Sort a list in python and then return the index of elements in sorted order.
Examples:
Input : [2, 3, 1, 4, 5] Output : [2, 0, 1, 3, 4] After sorting list becomes [1, 2, 3, 4, 5] and their index as [2, 0, 1, 3, 4] Input : [6, 4, 7, 8, 1] Output : [4, 1, 0, 2, 3] After sorting the list becomes [1, 4, 6, 7, 8] and their index as [4, 1, 0, 2, 3].
Method 1
Python
import numpys = numpy.array([2, 3, 1, 4, 5])sort_index = numpy.argsort(s)print(sort_index) |
Output :
[2, 0, 1, 3, 4]
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
Method 2
Python
s = [2, 3, 1, 4, 5]li=[]for i in range(len(s)): li.append([s[i],i])li.sort()sort_index = []for x in li: sort_index.append(x[1])print(sort_index) |
Output:
[2, 0, 1, 3, 4]
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
Method #3: Using the built-in sorted() function along with enumerate() function
Python3
s = [2, 3, 1, 4, 5]# Using the built-in sorted() function along with enumerate() function# enumerate() returns a tuple (index, value) for each element in the list# and sorted() sorts the list based on the valuesort_index = [i for i, x in sorted(enumerate(s), key=lambda x: x[1])]print(sort_index)#This code is contributed by Edula Vinay Kumar Reddy |
[2, 0, 1, 3, 4]
This method uses the built-in sorted() function which takes an iterable (in this case, the enumerate of the original list) and sorts it based on a key function (in this case, the lambda function that accesses the second element of each tuple returned by enumerate). The [i for i, x in …] list comprehension then iterates through the sorted enumerate and appends the first element (the index) of each tuple to the sort_index list.
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
Method #3: Using the heapq module
The heapq module provides functions for creating and manipulating heaps, which are data structures that can be used to efficiently find the smallest or largest elements in a collection. We can use the nlargest() function from the heapq module to find the indices of the k largest elements in the list s.
Python3
import heapqdef sort_indices(s, k): # Use the nlargest() function from the heapq module to get the indices of the k largest elements in s # Pass a lambda function to nlargest() that returns the negative value of each element in s # This will effectively find the indices of the k smallest elements in s return [i for i, x in heapq.nlargest(k, enumerate(s), key=lambda x: -x[1])]s = [2, 3, 1, 4, 5]k = 3sort_index = sort_indices(s, k)print(sort_index) |
[2, 0, 1]
Time Complexity: O(n log n), where n is the length of the input list s.
Auxiliary Space: O(n), where n is the length of the input list s.
