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20250424.py
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'''
https://leetcode.com/problems/count-complete-subarrays-in-an-array/
Count Complete Subarrays in an Array
You are given an array nums consisting of positive integers.
We call a subarray of an array complete if the following condition is satisfied:
The number of distinct elements in the subarray is equal to the number of distinct elements in the whole array.
Return the number of complete subarrays.
A subarray is a contiguous non-empty part of an array.
Example 1:
Input: nums = [1,3,1,2,2]
Output: 4
Explanation: The complete subarrays are the following: [1,3,1,2], [1,3,1,2,2], [3,1,2] and [3,1,2,2].
Example 2:
Input: nums = [5,5,5,5]
Output: 10
Explanation: The array consists only of the integer 5, so any subarray is complete. The number of subarrays that we can choose is 10.
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 2000
'''
from typing import List
from collections import Counter, defaultdict
class Solution:
def countCompleteSubarrays(self, nums: List[int]) -> int:
occurrences = Counter(nums)
result = 0
n = len(nums)
for left in range(n):
window_counter = defaultdict(int)
for right in range(left, n):
window_counter[nums[right]] += 1
if len(window_counter) == len(occurrences):
result += 1
return result
solution = Solution()
print(solution.countCompleteSubarrays([1,3,1,2,2])) # -> 4
print(solution.countCompleteSubarrays([5,5,5,5])) # -> 10
'''
Pseudocode:
- Create an occurrences dict with keys of numbers in nums and their occurrences as values
- Initialize result as zero
- Declare n as length of nums
- Iterate through nums (outer loop)
- Declare window_counter as an integer defaultdict
- Iterate through the window (inner loop)
- Increment window_counter[nums[right]] by 1
- If length of window_counter and length of occurrences is equal:
- Increment result by 1
- Return result
- Time complexity: O(n^2)
- Space complexity: O(n)
'''