Python | Check if a list is contained in another list

Given two lists A and B, write a Python program to Check if list A is contained in list B without breaking A’s order. 

Examples:

Input : A = [1, 2], B = [1, 2, 3, 1, 1, 2, 2]
Output : True

Input : A = ['x', 'y', 'z'], B = ['x', 'a', 'y', 'x', 'b', 'z']
Output : False

Approach #1 : Naive Approach A simple naive approach is to use two for loops and check if the whole list A is contained within list B or not. If such a position is met in list A, then break the loop and return true, otherwise false 

Python3

# Python3 program to Check if a list is
# contained in another list without breaking order

def removeElements(A, B):
    for i in range(len(B)-len(A)+1):
        for j in range(len(A)):
            if B[i + j] != A[j]:
                break
        else:
            return True
    return False

# Driver code
A = [1, 2]
B = [1, 2, 3, 1, 1, 2, 2]
print(removeElements(A, B))

Output:

True

Time Complexity: O(n*n)
Auxiliary Space: O(1)

Approach #2 : List comprehension A more efficient approach is to use List comprehension. We first initialize ‘n’ with length of A. Now use a for loop till len(B)-n and check in each iteration if A == B[i:i+n] or not. 

Python3

# Python3 program to Remove elements of
# list that repeated less than k times

def removeElements(A, B):
    n = len(A)
    return any(A == B[i:i + n] for i in range(len(B)-n + 1))


# Driver code
A = [1, 2]
B = [1, 2, 3, 1, 1, 2, 2]
print(removeElements(A, B))

Output:

True

Time complexity: O(n^2). where A and B are both of length n,
Auxiliary space: O(n), where n is the length of B.

Approach #3 : Using join and map module Here we use join to join both lists to strings and then use in operator to check if list A is contained in B or not. 

Python3

# Python3 program to Remove elements of
# list that repeated less than k times


def removeElements(A, B):
    return ', '.join(map(str, A)) in ', '.join(map(str, B))

# Driver code
A = ['x', 'y', 'z']
B = ['x', 'a', 'y', 'x', 'b', 'z']
print(removeElements(A, B))

Output:

False

Time complexity : O(m*n) , where m is length of A and n is length of B.
Auxiliary space : O(1),where m is length of A and n is length of B.

 Approach #4:  Using re To check if a list is contained in another list using the Python re (regular expression) module, you can use the re.findall() function to find all instances of the list A within the list B as a string. If the number of instances found is greater than 0, it means that list A is contained within list B.

Here is an example of how this can be done:

Python3

import re

def check_list_contained(A, B):
    A_str = ' '.join(map(str, A))  # convert list A to string
    B_str = ' '.join(map(str, B))  # convert list B to string
    instances = re.findall(A_str, B_str)  # find all instances of A within B
    return len(instances) > 0  # return True if any instances were found, False otherwise

# test the function
A = ['x', 'y', 'z']
B = ['x', 'a', 'y', 'x', 'b', 'z']
print(check_list_contained(A, B))  # False
#This code is contributed by Edula Vinay Kumar Reddy

Output

False

Time complexity : O(m*n) , where m is length of A and n is length of B.
Auxiliary space : O(1),where m is length of A and n is length of B.

Approach #5: Using Recursive method.

Sure, here’s a step-by-step algorithm for the is_sublist function:

1. If list A is empty, return True, as an empty list is always a sublist of any list.
2. If list B is empty, return False, as an empty list cannot contain any sublist.
3. If A[0] matches B[0], recursively check the rest of the elements of A and B by calling the is_sublist method with A[1:] and B[1:].
4. If A[0] does not match B[0], recursively check if A is a sublist of the remaining elements of B by calling the is_sublist method with A and B[1:]
5. Return True if the previous recursive call returned True, indicating that A is a sublist of B, or False otherwise.

Python3

def is_sublist(A, B):
    if not A:
        return True
    if not B:
        return False
    if A[0] == B[0]:
        return is_sublist(A[1:], B[1:])
    return is_sublist(A, B[1:])

A = [1, 2]
B = [1, 2, 3, 1, 1, 2, 2]

res=is_sublist(A, B)
print(res)
#this code contributed by tvsk

Output

True

Time complexity: O(n * m), where n is the length of list A and m is the length of list B. This is because the function needs to traverse both lists in the worst case scenario, and the length of list B may need to be traversed multiple times depending on the position of the matching element of A.

Auxiliary space: O(max(n, m)), where n is the length of list A and m is the length of list B. This is because the function makes recursive calls, and each call adds a new stack frame to the call stack, potentially using up additional memory. However, the maximum depth of the call stack is bounded by the maximum length of A or B, so the space used is proportional to the maximum of the two lengths.