# Python: Check if the elements of the first list are contained in the second one regardless of order

## Python List: Exercise - 270 with Solution

Write a Python program to check if the elements of the first list are contained in the second one regardless of order.

- Use count() to check if any value in a has more occurrences than it has in b.
- Return False if any such value is found, True otherwise.

**Sample Solution:**

**Python Code:**

```
# Define a function named 'is_contained_in' to check if all elements in l1 are contained in l2.
# It takes two parameters: 'l1' (the first list) and 'l2' (the second list).
def is_contained_in(l1, l2):
# Iterate through the set of unique elements in l1.
for x in set(l1):
# Check if the count of the element in l1 is greater than the count in l2.
if l1.count(x) > l2.count(x):
# If any element is more frequent in l1 than in l2, return False.
return False
# If all elements in l1 are contained in l2, return True.
return True
# Test the 'is_contained_in' function with different lists.
print(is_contained_in([1, 2], [2, 4, 1])) # Check if [1, 2] is contained in [2, 4, 1].
print(is_contained_in([1], [2, 4, 1])) # Check if [1] is contained in [2, 4, 1].
print(is_contained_in([1, 1], [4, 2, 1])) # Check if [1, 1] is contained in [4, 2, 1].
print(is_contained_in([1, 1], [3, 2, 4, 1, 5, 1])) # Check if [1, 1] is contained in [3, 2, 4, 1, 5, 1].
```

Sample Output:

True True False True

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**Python Code Editor:**

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