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Python | Inserting item in sorted list maintaining order

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  • Difficulty Level : Basic
  • Last Updated : 01 Feb, 2023
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Working with sorted list is essential because mostly the data we get is ordered. Any query can come to insert the new data into its appropriate position. Hence requires to know how to perform these dynamic queries. Lets discuss certain ways in which this can be performed. 

Method #1 : Naive Method In this method, we just simply test for the value and check if next element is greater than the input element and store the index which is then used to slice in that element at that position. 
 

Python3




# Python3 code to demonstrate
# insertion in sorted list
# using naive method
 
# initializing list
test_list = [1, 2, 3, 6, 7]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# insert element
k = 5
 
# using naive method
# insertion in sorted list
# using naive method
for i in range(len(test_list)):
    if test_list[i] > k:
        index = i
        break
res = test_list[: i] + [k] + test_list[i :]
 
# printing result
print ("The list after insertion is : " + str(res))

Output

The original list is : [1, 2, 3, 6, 7]
The list after insertion is : [1, 2, 3, 5, 6, 7]

Time complexity: O(n) where n is size of  list
Auxiliary Space: O(n)

Method #2 : Using bisect.insort() This is more concise and recommended method to perform this particular task. This method is designed for this particular task and performs this task in most efficient manner. 
 

Python3




# Python3 code to demonstrate
# insertion in sorted list
# using bisect.insort()
import bisect
 
# initializing list
test_list = [1, 2, 3, 6, 7]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# insert element
k = 5
 
# using bisect.insort()
# insertion in sorted list
# using naive method
bisect.insort(test_list, k)
 
# printing result
print ("The list after insertion is : " + str(test_list))

Output

The original list is : [1, 2, 3, 6, 7]
The list after insertion is : [1, 2, 3, 5, 6, 7]

Time complexity: O(n) where n is size of given list
Auxiliary space: O(1)

 Method #3 : Using heapq

To insert an element into a sorted list using a min heap , you can use the following code:

Python3




import heapq
 
test_list = [1, 2, 3, 6, 7]
k = 5
heap = test_list + [k]
heapq.heapify(heap)
sorted_list = [heapq.heappop(heap) for _ in range(len(heap))]
print(sorted_list)
#This code is contributed by Edula Vinay Kumar Reddy

Output

[1, 2, 3, 5, 6, 7]

This code first creates a min heap from the elements of test_list and the element k using the heapq.heapify() function. Then, it extracts the elements from the heap one by one using a list comprehension and the heapq.heappop() function, and stores them in a new list sorted_list, which is then printed.

This approach has a time complexity of O(n log n), since the heapq.heapify() and heapq.heappop() functions have a time complexity of O(n) and O(log n), respectively. It also has a space complexity of O(n), since it requires an additional heap data structure to store the elements.

Method #4 : Using append() and sort() methods

Python3




# Python3 code to demonstrate
# insertion in sorted list
 
# initializing list
test_list = [1, 2, 3, 6, 7]
 
# printing original list
print ("The original list is : " + str(test_list))
 
# insert element
k = 5
 
test_list.append(k)
test_list.sort()
# printing result
print ("The list after insertion is : " + str(test_list))

Output

The original list is : [1, 2, 3, 6, 7]
The list after insertion is : [1, 2, 3, 5, 6, 7]

Time Complexity : O(N log N)

Auxiliary Space : O(1)

Method #5 : .insert() 

Python3




from bisect import bisect_left
test_list = [1, 2, 3, 6, 7]
k = 5
index = bisect_left(test_list, k)
test_list.insert(index, k)
print(test_list)

Output

[1, 2, 3, 5, 6, 7]

Time Complexity : O(N)

Auxiliary Space : O(1)

Method #6: Using list comprehension

Python3




test_list = [1, 2, 3, 6, 7]
# printing original list
print ("The original list is : " + str(test_list))
k = 5
res = [x for x in test_list if x < k] + [k] + [x for x in test_list if x >= k]
print(res)
#This code is contributed by Vinay pinjala.

Output

The original list is : [1, 2, 3, 6, 7]
[1, 2, 3, 5, 6, 7]

Time Complexity : O(N)

Auxiliary Space : O(N)


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