Given a sorted array of positive integers, rearrange the array alternately i.e first element should be maximum value, second minimum value, third second max, fourth second min and so on.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 6, 7}
Output : arr[] = {7, 1, 6, 2, 5, 3, 4}
Input : arr[] = {1, 2, 3, 4, 5, 6}
Output : arr[] = {6, 1, 5, 2, 4, 3}
Expected time complexity is O(n).
The idea is use an auxiliary array. We maintain two pointers one to leftmost or smallest element and other to rightmost or largest element. We more both pointers toward each other and alternatively copy elements at these pointers to an auxiliary array. Finally we copy auxiliary array back to original array.
Below are programs based on above facts.
C++
// C++ program to rearrange an array in minimum // maximum form #include <bits/stdc++.h> using namespace std; // Prints max at first position, min at second position // second max at third position, second min at fourth // position and so on. void rearrange(int arr[], int n) { // Auxiliary array to hold modified array int temp[n]; // Indexes of smallest and largest elements // from remaining array. int small=0, large=n-1; // To indicate whether we need to copy rmaining // largest or remaining smallest at next position int flag = true; // Store result in temp[] for (int i=0; i<n; i++) { if (flag) temp[i] = arr[large--]; else temp[i] = arr[small++]; flag = !flag; } // Copy temp[] to arr[] for (int i=0; i<n; i++) arr[i] = temp[i]; } // Driver program to test above function int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Original Arrayn"; for (int i=0; i<n; i++) cout << arr[i] << " "; rearrange(arr, n); cout << "nModified Arrayn"; for (int i=0; i<n; i++) cout << arr[i] << " "; return 0; } |
Java
// Java program to rearrange an array in minimum // maximum form import java.util.Arrays; public class GFG { // Prints max at first position, min at second position // second max at third position, second min at fourth // position and so on. static void rearrange(int[] arr, int n) { // Auxiliary array to hold modified array int temp[] = new int[n]; // Indexes of smallest and largest elements // from remaining array. int small=0, large=n-1; // To indicate whether we need to copy rmaining // largest or remaining smallest at next position boolean flag = true; // Store result in temp[] for (int i=0; i<n; i++) { if (flag) temp[i] = arr[large--]; else temp[i] = arr[small++]; flag = !flag; } // Copy temp[] to arr[] arr = temp.clone(); } // Driver method to test the above function public static void main(String[] args) { int arr[] = new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9}; System.out.println("Original Array "); System.out.println(Arrays.toString(arr)); rearrange(arr,arr.length); System.out.println("Modified Array "); System.out.println(Arrays.toString(arr)); } } |
Python
# Python program to rearrange an array in minimum # maximum form # Prints max at first position, min at second position # second max at third position, second min at fourth # position and so on. def rearrange(arr, n): # Auxiliary array to hold modified array temp = n*[None] # Indexes of smallest and largest elements # from remaining array. small,large =0,n-1 # To indicate whether we need to copy rmaining # largest or remaining smallest at next position flag = True # Store result in temp[] for i in range(n): if flag is True: temp[i] = arr[large] large -= 1 else: temp[i] = arr[small] small += 1 flag = bool(1-flag) # Copy temp[] to arr[] for i in range(n): arr[i] = temp[i] return arr # Driver program to test above function arr = [1, 2, 3, 4, 5, 6, 7, 8, 9] n = len(arr) print("Original Array") print(arr) print("Modified Array") print(rearrange(arr, n)) # This code is contributed by Pratik Chhajer |
C#
// C# program to rearrange // an array in minimum // maximum form using System; class GFG { // Prints max at first position, // min at second position second // max at third position, second // min at fourth position and so on. static void rearrage(int[] arr, int n) { // Auxiliary array to // hold modified array int []temp = new int[n]; // Indexes of smallest // and largest elements // from remaining array. int small = 0, large = n - 1; // To indicate whether we // need to copy rmaining // largest or remaining // smallest at next position bool flag = true; // Store result in temp[] for (int i = 0; i < n; i++) { if (flag) temp[i] = arr[large--]; else temp[i] = arr[small++]; flag = !flag; } // Copy temp[] to arr[] for (int i = 0; i < n; i++) arr[i] = temp[i]; } // Driver Code static void Main() { int[] arr = {1, 2, 3, 4, 5, 6, 7, 8, 9}; Console.WriteLine("Original Array"); for(int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); rearrage(arr, arr.Length); Console.WriteLine("
Modified Array"); for(int i = 0; i < arr.Length; i++) Console.Write(arr[i] + " "); } } // This code is contributed // by Sam007 |
PHP
<?php // PHP program to rearrange an array in // minimum-maximum form // Prints max at first position, min at // second position second max at third // position, second min at fourth // position and so on. function rearrange(&$arr, $n) { // Auxiliary array to hold modified array $temp = array(); // Indexes of smallest and largest elements // from remaining array. $small = 0; $large = $n - 1; // To indicate whether we need to copy // remaining largest or remaining smallest // at next position $flag = true; // Store result in temp[] for ($i = 0; $i < $n; $i++) { if ($flag) $temp[$i] = $arr[$large--]; else $temp[$i] = $arr[$small++]; $flag = !$flag; } // Copy temp[] to arr[] for ($i = 0; $i < $n; $i++) $arr[$i] = $temp[$i]; } // Driver Code $arr = array(1, 2, 3, 4, 5, 6, 7, 8, 9); $n = count($arr); echo "Original Arrayn
"; for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; rearrange($arr, $n); echo "
Modified Arrayn
"; for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; // This code is contributed by // Rajput-Ji ?> |
Output:
Original Array 1 2 3 4 5 6 7 8 9 Modified Array 9 1 8 2 7 3 6 4 5
Time Complexity : O(n)
Auxiliary Space : O(n)
Exercise : How to solve this problem if extra space is not allowed?
Rearrange an array in maximum minimum form | Set 2 (O(1) extra space)
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