.out.print(shift[i] + " "); } // Driver code public static void main(String[] args) { int arr[] = { 1, 4, 3, 2, 5 }; int N = arr.length; findShifts(arr, N); } } // This code is contributed by divyeshrabadiya07 |
Python3
# Python3 Program to implement # the approach # Function to find the right # shifts required for each # element to reach its sorted # array position in A[] def findShifts(A, N): # Stores required number of # shifts for each element shift = [0 for i in range(N)] for i in range(N): # 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 i in range(N): print(shift[i], end = " ") # Driver Code if __name__ == '__main__': arr = [ 1, 4, 3, 2, 5 ] N = len(arr) findShifts(arr, N) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement // the approach using System; class GFG{ // Function to find the right // shifts required for each // element to reach its sorted // array position in []A public static void findShifts(int[] A, int N) { // Stores required number of // shifts for each element int[] shift = new int[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++) Console.Write(shift[i] + " "); } // Driver code public static void Main(String[] args) { int []arr = { 1, 4, 3, 2, 5 }; int N = arr.Length; findShifts(arr, N); } } // This code is contributed by amal kumar choubey |
0 2 0 3 0
Time Complexity: O(N)
Auxiliary Space: O(N)
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