Given an array, arr[] of N integers, the task is to rearrange the array elements such that the smallest element is at the 0th position, the second smallest element is at the (N-1)th position, the third smallest element at 1st position, 4th smallest element is at the (N-2)th position, and so on for all integers in arr[].
Examples:
Input: arr[] = {10, 23, 12, 17, 9}
Output: 9 12 23 17 10
Explanation:
The smallest element is 9 which is put in the index 0.
Then the second smallest element is 10 which is put in the last position.
The third smallest element is put in the second position from the start.
The fourth smallest in the second position from the last and so on.
Input: arr[] = {1, 3, 3, 4, 5}
Output: 1 3 5 4 3
Approach: We will be using the Two Pointer Technique. The idea is to iterate from the start marked by variable i to the end marked by variable j of the array alternatively until they meet in the middle and keep updating the minimum values at these indices. Below are the steps:
- Set the element at index i as the minimum value. Iterate over [i, j] and compare each value in this range with arr[i], and if the value in the range is less than arr[i] then swap the value between arr[i] and the current element. Increment the value of i.
- Set the element at index j as the minimum value. Iterate over [i, j] and compare each value in this range with arr[j], and if the value in the range is less than arr[j] then swap the value between arr[j] and the current element. Decrement the value of j.
- Place the smallest element in the ith position at the first iteration and the next smallest element in the jth position. Do this until both i and j point to the same position.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h>using namespace std;// Function to perform the rearrangement void rearrange(int a[], int N) { // Initialize variables int i = 0, j = N - 1; int min = 0, k, x = 0, temp; // Loop until i crosses j while (i < j) { // This check is to find the // minimum values in the // ascending order for(k = i; k <= j; k++) { if (a[k] < a[min]) min = k; } // Condition to alternatively // iterate variable i and j if (x % 2 == 0) { // Perform swap operation temp = a[i]; a[i] = a[min]; a[min] = temp; // Increment i i++; // Assign the value of min min = i; } else { // Perform swap temp = a[j]; a[j] = a[min]; a[min] = temp; // Decrement i j--; // Assign the value of min min = j; } x++; } // Print the array for(i = 0; i < N; i++) cout << a[i] << " "; }// Driver Codeint main(){ // Given Array arr[] int arr[] = { 1, 3, 3, 4, 5 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call rearrange(arr, N); return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java program for the above approach import java.io.*; import java.util.*; public class GFG { // Function to perform the rearrangement static void rearrange(int[] a) { // Initialize variables int i = 0, j = a.length - 1; int min = 0, k, x = 0, temp; // Loop until i crosses j while (i < j) { // This check is to find the // minimum values in the // ascending order for (k = i; k <= j; k++) { if (a[k] < a[min]) min = k; } // Condition to alternatively // iterate variable i and j if (x % 2 == 0) { // Perform swap operation temp = a[i]; a[i] = a[min]; a[min] = temp; // Increment i i++; // Assign the value of min min = i; } else { // Perform swap temp = a[j]; a[j] = a[min]; a[min] = temp; // Decrement i j--; // Assign the value of min min = j; } x++; } // Print the array for (i = 0; i < a.length; i++) System.out.print(a[i] + " "); } // Driver Code public static void main(String[] args) { // Given Array arr[] int arr[] = { 1, 3, 3, 4, 5 }; // Function Call rearrange(arr); } } |
Python3
# Python3 program for # the above approach # Function to perform # the rearrangement def rearrange(a, N): # Initialize variables i = 0 j = N - 1 min = 0 x = 0 # Loop until i crosses j while (i < j): # This check is to find the # minimum values in the # ascending order for k in range (i, j + 1): if (a[k] < a[min]): min = k # Condition to alternatively # iterate variable i and j if (x % 2 == 0): # Perform swap operation temp = a[i] a[i] = a[min] a[min] = temp # Increment i i += 1 # Assign the value of min min = i else: # Perform swap temp = a[j] a[j] = a[min] a[min] = temp # Decrement i j -= 1 # Assign the value of min min = j x += 1 # Print the array for i in range (N): print (a[i] ,end = " ")# Driver Codeif __name__ == "__main__": # Given Array arr[] arr = [1, 3, 3, 4, 5] N = len(arr) # Function call rearrange(arr, N) # This code is contributed by Chitranayal |
C#
// C# program for the above approachusing System;class GFG{// Function to perform the rearrangementstatic void rearrange(int[] a){ // Initialize variables int i = 0, j = a.Length - 1; int min = 0, k, x = 0, temp; // Loop until i crosses j while (i < j) { // This check is to find the // minimum values in the // ascending order for(k = i; k <= j; k++) { if (a[k] < a[min]) min = k; } // Condition to alternatively // iterate variable i and j if (x % 2 == 0) { // Perform swap operation temp = a[i]; a[i] = a[min]; a[min] = temp; // Increment i i++; // Assign the value of min min = i; } else { // Perform swap temp = a[j]; a[j] = a[min]; a[min] = temp; // Decrement i j--; // Assign the value of min min = j; } x++; } // Print the array for(i = 0; i < a.Length; i++) Console.Write(a[i] + " ");}// Driver Codepublic static void Main(string[] args){ // Given array arr[] int []arr = { 1, 3, 3, 4, 5 }; // Function call rearrange(arr);}}// This code is contributed by rutvik_56 |
Javascript
<script>// javascript program for the above approach // Function to perform the rearrangement function rearrange(a) { // Initialize variables var i = 0, j = a.length - 1; var min = 0, k, x = 0, temp; // Loop until i crosses j while (i < j) { // This check is to find the // minimum values in the // ascending order for (k = i; k <= j; k++) { if (a[k] < a[min]) min = k; } // Condition to alternatively // iterate variable i and j if (x % 2 == 0) { // Perform swap operation temp = a[i]; a[i] = a[min]; a[min] = temp; // Increment i i++; // Assign the value of min min = i; } else { // Perform swap temp = a[j]; a[j] = a[min]; a[min] = temp; // Decrement i j--; // Assign the value of min min = j; } x++; } // Print the array for (i = 0; i < a.length; i++) document.write(a[i] + " "); } // Driver Code // Given Array arr var arr = [ 1, 3, 3, 4, 5 ]; // Function Call rearrange(arr);// This code is contributed by todaysgaurav</script> |
Output
1 3 5 4 3
Time Complexity: O(N2) since two nested loops are used the time taken by the algorithm to complete all operations is quadratic.
Auxiliary Space: O(1) since no extra array is used so the space taken by the algorithm is constant
Efficient Approach
First sort the array and then using two pointers we will assign value as required by the question in the newArr array and then assign this newArr array value to the original array.
As mentioned in the question the first element of the new array is the smallest element in the 0th position, so newArr[0] = arr[0] as arr is already sorted, then the second smallest element to (n-1)th position means newArr[n-1] = arr[1], then newArr[1] = arr[2], newArr[n-2] = arr[3], so for this, we use two variable, where small pointing 0th position of newArr and large pointing (n-1)th position of newArr. on every fill-up of value in newArr, we increase small by 1 and decrease large by 1. so this way we achieve newArr as required in the question.
Step-by-step approach:
1. Sort given array arr
2. create n size newArr array
3. small = 0, large = 0
4. iterate from i = 0 to n-1
if(i%2 == 0)
newArr[small] = arr[i]
small++
else
newArr[large] = arr[i]
large--
5. copy all element from newArr to arr
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>using namespace std;void rearrangeArray(int arr[], int n) { // Sort the array sort(arr, arr + n); int newArr[n]; // Temporary array to store rearranged array int small = 0, large = n - 1; // Traverse the sorted array and rearrange elements for (int i = 0; i < n; i++) { // If i is even, pick smallest element if (i % 2 == 0) { newArr[small] = arr[i]; small++; } // If i is odd, pick largest element else { newArr[large] = arr[i]; large--; } } // Copy rearranged array to original array for (int i = 0; i < n; i++) { arr[i] = newArr[i]; }}int main() { int arr[] = { 1, 3, 3, 4, 5}; int n = sizeof(arr) / sizeof(arr[0]); rearrangeArray(arr, n); for (int i = 0; i < n; i++) { cout << arr[i] << " "; } cout << endl; return 0;} |
Java
import java.util.Arrays;public class GFG { // Function to rearrange the array alternatively static void rearrangeArray(int[] arr, int n) { // Sort the array Arrays.sort(arr); int[] newArr = new int[n]; // Temporary array to store rearranged array int small = 0, large = n - 1; // Traverse the sorted array and rearrange elements for (int i = 0; i < n; i++) { // If i is even, pick the smallest element if (i % 2 == 0) { newArr[small] = arr[i]; small++; } // If i is odd, pick the largest element else { newArr[large] = arr[i]; large--; } } // Copy rearranged array to the original array for (int i = 0; i < n; i++) { arr[i] = newArr[i]; } } public static void main(String[] args) { int[] arr = {1, 3, 3, 4, 5}; int n = arr.length; rearrangeArray(arr, n); // Print the rearranged array for (int i = 0; i < n; i++) { System.out.print(arr[i] + " "); } System.out.println(); }} |
Python3
def rearrange_array(arr): # Sort the array arr.sort() n = len(arr) new_arr = [0] * n # Temporary array to store rearranged array small = 0 large = n - 1 # Traverse the sorted array and rearrange elements for i in range(n): # If i is even, pick smallest element if i % 2 == 0: new_arr[small] = arr[i] small += 1 # If i is odd, pick largest element else: new_arr[large] = arr[i] large -= 1 # Copy rearranged array to original array for i in range(n): arr[i] = new_arr[i] return arrarr = [1, 3, 3, 4, 5]rearranged_arr = rearrange_array(arr)print(rearranged_arr) |
C#
using System;class Program { static void rearrangeArray(int[] arr, int n) { // Sort the array Array.Sort(arr); int[] newArr = new int[n]; // Temporary array to // store rearranged array int small = 0, large = n - 1; // Traverse the sorted array and rearrange elements for (int i = 0; i < n; i++) { // If i is even, pick the smallest element if (i % 2 == 0) { newArr[small] = arr[i]; small++; } // If i is odd, pick the largest element else { newArr[large] = arr[i]; large--; } } // Copy rearranged array to the original array for (int i = 0; i < n; i++) { arr[i] = newArr[i]; } } static void Main() { int[] arr = { 1, 3, 3, 4, 5 }; int n = arr.Length; rearrangeArray(arr, n); for (int i = 0; i < n; i++) { Console.Write(arr[i] + " "); } Console.WriteLine(); }} |
Javascript
function rearrangeArray(arr) { // Step 1: Sort the input array in ascending order arr.sort((a, b) => a - b); // Step 2: Create a new array to hold the rearranged elements const newArr = new Array(arr.length); let small = 0, large = arr.length - 1; // Step 3: Iterate through the sorted array and rearrange elements for (let i = 0; i < arr.length; i++) { // If the current index is even, place the smallest remaining element in newArr if (i % 2 === 0) { newArr[small] = arr[i]; small++; } // If the current index is odd, place the largest remaining element in newArr else { newArr[large] = arr[i]; large--; } } // Step 4: Copy the rearranged elements back to the original array for (let i = 0; i < arr.length; i++) { arr[i] = newArr[i]; }}const arr = [1, 3, 3, 4, 5];rearrangeArray(arr);// Step 5: Print the rearranged array elementsconsole.log(arr.join(" ")); |
Output
1 3 5 4 3
Time Complexity: O(N*log(N)) sort operation take nlog(n) time.
Auxiliary Space: O(N) since we use extra temp array
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