LeetCode Python Solutions: 350. Intersection of Two Arrays II - An Alternative Solution #coding

ZeroStress LeetCode Python Solutions: 350. Intersection of Two Arrays II - Alternative Solution #coding #python #leetcode

The approach to solving the "Intersection of Two Arrays II" problem can be summarized as follows:

Import the Counter class from the collections module.

Create two Counter objects, counter1 and counter2, by passing in the respective arrays, nums1 and nums2, to the Counter constructor. These counters will count the frequency of each number in the arrays.

Initialize an empty list called intersection to store the common elements.

Iterate through the elements in counter1 using a for loop.

For each element, check if it exists in counter2 using the in operator.

If the element is present in both counters, calculate the minimum count of that element using the min function with counter1[num] and counter2[num].

Extend the intersection list by adding the current element repeated count times. This is done by using the extend method and multiplying [num] by count.

After iterating through all elements in counter1, return the intersection list as the final result.
The approach utilizes the Counter class to count the frequencies of elements in both arrays. By iterating through one counter and checking for the existence of each element in the other counter, we can find the common elements. The min function helps determine the minimum count of each element, ensuring that it appears in the intersection the correct number of times.

By following this approach, we can efficiently find the intersection of two arrays while considering the frequencies of elements.

The time complexity of the given solution is O(m + n), where m is the length of nums1 and n is the length of nums2. Here's the breakdown:

Creating the Counter objects for nums1 and nums2 takes O(m + n) time. This is because we iterate through each array once to count the frequencies of the elements.

Iterating through counter1 takes O(m) time, as it has at most m unique elements.

For each element in counter1, we perform a lookup in counter2, which takes O(1) time on average. This is because dictionaries (which Counter is based on) have a constant-time average lookup.

Extending the intersection list takes O(min(m, n)) time, as we add the current element a number of times equal to the minimum count between counter1[num] and counter2[num].

Returning the intersection list at the end of the function does not contribute to the time complexity.
Therefore, the overall time complexity is O(m + n) for this solution.

The space complexity of the solution is O(m + n) as well. This is because we create two Counter objects, counter1 and counter2, which store the frequencies of elements in nums1 and nums2 respectively. In the worst case, these counters can store all the unique elements from both arrays, resulting in a space complexity of O(m + n). Additionally, the intersection list will store the common elements, but its space requirements are limited by the maximum number of unique elements between nums1 and nums2. Therefore, the space complexity of the solution is O(m + n).
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03:36 DE-CODING
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[NeedCodeLeetCode]

from collections import Counter # Counter is a convenient data structure in Python for counting the occurrences of elements in a list.

class Solution:
def intersect(self, nums1, nums2):
counter1 = Counter(nums1) # count the occurrences of elements in nums1: nums1 = [1, 2, 2, 1, 3] -> {1:2, 2:2, 3:1}
counter2 = Counter(nums2) # nums2 = [2, 2] -> {2:2}
intersection = [] # initialize an empty list called intersection to store the common elements between nums1 and nums2.

for num in counter1: # This loop iterates over the elements in counter1 (which represents the unique elements in nums1).
if num in counter2: # and checks if each element is present in counter2 (which represents the unique elements in nums2).
count = min(counter1[num], counter2[num]) #If an element is found in both lists, it calculates the minimum count of that element in counter1 and counter2 using the min() function.
# [num] * count creates a new list that contains the element num repeated count times.
# For example, if num is 2 and count is 3, then [num] * count would result in [2, 2, 2]
intersection.extend([num] * count) # Extend the intersection list with [num] repeated count times. This ensures that the common element appears the same number of times as its minimum count in both lists.

return intersection

[NeedCodeLeetCode]

The space complexity of the solution is O(m + n) as well. This is because we create two Counter objects, counter1 and counter2, which store the frequencies of elements in nums1 and nums2 respectively. In the worst case, these counters can store all the unique elements from both arrays, resulting in a space complexity of O(m + n). Additionally, the intersection list will store the common elements, but its space requirements are limited by the maximum number of unique elements between nums1 and nums2. Therefore, the space complexity of the solution is O(m + n).

[NeedCodeLeetCode]

The time complexity of the given solution is O(m + n), where m is the length of nums1 and n is the length of nums2. Here's the breakdown:

Creating the Counter objects for nums1 and nums2 takes O(m + n) time. This is because we iterate through each array once to count the frequencies of the elements.
Iterating through counter1 takes O(m) time, as it has at most m unique elements.
For each element in counter1, we perform a lookup in counter2, which takes O(1) time on average. This is because dictionaries (which Counter is based on) have a constant-time average lookup.
Extending the intersection list takes O(min(m, n)) time, as we add the current element a number of times equal to the minimum count between counter1[num] and counter2[num].
Returning the intersection list at the end of the function does not contribute to the time complexity.