Given an array b. The array b is obtained initially by array a, by doing the following operations.

• Choose a fixed point x from array a.
• Cyclically rotate the array a to the left exactly x times.

Fixed point x is that point where (a[i] = i ). These operations are performed on the array a k times, determine if you will be able to restore the array a given array b.

Examples:

Input : n = 5 , k = 3 , b = { 4,3, 3, 2, 3 }
Output: Yes
Explanation: The array 'a' was initially given as [3, 2, 3, 4, 3]. In the first operation, a specific position, denoted as x=2, was selected. After performing 2 left shifts at this position, the array transformed into [3, 4, 3, 3, 2]. For the second operation, a different fixed point x=3 was chosen, and executing 3 left shifts resulted in the array [3, 2, 3, 4, 3]. Subsequently, in the third operation, the same fixed point x=3 was selected, and after 3 left shifts, the array transformed into [4, 3, 3, 2, 3]. Notably, this final array is equivalent to the array 'b'.

Input : n = 5 , k = 5 , b = { 6, 1, 1, 1, 1 }
Output : No
Explanation : the size of the array is 5 and we are given with 6 as our array element, and our condition specifically says a[i] = i , hence 6 can never satisfy our condition for the cyclic rotation of array.

Approach:

The Idea here is that the last element will always be the one which was our fixed point one step before and now is at the last because we cyclically rotated the array towards left which results in the fixed point array element being at the last of the array. If we try to right rotate the array k times then we will surely be getting our original array a, but if we found and element at the end whose value is greater then the size of the array , then we stop and report that we cannot generate our array a, else we continue right shifting k times. Since right shifting can be a time consuming task we just shift the last index pointer in our array, and % remainder modulus it so that it does not goes out of bounds from the indexes.

Follow the steps to solve the above problem:

• Begin by setting a boolean flag to true and initializing the variable lastindex to the index of the last element in the array 'b'.
• Iterate for a maximum of min(n, k) times.
  • Inside the loop, check if the element at the current lastindex is greater than n. If true, set the flag to false and break out of the loop.
  • Otherwise, calculate the distance needed to move lastindex to a new position based on the current element and update lastindex accordingly, moving the calculated distance steps clockwise in the array.
• Finally, based on the value of the flag, output "Yes" if the array 'a' can be restored or "No" otherwise.

Below is the implementation of above approach:

#include <bits/stdc++.h>
#include <math.h>

using namespace std;

void solve(int n, int k, vector<int> arr)
{
    bool flag = true;

    // Initialize 'lastindex' to the index of the last
    // element in the array
    int lastindex = n - 1;

    // Iterate for a maximum of min(n, k) times to prevent
    // exceeding array bounds
    for (int i = 0; i < min(n, k); i++) {
        // Check if the element at the current 'lastindex'
        // is greater than n
        if (arr[lastindex] > n) {
            // If true, set 'flag' to false and break out of
            // the loop
            flag = false;
            break;
        }

        // Calculate the distance needed to move 'lastindex'
        // to a new position based on the current element
        int dist = n - arr[lastindex];

        // Update 'lastindex' accordingly, moving dist steps
        // clockwise in the array
        lastindex = (lastindex + dist) % n;
    }

    // Output the result based on the value of 'flag'
    if (flag)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}

// Driver Code

int main()
{

    int n = 5, k = 3;

    vector<int> arr = { 4, 3, 3, 2, 3 };

    solve(n, k, arr);
}

#include <bits/stdc++.h>

#include <math.h>

​

using namespace std;

​

void solve(int n, int k, vector<int> arr)

{

    bool flag = true;

​

    // Initialize 'lastindex' to the index of the last

    // element in the array

    int lastindex = n - 1;

​

    // Iterate for a maximum of min(n, k) times to prevent

    // exceeding array bounds

    for (int i = 0; i < min(n, k); i++) {

        // Check if the element at the current 'lastindex'

        // is greater than n

        if (arr[lastindex] > n) {

            // If true, set 'flag' to false and break out of

            // the loop

            flag = false;

            break;

        }

​

        // Calculate the distance needed to move 'lastindex'

        // to a new position based on the current element

        int dist = n - arr[lastindex];

​

        // Update 'lastindex' accordingly, moving dist steps

        // clockwise in the array

        lastindex = (lastindex + dist) % n;

    }

​

    // Output the result based on the value of 'flag'

    if (flag)

        cout << "Yes" << endl;

    else

        cout << "No" << endl;

}

​

// Driver Code

​

int main()

{

​

    int n = 5, k = 3;

​

    vector<int> arr = { 4, 3, 3, 2, 3 };

​

    solve(n, k, arr);

}

public class SolveProblem {

    static void solve(int n, int k, int[] arr) {
        // Initialize 'flag' to true
        boolean flag = true;

        // Initialize 'lastIndex' to the index of the last
        // element in the array
        int lastIndex = n - 1;

        // Iterate for a maximum of min(n, k) times to prevent
        // exceeding array bounds
        for (int i = 0; i < Math.min(n, k); i++) {
            // Check if the element at the current 'lastIndex'
            // is greater than n
            if (arr[lastIndex] > n) {
                // If true, set 'flag' to false and break out of
                // the loop
                flag = false;
                break;
            }

            // Calculate the distance needed to move 'lastIndex'
            // to a new position based on the current element
            int dist = n - arr[lastIndex];

            // Update 'lastIndex' accordingly, moving dist steps
            // clockwise in the array
            lastIndex = (lastIndex + dist) % n;
        }

        // Output the result based on the value of 'flag'
        if (flag) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }

    // Driver Code
    public static void main(String[] args) {
        int n = 5;
        int k = 3;
        int[] arr = {4, 3, 3, 2, 3};

        solve(n, k, arr);
    }
}

def solve(n, k, arr):
    # Initialize 'flag' to True
    flag = True
    
    # Initialize 'lastindex' to the index of the last 
    # element in the array 
    lastindex = n - 1

    # Iterate for a maximum of min(n, k) times to prevent 
    # exceeding array bounds 
    for i in range(min(n, k)):
        # Check if the element at the current 'lastindex' 
        # is greater than n 
        if arr[lastindex] > n:
            # If true, set 'flag' to False and break out of 
            # the loop
            flag = False
            break

        # Calculate the distance needed to move 'lastindex' 
        # to a new position based on the current element 
        dist = n - arr[lastindex]
        
        # Update 'lastindex' accordingly, moving dist steps 
        # clockwise in the array 
        lastindex = (lastindex + dist) % n
    
    # Output the result based on the value of 'flag'
    if flag:
        print("Yes")
    else:
        print("No")

# Driver Code 
n = 5
k = 3
arr = [4, 3, 3, 2, 3]

solve(n, k, arr)

using System;
using System.Collections.Generic;

class Program
{
    static void Solve(int n, int k, List<int> arr)
    {
        bool flag = true;

        // Initialize 'lastindex' to the index of the last
        // element in the array
        int lastindex = n - 1;

        // Iterate for a maximum of Math.Min(n, k) times to prevent
        // exceeding array bounds
        for (int i = 0; i < Math.Min(n, k); i++)
        {
            // Check if the element at the current 'lastindex'
            // is greater than n
            if (arr[lastindex] > n)
            {
                // If true, set 'flag' to false and break out of
                // the loop
                flag = false;
                break;
            }

            // Calculate the distance needed to move 'lastindex'
            // to a new position based on the current element
            int dist = n - arr[lastindex];

            // Update 'lastindex' accordingly, moving dist steps
            // clockwise in the array
            lastindex = (lastindex + dist) % n;
        }

        // Output the result based on the value of 'flag'
        if (flag)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }

    // Driver Code
    static void Main()
    {
        int n = 5, k = 3;

        List<int> arr = new List<int> { 4, 3, 3, 2, 3 };

        Solve(n, k, arr);
    }
}

function solve(n, k, arr) {
    let flag = true;
    
    // Initialize 'lastindex' to the index of the last 
    // element in the array 
    let lastindex = n - 1;

    // Iterate for a maximum of min(n, k) times to prevent 
    // exceeding array bounds 
    for (let i = 0; i < Math.min(n, k); i++) {
    
        // Check if the element at the current 'lastindex' 
        // is greater than n 
        if (arr[lastindex] > n) {
        
            // If true, set 'flag' to false and break out of 
            // the loop
            flag = false;
            break;
        }

        // Calculate the distance needed to move 'lastindex' 
        // to a new position based on the current element 
        let dist = n - arr[lastindex];
        
        // Update 'lastindex' accordingly, moving dist steps 
        // clockwise in the array 
        lastindex = (lastindex + dist) % n;
    }
    
    // Output the result based on the value of 'flag'
    if (flag)
        console.log("Yes");
    else
        console.log("No");
}

// Driver Code 
let n = 5, k = 3;
let arr = [4, 3, 3, 2, 3];

solve(n, k, arr);

Yes

Time Complexity : O(n), where n is the length of the array.

Auxiliary space : O(1).

sundaransinha

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