Given an array arr[] of integers, the task is to arrange them in a way similar to the to-and-fro movement of a Pendulum without using any extra space.
Pendulum Arrangement:
- The minimum element out of the list of integers must come in the center position of the array.
- The number in the ascending order next to the minimum, goes to the right, the next higher number goes to the left of minimum number and it continues.
- As higher numbers are reached, one goes to one side in a to-and-fro manner similar to that of a Pendulum.
Examples:
Input: arr[] = {2, 3, 5, 1, 4}
Output: 5 3 1 2 4
The minimum element is 1, so it is moved to the middle.
The next higher element 2 is moved to the right of the
middle element while the next higher element 3 is
moved to the left of the middle element and
this process is continued.
Input: arr[] = {11, 2, 4, 55, 6, 8}
Output: 11 6 2 4 8 55
Approach: An approach which uses an auxiliary array has been discussed in this article. Here’s an approach without using extra space:
- Sort the given array.
- Move all the odd position element in the right side of the array.
- Reverse the element from 0 to (n-1)/2 position of the array.
For example, let arr[] = {2, 3, 5, 1, 4}
Sorted array will be arr[] = {1, 2, 3, 4, 5}.
After moving all odd index position elements to the right,
arr[] = {1, 3, 5, 2, 4} (1 and 3 are the odd index positions)
After reversing elements from 0 to (n – 1) / 2,
arr[] = {5, 3, 1, 2, 4} which is the required array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print the Pendulum// arrangement of the given arrayvoid pendulumArrangement(int arr[], int n){ // Sort the array sort(arr, arr + n); int odd, temp, in, pos; // pos stores the index of // the last element of the array pos = n - 1; // odd stores the last odd index in the array if (n % 2 == 0) odd = n - 1; else odd = n - 2; // Move all odd index positioned // elements to the right while (odd > 0) { temp = arr[odd]; in = odd; // Shift the elements by one position // from odd to pos while (in != pos) { arr[in] = arr[in + 1]; in++; } arr[in] = temp; odd = odd - 2; pos = pos - 1; } // Reverse the element from 0 to (n - 1) / 2 int start = 0, end = (n - 1) / 2; for (; start < end; start++, end--) { temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; } // Printing the pendulum arrangement for (int i = 0; i < n; i++) cout << arr[i] << " ";}// Driver codeint main(){ int arr[] = { 11, 2, 4, 55, 6, 8 }; int n = sizeof(arr) / sizeof(arr[0]); pendulumArrangement(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.Arrays;import java.io.*;class GFG { // Function to print the Pendulum // arrangement of the given array static void pendulumArrangement(int arr[], int n) { // Sort the array // sort(arr, arr + n); Arrays.sort(arr); int odd, temp, in, pos; // pos stores the index of // the last element of the array pos = n - 1; // odd stores the last odd index in the array if (n % 2 == 0) odd = n - 1; else odd = n - 2; // Move all odd index positioned // elements to the right while (odd > 0) { temp = arr[odd]; in = odd; // Shift the elements by one position // from odd to pos while (in != pos) { arr[in] = arr[in + 1]; in++; } arr[in] = temp; odd = odd - 2; pos = pos - 1; } // Reverse the element from 0 to (n - 1) / 2 int start = 0, end = (n - 1) / 2; for (; start < end; start++, end--) { temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; } // Printing the pendulum arrangement for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); } // Driver code public static void main(String[] args) { int arr[] = { 11, 2, 4, 55, 6, 8 }; int n = arr.length; pendulumArrangement(arr, n); }}// This code is contributed by akt_mit |
Python3
# Python 3 implementation of the approach# Function to print the Pendulum# arrangement of the given arraydef pendulumArrangement(arr, n): # Sort the array arr.sort(reverse = False) # pos stores the index of # the last element of the array pos = n - 1 # odd stores the last odd index in the array if (n % 2 == 0): odd = n - 1 else: odd = n - 2 # Move all odd index positioned # elements to the right while (odd > 0): temp = arr[odd] in1 = odd # Shift the elements by one position # from odd to pos while (in1 != pos): arr[in1] = arr[in1 + 1] in1 += 1 arr[in1] = temp odd = odd - 2 pos = pos - 1 # Reverse the element from 0 to (n - 1) / 2 start = 0 end = int((n - 1) / 2) while(start < end): temp = arr[start] arr[start] = arr[end] arr[end] = temp start += 1 end -= 1 # Printing the pendulum arrangement for i in range(n): print(arr[i], end = " ")# Driver codeif __name__ == '__main__': arr = [11, 2, 4, 55, 6, 8] n = len(arr) pendulumArrangement(arr, n)# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System; class GFG { // Function to print the Pendulum // arrangement of the given array static void pendulumArrangement(int[] arr, int n) { // Sort the array // sort(arr, arr + n); Array.Sort(arr); int odd, temp, p, pos; // pos stores the index of // the last element of the array pos = n - 1; // odd stores the last odd index in the array if (n % 2 == 0) odd = n - 1; else odd = n - 2; // Move all odd index positioned // elements to the right while (odd > 0) { temp = arr[odd]; p = odd; // Shift the elements by one position // from odd to pos while (p != pos) { arr[p] = arr[p + 1]; p++; } arr[p] = temp; odd = odd - 2; pos = pos - 1; } // Reverse the element from 0 to (n - 1) / 2 int start = 0, end = (n - 1) / 2; for (; start < end; start++, end--) { temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; } // Printing the pendulum arrangement for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); } // Driver code public static void Main() { int[] arr = { 11, 2, 4, 55, 6, 8 }; int n = arr.Length; pendulumArrangement(arr, n); }}// This code is contributed by ChitraNayal |
PHP
<?php// PHP implementation of the approach // Function to print the Pendulum // arrangement of the given array function pendulumArrangement($arr, $n) { // Sort the array sort($arr) ; // pos stores the index of // the last element of the array $pos = $n - 1; // odd stores the last odd index in the array if ($n % 2 == 0) $odd = $n - 1; else $odd = $n - 2; // Move all odd index positioned // elements to the right while ($odd > 0) { $temp = $arr[$odd]; $in = $odd; // Shift the elements by one position // from odd to pos while ($in != $pos) { $arr[$in] = $arr[$in + 1]; $in++; } $arr[$in] = $temp; $odd = $odd - 2; $pos = $pos - 1; } // Reverse the element from 0 to (n - 1) / 2 $start = 0; $end = floor(($n - 1) / 2); for (; $start < $end; $start++, $end--) { $temp = $arr[$start]; $arr[$start] = $arr[$end]; $arr[$end] = $temp; } // Printing the pendulum arrangement for ($i = 0; $i < $n; $i++) echo $arr[$i], " "; } // Driver code $arr = array( 11, 2, 4, 55, 6, 8 ); $n = count($arr); pendulumArrangement($arr, $n); // This code is contributed by AnkitRai01?> |
Javascript
<script> // Javascript implementation of the approach // Function to print the Pendulum // arrangement of the given array function pendulumArrangement(arr, n) { // Sort the array // sort(arr, arr + n); arr.sort(function(a, b){return a - b}); let odd, temp, p, pos; // pos stores the index of // the last element of the array pos = n - 1; // odd stores the last odd index in the array if (n % 2 == 0) odd = n - 1; else odd = n - 2; // Move all odd index positioned // elements to the right while (odd > 0) { temp = arr[odd]; p = odd; // Shift the elements by one position // from odd to pos while (p != pos) { arr[p] = arr[p + 1]; p++; } arr[p] = temp; odd = odd - 2; pos = pos - 1; } // Reverse the element from 0 to (n - 1) / 2 let start = 0, end = parseInt((n - 1) / 2, 10); for (; start < end; start++, end--) { temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; } // Printing the pendulum arrangement for (let i = 0; i < n; i++) document.write(arr[i] + " "); } let arr = [ 11, 2, 4, 55, 6, 8 ]; let n = arr.length; pendulumArrangement(arr, n); </script> |
Output:
11 6 2 4 8 55
Time Complexity: O(n2)
Auxiliary Space: O(1)
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