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# Functiondef 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# Initializing lists A = [1, 2]B = [1, 2, 3, 1, 1, 2, 2]# Printing resultprint(removeElements(A, B)) |
True
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
# Python3 program to Remove elements of# list that repeated less than k timesdef removeElements(A, B): n = len(A) return any(A == B[i:i + n] for i in range(len(B)-n + 1))# Initializing listsA = [1, 2]B = [1, 2, 3, 1, 1, 2, 2]print(removeElements(A, B)) |
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 timesdef removeElements(A, B): return ', '.join(map(str, A)) in ', '.join(map(str, B))# Initializing listsA = ['x', 'y', 'z']B = ['x', 'a', 'y', 'x', 'b', 'z']print(removeElements(A, B)) |
False
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 redef check_list_contained(A, B): # convert list A to string A_str = ' '.join(map(str, A)) # convert list B to string B_str = ' '.join(map(str, B)) # find all instances of A within B instances = re.findall(A_str, B_str) # return True if any instances were found, False otherwise return len(instances) > 0# Initializing listsA = ['x', 'y', 'z']B = ['x', 'a', 'y', 'x', 'b', 'z']print(check_list_contained(A, B))# This code is contributed by Edula Vinay Kumar Reddy |
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
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:])# Initializing listsA = [1, 2]B = [1, 2, 3, 1, 1, 2, 2]res = is_sublist(A, B)# Printing the resultprint(res)# This code contributed by tvsk |
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.
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 npdef check_list_contained(A, B): # convert list A to numpy array A_arr = np.array(A) # convert list B to numpy array 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# Initializing listsA = [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.
