Python – First occurrence of one list in another

Given two lists, the task is to write a Python program to extract the first element that occurs in list 1 from list 2.

Examples:

Input : test_list1 = [1, 6, 3, 7, 8, 9, 2], test_list2 = [4, 10, 8, 2, 0, 11]

Output : 8

Explanation : 8 is first element from list 2, that occurs in list 1, in 5th index.

Input : test_list1 = [1, 6, 3, 7, 8, 9, 2], test_list2 = [4, 10, 18, 12, 0, 11]

Output : None

Explanation : No element of list 2 found in list 1.

Approach 1: Using set() + next()

In this, initially, the check container is converted to set, and each element is checked using next() and generator expression. The next() function returns the first element matching, else if no match element is found, None is returned.

Python3

# Python3 code to demonstrate working of
# First occurrence of one list in another
# Using next() + set()

# initializing lists
test_list1 = [1, 6, 3, 7, 8, 9, 2]
test_list2 = [4, 10, 8, 2, 0, 11]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))

# converting test list to sets
test_list2 = set(test_list2)

# stops when 1st match element is found
res = next((ele for ele in test_list1 if ele in test_list2), None)

# printing result
print("First element in list 1 from 2 : " + str(res))

Output

The original list 1 is : [1, 6, 3, 7, 8, 9, 2]
The original list 2 is : [4, 10, 8, 2, 0, 11]
First element in list 1 from 2 : 8

Time complexity: O(n), where n is the length of the test_list. The set() + next() takes O(n) time
Auxiliary Space: O(1), extra space required is not required

Approach 2: Using for loops

Python3

# Python3 code to demonstrate working of
# First occurrence of one list in another

# initializing lists
test_list1 = [1, 6, 3, 7, 8, 9, 2]
test_list2 = [4, 10, 18, 12, 0, 11]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))


# stops when 1st match element is found
res=None
for i in test_list2:
    if i in test_list1:
        res=i
        break

# printing result
print("First element in list 1 from 2 : " + str(res))

Output

The original list 1 is : [1, 6, 3, 7, 8, 9, 2]
The original list 2 is : [4, 10, 18, 12, 0, 11]
First element in list 1 from 2 : None

Time Complexity: O(n)
Auxiliary Space: O(n)

Approach 3:  using Counter() function

Python3

# Python3 code to demonstrate working of
# First occurrence of one list in another
from collections import Counter
# initializing lists
test_list1 = [1, 6, 3, 7, 8, 9, 2]
test_list2 = [4, 10, 18, 12, 0, 11]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
freq1= Counter(test_list1)

# stops when 1st match element is found
res=None
for i in test_list2:
    if i in freq1.keys():
        res=i
        break

# printing result
print("First element in list 1 from 2 : " + str(res))

Output

The original list 1 is : [1, 6, 3, 7, 8, 9, 2]
The original list 2 is : [4, 10, 18, 12, 0, 11]
First element in list 1 from 2 : None

Time Complexity: O(N)

Auxiliary Space : O(N)

Approach 4: Using recursive method.

Python3

# Python3 code to demonstrate working of
# First occurrence of one list in another
#using recursive approach
def first_match(start,lst1,lst2):
  if start==len(lst1):  #base condition
    return None
  if lst1[start] in lst2:   #checking if lst1[start] present in lst2 or not
    return lst1[start]
  return first_match(start+1,lst1,lst2)  #calling recursive call

# initializing lists
test_list1 = [1, 6, 3, 7, 8, 9, 2]
test_list2 = [4, 10, 8, 12, 0, 11]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))



res=first_match(0,test_list1,test_list2)

# printing result
print("First element in list 1 from 2 : " + str(res))

#this code contributed by tvsk

Output

The original list 1 is : [1, 6, 3, 7, 8, 9, 2]
The original list 2 is : [4, 10, 8, 12, 0, 11]
First element in list 1 from 2 : 8

Time Complexity: O(N)

Auxiliary Space : O(N)

Approach 5: Using intersection method.

Python3

test_list1 = [1, 6, 3, 7, 8, 9, 2]
test_list2 = [4, 10, 8, 2, 0, 11]
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
result = set(test_list1).intersection(test_list2)
if result:
    res = result.pop()
else:
    res = None
print("First element in list 1 from 2 : " + str(res))
#This code is contributed by Jyothi pinjala.

Output

The original list 1 is : [1, 6, 3, 7, 8, 9, 2]
The original list 2 is : [4, 10, 8, 2, 0, 11]
First element in list 1 from 2 : 8

Time Complexity: O(N)

Auxiliary Space : O(N)

Approach 6: Using list comprehension

Python3

# Python3 code to demonstrate working of
# First occurrence of one list in another

# initializing lists
test_list1 = [1, 6, 3, 7, 8, 9, 2]
test_list2 = [4, 10, 8, 2, 0, 11]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))

# using list comprehension
try:
    # Find the first element in test_list2 that is also in test_list1
    res = next(i for i in [i for i in test_list2 if i in test_list1])
except StopIteration:
    # If there are no matching elements, set res to None
    res = None

# printing result
print("First element in list 1 from 2 : " + str(res))
#This code is contributed by Vinay Pinjala.

Output

The original list 1 is : [1, 6, 3, 7, 8, 9, 2]
The original list 2 is : [4, 10, 8, 2, 0, 11]
First element in list 1 from 2 : 8

Time Complexity: O(N)

Auxiliary Space : O(N)

Approach 7: Using filter method

Algorithm: 

• InitaInitializelize list1. 
• Initialize list2. 
• Use the filter method on list1 that filter out element of list1 present in list2.
• Print the first element. 

Python

# Python3 code to demonstrate working of
# First occurrence of one list in another

# initializing lists
test_list1 = [1, 6, 3, 7, 8, 9, 2]
test_list2 = [4, 10, 8, 12, 0, 11]

# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))


# filter all comman element from list.
res = [filter(lambda a : ( a in test_list2 ), test_list1  )][0]

# printing result
print("First element in list 1 from 2 : " + str(res))

Output

The original list 1 is : [1, 6, 3, 7, 8, 9, 2]
The original list 2 is : [4, 10, 8, 12, 0, 11]
First element in list 1 from 2 : [8]

Time Complexity: O(N) where N is the length of list1. 

Auxiliary Space: O(M) Where M is the length of the new filter list.