LeetCode Python Solutions: 349 Intersection of Two Arrays #python #coding #leetcode

ZeroStress LeetCode Python Solutions: 349 Intersection of Two Arrays Problem #python #leetcode

The algorithm used in this code is straightforward. By converting the input lists to sets, duplicate elements are eliminated, and the & operator is used to find the common elements between the two sets. Converting the set back to a list ensures the desired output format.

The time complexity of this solution is determined by the set operations and list conversion. Let's analyze each step:

Converting nums1 and nums2 into sets using set() takes O(n) time complexity, where n is the length of the longer array.

Performing the set intersection using the & operator takes O(min(n1, n2)) time complexity, where n1 and n2 are the lengths of the two sets obtained in the previous step.

Converting the resulting set back to a list using list() takes O(k) time complexity, where k is the size of the resulting set.

Therefore, the overall time complexity is O(n + min(n1, n2) + k), where n is the length of the longer array, n1 and n2 are the lengths of the two sets, and k is the size of the resulting set.

As for space complexity:

Converting nums1 and nums2 into sets using set() requires additional space to store the unique elements. The space complexity for this step is O(n1 + n2), where n1 and n2 are the lengths of the two arrays.

The space complexity for the resulting set obtained from the set intersection is O(k), where k is the size of the resulting set.

Converting the set back to a list using list() requires additional space to store the list of elements. The space complexity for this step is O(k).

Therefore, the overall space complexity is O(n1 + n2 + k), where n1 and n2 are the lengths of the two arrays, and k is the size of the resulting list.

In summary, the time complexity of this solution is determined by the lengths of the arrays and the size of the resulting list, while the space complexity depends on the lengths of the arrays and the size of the resulting set/list.

Understanding and applying the concept of sets, set operations, and list conversion is crucial in solving this problem efficiently.

00:00 Code
02:00 Main
08:20 End

[NeedCodeLeetCode]

class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
return list(set(nums1) & set(nums2))

The code defines a class named Solution that encapsulates the solution to the problem. The class has a single method called intersection.

The intersection method takes two input parameters: nums1 and nums2, both of which are lists of integers. The method is expected to return a list of integers.

set(nums1) converts nums1 into a set, removing any duplicate elements. Similarly, set(nums2) converts nums2 into a set.

The & operator is used to perform the set intersection operation between set(nums1) and set(nums2). It returns a new set containing the common elements present in both sets.

list() is used to convert the resulting set back to a list. This is done because the problem requires the result to be returned as a list.

Finally, the resulting list is returned as the solution to the problem.

The algorithm used in this code is straightforward. By converting the input lists to sets, duplicate elements are eliminated, and the & operator is used to find the common elements between the two sets. Converting the set back to a list ensures the desired output format.

The time complexity of this solution is determined by the set operations, which typically have a time complexity of O(n) for each set creation and intersection, where n is the size of the input list. Thus, the overall time complexity is O(n), where n is the maximum size between nums1 and nums2.

The space complexity is determined by the space required for the sets and the resulting list. Since the sets store the unique elements from the input lists, the space complexity is proportional to the number of unique elements. In the worst case, if all elements are unique, the space complexity would be O(n), where n is the sum of the sizes of nums1 and nums2.

Understanding and applying the concept of sets, set operations, and list conversion is crucial in solving this problem efficiently.