Python | Check if a list is contained in another list
Last Updated :
10 May, 2023
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
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
A = [1, 2]
B = [1, 2, 3, 1, 1, 2, 2]
print(removeElements(A, B))
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Time Complexity: O(n*n)
Auxiliary Space: O(1)
Approach #2: Using list comprehension
A more efficient approach is to use List comprehension. We first initialize ‘n’ with the 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
def removeElements(A, B):
n = len(A)
return any(A == B[i:i + n] for i in range(len(B)-n + 1))
A = [1, 2]
B = [1, 2, 3, 1, 1, 2, 2]
print(removeElements(A, B))
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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
def removeElements(A, B):
return ', '.join(map(str, A)) in ', '.join(map(str, B))
A = ['x', 'y', 'z']
B = ['x', 'a', 'y', 'x', 'b', 'z']
print(removeElements(A, B))
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Time complexity: O(m*n) , where m is the length of A and n is the length of B.
Auxiliary space: O(1),where m is length of A and n is the length of B.
Approach #4: Using regular expressions
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 list A within 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))
B_str = ' '.join(map(str, B))
instances = re.findall(A_str, B_str)
return len(instances) > 0
A = ['x', 'y', 'z']
B = ['x', 'a', 'y', 'x', 'b', 'z']
print(check_list_contained(A, B))
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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
Steps:
- If list A is empty, return True, as an empty list is always a sublist of any list.
- If list B is empty, return False, as an empty list cannot contain any sublist.
- 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:].
- 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:]
- 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)
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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.
Approach #6: Using Numpy
Note: install numpy module using command “pip install numpy”
Steps:
- Import numpy module and define a function named “check_list_contained” that takes two parameters, “A” and “B”.
- Convert “A” and “B” to numpy arrays “A_arr” and “B_arr” respectively.
- Loop through “B_arr” using a for loop and range function for index “i” in range of length of “B_arr”.
- Check if the subarray of “B_arr” starting from index “i” and of length equal to the length of “A_arr” is equal to “A_arr” using the numpy array_equal() function.
- If the above condition is true, then return True, as the list “A” is contained in the list “B”.
- If the loop finishes execution without returning True, return False, as the list “A” is not contained in the list “B”.
- Test the function by passing two list parameters “A” and “B”.
- Print the result of the function call.
Python3
import numpy as np
def check_list_contained(A, B):
A_arr = np.array(A)
B_arr = np.array(B)
for i in range(len(B_arr)):
if np.array_equal(A_arr, B_arr[i:i+len(A_arr)]):
return True
return False
A = [1, 2]
B = [1, 2, 3, 1, 1, 2, 2]
print(check_list_contained(A, B))
A = ['x', 'y', 'z']
B = ['x', 'a', 'y', 'x', 'b', 'z']
print(check_list_contained(A, B))
|
Output:
True
False
Time complexity: O(n^2), where n is the length of B.
Auxiliary space: O(n), where n is the length of B.
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