Given an array arr[] of size N, the task is to perform the following operation exactly N times,
Create an empty list of integers b[] and in the ith operation,
- Append arr[i] to the end of b[].
- Reverse the elements in b[].
Finally, print the contents of the list b[] after the end of all operations.
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 4 2 1 3
Operation b[] 1 {1} 2 {2, 1} 3 {3, 1, 2} 4 {4, 2, 1, 3} Input: arr[] = {1, 2, 3}
Output: 3 1 2
Approach: We need some observations to solve this problem. Suppose the number of elements in the array is even. Say our array is {4, 8, 6, 1, 7, 9}.
| Operation | b[] |
|---|---|
| 1 | {4} |
| 2 | {8, 4} |
| 3 | {6, 4, 8} |
| 4 | {1, 8, 4, 6} |
| 5 | {7, 6, 4, 8, 1} |
| 6 | {9, 1, 8, 4, 6, 7} |
After carefully observing, we conclude that for even size of elements in the array the numbers which are at even positions (index 1 based) are reversed and added at the beginning and the numbers which are at the odd positions are kept in same order and added in the end.
While for odd-sized arrays, the elements at odd positions are reversed and added at the beginning while elements in the array at even positions are kept same and added in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that generates the// array b[] when n is evenvoid solveEven(int n, int* arr, int* b){ int left = n - 1; // Fill the first half of the final array // with reversed sequence for (int i = 0; i < (n / 2); ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half int right = 0; for (int i = n / 2; i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; }}// Function that generates the// array b[] when n is oddvoid solveOdd(int n, int* arr, int* b){ // Fill the first half of the final array // with reversed sequence int left = n - 1; for (int i = 0; i < (n / 2) + 1; ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half int right = 1; for (int i = (n / 2) + 1; i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; }}// Function to find the final array b[]// after n operations of given typevoid solve(int n, int* arr){ // Create the array b int b[n]; // If the array size is even if (n % 2 == 0) solveEven(n, arr, b); else solveOdd(n, arr, b); // Print the final array elements for (int i = 0; i <= n - 1; ++i) { cout << b[i] << " "; }}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4 }; int n = sizeof(arr) / sizeof(arr[0]); solve(n, arr); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {// Function that generates the// array b[] when n is evenstatic void solveEven(int n, int arr[], int b[]){ int left = n - 1; // Fill the first half of the final array // with reversed sequence for (int i = 0; i < (n / 2); ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half int right = 0; for (int i = n / 2; i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; }}// Function that generates the// array b[] when n is oddstatic void solveOdd(int n, int arr[], int b[]){ // Fill the first half of the final array // with reversed sequence int left = n - 1; for (int i = 0; i < (n / 2) + 1; ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half int right = 1; for (int i = (n / 2) + 1; i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; }}// Function to find the final array b[]// after n operations of given typestatic void solve(int n, int arr[]){ // Create the array b int b[] = new int[n]; // If the array size is even if (n % 2 == 0) solveEven(n, arr, b); else solveOdd(n, arr, b); // Print the final array elements for (int i = 0; i <= n - 1; ++i) { System.out.print( b[i] + " "); }}// Driver codepublic static void main (String[] args) { int []arr = { 1, 2, 3, 4 }; int n = arr.length; solve(n, arr);}}// This code is contributed by anuj_67.. |
Python3
# Python 3 implementation of the approach# Function that generates the# array b[] when n is evendef solveEven(n, arr, b): left = n - 1 # Fill the first half of the final array # with reversed sequence for i in range((n // 2)): b[i] = arr[left] left = left - 2 if (left < 0): break # Fill the second half right = 0 for i in range(n // 2, n, 1): b[i] = arr[right] right = right + 2 if (right > n - 2): break# Function that generates the# array b[] when n is odddef solveOdd(n, arr, b): # Fill the first half of the final array # with reversed sequence left = n - 1 for i in range(n // 2 + 1): b[i] = arr[left] left = left - 2 if (left < 0): break # Fill the second half right = 1 for i in range(n // 2 + 1, n, 1): b[i] = arr[right] right = right + 2 if (right > n - 2): break# Function to find the final array b[]# after n operations of given typedef solve(n, arr): # Create the array b b = [0 for i in range(n)] # If the array size is even if (n % 2 == 0): solveEven(n, arr, b) else: solveOdd(n, arr, b) # Print the final array elements for i in range(n): print(b[i], end = " ")# Driver codeif __name__ == '__main__': arr = [1, 2, 3, 4] n = len(arr) solve(n, arr)# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG {// Function that generates the// array b[] when n is evenstatic void solveEven(int n, int []arr, int []b){ int left = n - 1; // Fill the first half of the final array // with reversed sequence for (int i = 0; i < (n / 2); ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half int right = 0; for (int i = n / 2; i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; }}// Function that generates the// array b[] when n is oddstatic void solveOdd(int n, int []arr, int []b){ // Fill the first half of the final array // with reversed sequence int left = n - 1; for (int i = 0; i < (n / 2) + 1; ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half int right = 1; for (int i = (n / 2) + 1; i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; }}// Function to find the final array b[]// after n operations of given typestatic void solve(int n, int []arr){ // Create the array b int []b = new int[n]; // If the array size is even if (n % 2 == 0) solveEven(n, arr, b); else solveOdd(n, arr, b); // Print the final array elements for (int i = 0; i <= n - 1; ++i) { Console.Write( b[i] + " "); }}// Driver codepublic static void Main () { int []arr = { 1, 2, 3, 4 }; int n = arr.Length; solve(n, arr);}}// This code is contributed by anuj_67.. |
Javascript
<script> // Javascript implementation of the approach // Function that generates the // array b[] when n is even function solveEven(n, arr, b) { let left = n - 1; // Fill the first half of the final array // with reversed sequence for (let i = 0; i < parseInt(n / 2, 10); ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half let right = 0; for (let i = parseInt(n / 2, 10); i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; } } // Function that generates the // array b[] when n is odd function solveOdd(n, arr, b) { // Fill the first half of the final array // with reversed sequence let left = n - 1; for (let i = 0; i < parseInt(n / 2, 10) + 1; ++i) { b[i] = arr[left]; left = left - 2; if (left < 0) break; } // Fill the second half let right = 1; for (let i = parseInt(n / 2, 10) + 1; i <= n - 1; ++i) { b[i] = arr[right]; right = right + 2; if (right > n - 2) break; } } // Function to find the final array b[] // after n operations of given type function solve(n, arr) { // Create the array b let b = new Array(n); // If the array size is even if (n % 2 == 0) solveEven(n, arr, b); else solveOdd(n, arr, b); // Print the final array elements for (let i = 0; i <= n - 1; ++i) { document.write( b[i] + " "); } } let arr = [ 1, 2, 3, 4 ]; let n = arr.length; solve(n, arr);</script> |
Output:
4 2 1 3
Time Complexity: O(n)
Auxiliary Space: O(n)
Efficient Approach:
The last Element of Array will be reversed only once. Last but one element will be reversed twice. Hence it goes to the last position in final result array ie b. Hence we can fill b array by iterating the original array from the end and placing elements at the not filled first index and not filled the last index. The same idea is implemented below.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std;int* solve(int arr[], int n) { static int b[4]; int p = 0; for (int i = n - 1; i >= 0; i--) { b[p] = arr[i--]; if (i >= 0) b[n - 1 - p] = arr[i]; p++; } return b;}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4 }; int n = sizeof(arr)/sizeof(arr[0]); int *b ; b = solve(arr, n); for(int i = 0; i < n; i++) cout << b[i] << " ";}// This code is contributed by Rajput-Ji |
Java
// Java implementation of the approachimport java.util.*;import java.lang.*;import java.io.*;class GFG { static int[] solve(int[] arr, int n) { int[] b = new int[n]; int p = 0; for (int i = n - 1; i >= 0; i--) { b[p] = arr[i--]; if (i >= 0) b[n - 1 - p] = arr[i]; p++; } return b; } public static void main(String[] args) { int []arr = { 1, 2, 3, 4 }; int n = arr.length; int[] b = solve(arr, n); System.out.println(Arrays.toString(b)); }}// This code is contributed by Pramod Hosahalli |
C#
// C# implementation of the approachusing System; class GFG { static int[] solve(int[] arr, int n) { int[] b = new int[n]; int p = 0; for (int i = n - 1; i >= 0; i--) { b[p] = arr[i--]; if (i >= 0) b[n - 1 - p] = arr[i]; p++; } return b; } // Driver Code public static void Main(String[] args) { int []arr = { 1, 2, 3, 4 }; int n = arr.Length; int[] b = solve(arr, n); Console.WriteLine("[" + String.Join(",", b) + "]"); }}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approachdef solve(arr, n): b = [0 for i in range(n)] p = 0 i = n - 1 while i >= 0: b[p] = arr[i] i -= 1 if (i >= 0): b[n - 1 - p] = arr[i] p += 1 i -= 1 return b# Driver Codearr = [1, 2, 3, 4]n = len(arr)b = solve(arr, n)print(b)# This code is contributed by Mohit kumar |
Javascript
<script>// javascript implementation of the approach function solve(arr , n) { var b = Array(n).fill(0); var p = 0; for (i = n - 1; i >= 0; i--) { b[p] = arr[i--]; if (i >= 0) b[n - 1 - p] = arr[i]; p++; } return b; } var arr = [ 1, 2, 3, 4 ]; var n = arr.length; var b = solve(arr, n); document.write("["+ b.toString()+ "]");// This code contributed by aashish1995 </script> |
Output:
[4, 2, 1, 3]
Time Complexity: O(n)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
