Given an array of n elements. Our task is to write a program to rearrange the array such that elements at even positions are greater than all elements before it and elements at odd positions are less than all elements before it.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 6, 7}
Output : 4 5 3 6 2 7 1
Input : arr[] = {1, 2, 1, 4, 5, 6, 8, 8}
Output : 4 5 2 6 1 8 1 8
The idea to solve this problem is to first create an auxiliary copy of the original array and sort the copied array. Now total number of even position in array with n elements will be floor(n/2) and remaining is the number of odd positions. Now fill the odd and even positions in the original array using the sorted array in the below manner:
- Total odd positions will be n – floor(n/2). Start from (n-floor(n/2))th position in the sorted array and copy the element to 1st position of sorted array. Start traversing the sorted array from this position towards left and keep filling the odd positions in the original array towards right.
- Start traversing the sorted array starting from (n-floor(n/2)+1)th position towards right and keep filling the original array starting from 2nd position.
Below is the implementation of above idea:
Python3
# Python3 code to rearrange the array# as per the given conditionimport array as aimport numpy as np# function to rearrange the arraydef rearrangeArr(arr, n): # total even positions evenPos = int(n / 2) # total odd positions oddPos = n - evenPos # initialising empty array in python tempArr = np.empty(n, dtype = object) # copy original array in an # auxiliary array for i in range(0, n): tempArr[i] = arr[i] # sort the auxiliary array tempArr.sort() j = oddPos - 1 # fill up odd position in original # array for i in range(0, n, 2): arr[i] = tempArr[j] j = j - 1 j = oddPos # fill up even positions in original # array for i in range(1, n, 2): arr[i] = tempArr[j] j = j + 1 # display array for i in range(0, n): print (arr[i], end = ' ')# Driver codearr = a.array('i', [ 1, 2, 3, 4, 5, 6, 7 ])rearrangeArr(arr, 7)# This code is contributed by saloni1297 |
Output:
4 5 3 6 2 7 1
Time Complexity: O(N*logN), as we are using a sort function.
Auxiliary Space: O(N), as we are using extra space.
Please refer complete article on Rearrange array such that arr[i] >= arr[j] if i is even and arr[i]<=arr[j] if i is odd and j < i for more details!
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