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Publish：Oct 27, 2024
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Codeblock

hello_sql_consecutive_days.sql
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-- SQL1：user who has logged in for more than 3 consecutive days
grouped_data AS (
    SELECT 
        user_id, 
        COUNT(*) AS consecutive_days,
        MIN(login_date) AS start_date,
        MAX(login_date) AS end_date
    FROM 
        (
	    SELECT 
             user_id, 
             login_date, 
             DATEDIFF(login_date, '1970-01-01') AS date_diff,
             ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY login_date) AS rn1,
             DATEDIFF(login_date, '1970-01-01') - ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY login_date) AS rn2
         FROM 
             user_info
		) login_data -- 
    GROUP BY 
        user_id, rn2  # Map consecutive dates to the same rn2 value
    HAVING 
        COUNT(*) >= 3
)


SELECT 
    DISTINCT user_id
FROM 
    grouped_data;
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import heapq
from typing import List

def smallest_k_numbers(arr: List[int], k: int) -> List[int]:
    if k == 0 or not arr:
        return []

    # Build a max heap using negative numbers from the first k elements
    heap = [-x for x in arr[:k]]
    heapq.heapify(heap)

    # Traverse the remaining elements; if the current element is smaller than the heap's top element, replace the heap's top
    for num in arr[k:]:
        if -num > heap[0]:
            heapq.heappop(heap)
            heapq.heappush(heap, -num)

    # Convert the elements in the heap back to positive numbers and return them
    return [-x for x in heap]

# Example usage
arr = [4, 5, 1, 6, 2, 7, 3, 8]
k = 4
output = smallest_k_numbers(arr, k)
print(output)  # Output: [1, 2, 3, 4] in any order