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 rearrangementvoid 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 approachimport 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 rearrangementdef 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++; } // Prvar 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> |
1 3 5 4 3
Time Complexity: O(N2)
Auxiliary Space: O(1)
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