feat: add solutions to lc problem: No.1384

No.1384: Total Sales Amount by Year

Author: yanglbme

solution/1300-1399/1384.Total Sales Amount by Year/README.md

@@ -93,7 +93,44 @@ LC Keychain 在 2019-12-01 至 2020-01-31 期间销售，该产品在2019
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    s.product_id,
+    p.product_name,
+    y.YEAR report_year,
+    s.average_daily_sales * (
+        IF (
+            YEAR (s.period_end) > y.YEAR,
+            y.days_of_year,
+            dayofyear(s.period_end)
+        ) - IF (
+            YEAR (s.period_start) < y.YEAR,
+            1,
+            dayofyear(s.period_start)
+        ) + 1
+    ) total_amount
+FROM
+    Sales s
+    INNER JOIN (
+        SELECT
+            '2018' YEAR,
+            365 days_of_year
+        UNION
+        ALL
+        SELECT
+            '2019' YEAR,
+            365 days_of_year
+        UNION
+        ALL
+        SELECT
+            '2020' YEAR,
+            366 days_of_year
+    ) y ON YEAR (s.period_start) <= y.YEAR
+    AND YEAR (s.period_end) >= y.YEAR
+    INNER JOIN Product p ON p.product_id = s.product_id
+ORDER BY
+    s.product_id,
+    y.YEAR
 ```
 
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solution/1300-1399/1384.Total Sales Amount by Year/README_EN.md

@@ -89,7 +89,44 @@ LC Keychain was sold for the period of 2019-12-01 to 2020-01-31, and there are 3
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    s.product_id,
+    p.product_name,
+    y.YEAR report_year,
+    s.average_daily_sales * (
+        IF (
+            YEAR (s.period_end) > y.YEAR,
+            y.days_of_year,
+            dayofyear(s.period_end)
+        ) - IF (
+            YEAR (s.period_start) < y.YEAR,
+            1,
+            dayofyear(s.period_start)
+        ) + 1
+    ) total_amount
+FROM
+    Sales s
+    INNER JOIN (
+        SELECT
+            '2018' YEAR,
+            365 days_of_year
+        UNION
+        ALL
+        SELECT
+            '2019' YEAR,
+            365 days_of_year
+        UNION
+        ALL
+        SELECT
+            '2020' YEAR,
+            366 days_of_year
+    ) y ON YEAR (s.period_start) <= y.YEAR
+    AND YEAR (s.period_end) >= y.YEAR
+    INNER JOIN Product p ON p.product_id = s.product_id
+ORDER BY
+    s.product_id,
+    y.YEAR
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1384.Total Sales Amount by Year/Solution.sql

@@ -0,0 +1,38 @@
+# Write your MySQL query statement below
+SELECT
+    s.product_id,
+    p.product_name,
+    y.YEAR report_year,
+    s.average_daily_sales * (
+        IF (
+            YEAR (s.period_end) > y.YEAR,
+            y.days_of_year,
+            dayofyear(s.period_end)
+        ) - IF (
+            YEAR (s.period_start) < y.YEAR,
+            1,
+            dayofyear(s.period_start)
+        ) + 1
+    ) total_amount
+FROM
+    Sales s
+    INNER JOIN (
+        SELECT
+            '2018' YEAR,
+            365 days_of_year
+        UNION
+        ALL
+        SELECT
+            '2019' YEAR,
+            365 days_of_year
+        UNION
+        ALL
+        SELECT
+            '2020' YEAR,
+            366 days_of_year
+    ) y ON YEAR (s.period_start) <= y.YEAR
+    AND YEAR (s.period_end) >= y.YEAR
+    INNER JOIN Product p ON p.product_id = s.product_id
+ORDER BY
+    s.product_id,
+    y.YEAR

solution/1300-1399/1385.Find the Distance Value Between Two Arrays/README.md

@@ -70,15 +70,15 @@
 
 由于 `arr1` 和 `arr2` 的长度不超过 500，因此可以直接暴力遍历。
 
-时间复杂度 $O(mn)$，其中 $m$ 为 `arr1` 的长度，$n$ 为 `arr2` 的长度。
+时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别为 `arr1` 和 `arr2` 的长度。
 
 **方法二：二分查找**
 
 对于 `arr1` 中的每个元素 `a`，若在 `arr2` 中存在 `b`，使得 `b ∈ [a - d, a + d]`，那么就符合距离要求，不进行累加。
 
 因此，可以先对 `arr2` 进行排序。然后对于每个元素 `a`，二分枚举 `arr2` 判断是否存在符合距离要求的 `b`。
 
-时间复杂度 $O((m + n)logn)$。
+时间复杂度 $O((m + n) \times \log n)$。
 
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