Python | Check if a list is contained in another list

Last Updated : 21 Nov, 2019

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 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

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 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

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 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

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