Given a circular array arr[] of size N and an index K, the task is to reverse all elements of the circular array starting from the index K.
Examples:
Input: arr[] = {3, 5, 2, 4, 1}, K = 2
Output: 4 2 5 3 1
Explanation:
After reversing the elements of the array from index K to K – 1, the modified arr[] is {4, 1, 2, 5, 3}.Input: arr[] = {1, 2, 3, 4, 5}, K = 4
Output: 3 2 1 5 4
Explanation:
After reversing the elements of the array from index K to K – 1, the modified arr[] is {3, 2, 1, 5, 4}.
Approach: To solve the given problem, the idea is to use Two Pointers Approach. Follow the steps below to solve the problem:
- Initialize three variables start as K and end as (K – 1), to keep track of the boundary using two pointer approach, and count as N / 2.
- Iterate until the value of count is positive and perform the following steps:
- Swap the elements arr[start % N] and arr[end % N].
- Increment start by 1 and decrement end by 1. If end is equal to -1, then update end as (N – 1).
- Decrement count by 1.
- After the above steps, print the updated array obtained after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print the array arr[]void printArray(int arr[], int N){ // Print the array for (int i = 0; i < N; i++) { cout << arr[i] << " "; }}// Function to reverse elements of given// circular array starting from index kvoid reverseCircularArray(int arr[], int N, int K){ // Initialize two variables as // start = k and end = k-1 int start = K, end = K - 1; // Initialize count = N/2 int count = N / 2; // Loop while count > 0 while (count--) { // Swap the elements at index // (start%N) and (end%N) int temp = arr[start % N]; arr[start % N] = arr[end % N]; arr[end % N] = temp; // Update the start and end start++; end--; // If end equals to -1 // set end = N-1 if (end == -1) { end = N - 1; } } // Print the circular array printArray(arr, N);}// Driver Codeint main(){ int arr[] = { 3, 5, 2, 4, 1 }; int K = 2; int N = sizeof(arr) / sizeof(arr[0]); // Function Call reverseCircularArray(arr, N, K); return 0;} |
Java
// Java program for the above approachclass GFG { // Function to print the array arr[] static void printArray(int arr[], int N) { // Print the array for (int i = 0; i < N; i++) { System.out.print(arr[i] + " "); } } // Function to reverse elements of given // circular array starting from index k static void reverseCircularArray(int arr[], int N, int K) { // Initialize two variables as // start = k and end = k-1 int start = K, end = K - 1; // Initialize count = N/2 int count = N / 2; // Loop while count > 0 while (count != 0) { // Swap the elements at index // (start%N) and (end%N) int temp = arr[start % N]; arr[start % N] = arr[end % N]; arr[end % N] = temp; // Update the start and end start++; end--; // If end equals to -1 // set end = N-1 if (end == -1) { end = N - 1; } count -= 1; } // Print the circular array printArray(arr, N); } // Driver Code public static void main (String[] args) { int arr[] = { 3, 5, 2, 4, 1 }; int K = 2; int N = arr.length; // Function Call reverseCircularArray(arr, N, K); }}// This code is contributed by AnkThon |
Python3
# Python3 program for the above approach# Function to print array arr[]def printArray(arr, N): # Print the array for i in range(N): print(arr[i], end = " ")# Function to reverse elements of given# circular array starting from index kdef reverseCircularArray(arr, N, K): # Initialize two variables as # start = k and end = k-1 start, end = K, K - 1 # Initialize count = N/2 count = N // 2 # Loop while count > 0 while (count): # Swap the elements at index # (start%N) and (end%N) temp = arr[start % N] arr[start % N] = arr[end % N] arr[end % N] = temp # Update the start and end start += 1 end -= 1 # If end equals to -1 # set end = N-1 if (end == -1): end = N - 1 count -= 1 # Print the circular array printArray(arr, N)# Driver Codeif __name__ == '__main__': arr = [ 3, 5, 2, 4, 1 ] K = 2 N = len(arr) # Function Call reverseCircularArray(arr, N, K)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG { // Function to print the array []arr static void printArray(int []arr, int N) { // Print the array for (int i = 0; i < N; i++) { Console.Write(arr[i] + " "); } } // Function to reverse elements of given // circular array starting from index k static void reverseCircularArray(int []arr, int N, int K) { // Initialize two variables as // start = k and end = k-1 int start = K, end = K - 1; // Initialize count = N/2 int count = N / 2; // Loop while count > 0 while (count != 0) { // Swap the elements at index // (start%N) and (end%N) int temp = arr[start % N]; arr[start % N] = arr[end % N]; arr[end % N] = temp; // Update the start and end start++; end--; // If end equals to -1 // set end = N-1 if (end == -1) { end = N - 1; } count -= 1; } // Print the circular array printArray(arr, N); } // Driver Code public static void Main(String[] args) { int []arr = { 3, 5, 2, 4, 1 }; int K = 2; int N = arr.Length; // Function Call reverseCircularArray(arr, N, K); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript program for the above approach// Function to print the array arr[]function printArray(arr, N){ // Print the array for (let i = 0; i < N; i++) { document.write(arr[i] + " "); }}// Function to reverse elements of given// circular array starting from index kfunction reverseCircularArray(arr, N, K){ // Initialize two variables as // start = k and end = k-1 let start = K, end = K - 1; // Initialize count = N/2 let count = Math.floor(N / 2); // Loop while count > 0 while (count--) { // Swap the elements at index // (start%N) and (end%N) let temp = arr[start % N]; arr[start % N] = arr[end % N]; arr[end % N] = temp; // Update the start and end start++; end--; // If end equals to -1 // set end = N-1 if (end === -1) { end = N - 1; } } // Print the circular array printArray(arr, N);}// Driver Code let arr = [ 3, 5, 2, 4, 1 ]; let K = 2; let N = arr.length; // Function Call reverseCircularArray(arr, N, K);// This code is contributed by Surbhi Tyagi.</script> |
Output:
4 2 5 3 1
Time Complexity: O(N)
Auxiliary Space: O(1)
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