Given an array arr[], the task is to print the array formed by traversing given array from first to the last index by flipping the whole array after printing every element.
Example:
Input: arr = {0, 1, 2, 3, 4, 5}
Output: 0 4 2 2 4 0
Explanation: On 1st iteration element on index 0 -> 0 is printed then the whole array is flipped: {0, 1, 2, 3, 4, 5} -> {5, 4, 3, 2, 1, 0}
On 2nd iteration element on index 1 -> 4 is printed then the whole array is flipped: : {5, 4, 3, 2, 1, 0} -> {0, 1, 2, 3, 4, 5}
On 3rd iteration element on index 2 -> 2 is printed then the whole array is flipped: {0, 1, 2, 3, 4, 5} -> {5, 4, 3, 2, 1, 0}
On 2nd iteration element on index 3 -> 2 is printed then the whole array is flipped: : {5, 4, 3, 2, 1, 0} -> {0, 1, 2, 3, 4, 5}
On 2nd iteration element on index 4 -> 4 is printed then the whole array is flipped: {0, 1, 2, 3, 4, 5} -> {5, 4, 3, 2, 1, 0}
On 2nd iteration element on index 5 -> 0 is printed then the whole array is flipped: : {5, 4, 3, 2, 1, 0} -> {0, 1, 2, 3, 4, 5}Input: arr = {0, 1, 2, 3, 4}
Output: 0 3 2 1 4
Approach: The given problem can be solved by using the two-pointer technique. The idea is to iterate the array from left to right starting from the first index, and from right to left starting from the second last index.
Below steps can be followed to solve the problem:
- Use pointer one to iterate the array from left to right, and use pointer two to traverse the array from right to left
- Print the elements pointed by both the pointers simultaneously and increment pointer one by 2 and decrement pointer two by 2
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to print array elements at// every index by flipping whole array// after printing every elementvoid printFlip(vector<int> arr){ // Initialize length of the array int N = arr.size(); // Initialize both the pointers int p1 = 0, p2 = N - 2; // Iterate until both pointers // are not out of bounds while (p1 < N || p2 >= 0) { // Print the elements cout << arr[p1] << " "; if (p2 > 0) cout << arr[p2] << " "; // Increment p1 by 2 p1 += 2; // Decrement p2 by 2 p2 -= 2; }}// Driver codeint main(){ // Initialize the array vector<int> arr = { 0, 1, 2, 3, 4 }; // Call the function // and print the array printFlip(arr); return 0;}// This code is contributed by Potta Lokesh. |
Java
// Java implementation for the above approachimport java.io.*;import java.util.*;class GFG { // Function to print array elements at // every index by flipping whole array // after printing every element public static void printFlip(int[] arr) { // Initialize length of the array int N = arr.length; // Initialize both the pointers int p1 = 0, p2 = N - 2; // Iterate until both pointers // are not out of bounds while (p1 < N || p2 >= 0) { // Print the elements System.out.print(arr[p1] + " "); if (p2 > 0) System.out.print(arr[p2] + " "); // Increment p1 by 2 p1 += 2; // Decrement p2 by 2 p2 -= 2; } } // Driver code public static void main(String[] args) { // Initialize the array int[] arr = { 0, 1, 2, 3, 4 }; // Call the function // and print the array printFlip(arr); }} |
Python3
# Python code for the above approach# Function to print array elements at# every index by flipping whole array# after printing every elementdef printFlip(arr): # Initialize length of the array N = len(arr); # Initialize both the pointers p1 = 0 p2 = N - 2; # Iterate until both pointers # are not out of bounds while (p1 < N or p2 >= 0): # Print the elements print(arr[p1], end=" "); if (p2 > 0): print(arr[p2], end=" "); # Increment p1 by 2 p1 += 2; # Decrement p2 by 2 p2 -= 2; # Driver Code# Initialize the arrayarr = [ 0, 1, 2, 3, 4 ];# Call the function# and print the arrayprintFlip(arr);# This code is contributed by gfgking. |
C#
// C# implementation for the above approachusing System;class GFG { // Function to print array elements at // every index by flipping whole array // after printing every element static void printFlip(int []arr) { // Initialize length of the array int N = arr.Length; // Initialize both the pointers int p1 = 0, p2 = N - 2; // Iterate until both pointers // are not out of bounds while (p1 < N || p2 >= 0) { // Print the elements Console.Write(arr[p1] + " "); if (p2 > 0) Console.Write(arr[p2] + " "); // Increment p1 by 2 p1 += 2; // Decrement p2 by 2 p2 -= 2; } } // Driver code public static void Main() { // Initialize the array int []arr = { 0, 1, 2, 3, 4 }; // Call the function // and print the array printFlip(arr); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script>// Javascript code for the above approach// Function to print array elements at// every index by flipping whole array// after printing every elementfunction printFlip(arr){ // Initialize length of the array let N = arr.length; // Initialize both the pointers let p1 = 0, p2 = N - 2; // Iterate until both pointers // are not out of bounds while (p1 < N || p2 >= 0) { // Print the elements document.write(arr[p1] + " "); if (p2 > 0) document.write(arr[p2] + " "); // Increment p1 by 2 p1 += 2; // Decrement p2 by 2 p2 -= 2; }}// Driver Code// Initialize the arraylet arr = [ 0, 1, 2, 3, 4 ];// Call the function// and print the arrayprintFlip(arr);// This code is contributed by Samim Hossain Mondal.</script> |
0 3 2 1 4
Time Complexity: O(N)
Auxiliary Space: O(1)
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