Example 1:
-Input: +Input: Accounts table: +----+----------+ | id | name | @@ -80,13 +80,13 @@ Logins table: | 1 | 2020-06-07 | | 7 | 2020-06-10 | +----+------------+ -Output: +Output: +----+----------+ | id | name | +----+----------+ | 7 | Jonathan | +----+----------+ -Explanation: +Explanation: User Winston with id = 1 logged in 2 times only in 2 different days, so, Winston is not an active user. User Jonathan with id = 7 logged in 7 times in 6 different days, five of them were consecutive days, so, Jonathan is an active user.@@ -100,7 +100,13 @@ User Jonathan with id = 7 logged in 7 times in 6 different days, five of them we -### Solution 1 +### Solution 1: Using Window Functions + +First, we join the `Logins` table and the `Accounts` table, and remove duplicates to get the temporary table `T`. + +Then, we use the window function `ROW_NUMBER()` to calculate the base login date `g` for each user `id`. If a user logs in for 5 consecutive days, their `g` values are the same. + +Finally, we group by `id` and `g` to count the number of logins for each user. If the number of logins is greater than or equal to 5, then the user is considered active. @@ -108,18 +114,30 @@ User Jonathan with id = 7 logged in 7 times in 6 different days, five of them we ```sql # Write your MySQL query statement below -WITH t AS - (SELECT *, - SUM(id) over(partition by id - ORDER BY login_date range interval 4 day preceding)/id cnt - FROM - (SELECT DISTINCT * - FROM Accounts - JOIN Logins using(id) ) tt ) - SELECT DISTINCT id, - name -FROM t -WHERE cnt=5; +WITH + T AS ( + SELECT DISTINCT * + FROM + Logins + JOIN Accounts USING (id) + ), + P AS ( + SELECT + *, + DATE_SUB( + login_date, + INTERVAL ROW_NUMBER() OVER ( + PARTITION BY id + ORDER BY login_date + ) DAY + ) g + FROM T + ) +SELECT DISTINCT id, name +FROM P +GROUP BY id, g +HAVING COUNT(*) >= 5 +ORDER BY 1; ``` diff --git a/solution/1400-1499/1454.Active Users/Solution.sql b/solution/1400-1499/1454.Active Users/Solution.sql index 719eff9fce0c8..36b04a638ef0d 100644 --- a/solution/1400-1499/1454.Active Users/Solution.sql +++ b/solution/1400-1499/1454.Active Users/Solution.sql @@ -1,13 +1,25 @@ # Write your MySQL query statement below -WITH t AS - (SELECT *, - SUM(id) over(partition by id - ORDER BY login_date range interval 4 day preceding)/id cnt - FROM - (SELECT DISTINCT * - FROM Accounts - JOIN Logins using(id) ) tt ) - SELECT DISTINCT id, - name -FROM t -WHERE cnt=5; +WITH + T AS ( + SELECT DISTINCT * + FROM + Logins + JOIN Accounts USING (id) + ), + P AS ( + SELECT + *, + DATE_SUB( + login_date, + INTERVAL ROW_NUMBER() OVER ( + PARTITION BY id + ORDER BY login_date + ) DAY + ) g + FROM T + ) +SELECT DISTINCT id, name +FROM P +GROUP BY id, g +HAVING COUNT(*) >= 5 +ORDER BY 1; diff --git a/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README.md b/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README.md index 62cb63f2eb800..b65cccfe7cadc 100644 --- a/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README.md +++ b/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README.md @@ -71,9 +71,9 @@ tags: ### 方法一：字符串分割 -将 $sentence$ 按空格分割为 $words$，然后遍历 $words$，检查 $words[i]$ 是否是 $searchWord$ 的前缀，是则返回 $i+1$。若遍历结束，所有单词都不满足，返回 $-1$。 +我们将 $\textit{sentence}$ 按空格分割为 $\textit{words}$，然后遍历 $\textit{words}$，检查 $\textit{words}[i]$ 是否是 $\textit{searchWord}$ 的前缀，是则返回 $i+1$。若遍历结束，所有单词都不满足，返回 $-1$。 -时间复杂度 $O(mn)$。其中 $m$ 是 $sentence$ 的长度，而 $n$ 是 $searchWord$ 的长度。 +时间复杂度 $O(m \times n)$，空间复杂度 $O(m)$。其中 $m$ 和 $n$ 分别是 $\textit{sentence}$ 和 $\textit{searchWord}$ 的长度。 @@ -176,9 +176,9 @@ class Solution { * @return Integer */ function isPrefixOfWord($sentence, $searchWord) { - $arr = explode(' ', $sentence); - for ($i = 0; $i < count($arr); $i++) { - if (strpos($arr[$i], $searchWord) === 0) { + $words = explode(' ', $sentence); + for ($i = 0; $i < count($words); ++$i) { + if (strpos($words[$i], $searchWord) === 0) { return $i + 1; } } diff --git a/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README_EN.md b/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README_EN.md index ff87165460c60..e97cf29ad7315 100644 --- a/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README_EN.md +++ b/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README_EN.md @@ -67,7 +67,11 @@ tags: -### Solution 1 +### Solution 1: String Splitting + +We split $\textit{sentence}$ by spaces into $\textit{words}$, then iterate through $\textit{words}$ to check if $\textit{words}[i]$ is a prefix of $\textit{searchWord}$. If it is, we return $i+1$. If the iteration completes and no words satisfy the condition, we return $-1$. + +The time complexity is $O(m \times n)$, and the space complexity is $O(m)$. Here, $m$ and $n$ are the lengths of $\textit{sentence}$ and $\textit{searchWord}$, respectively. @@ -170,9 +174,9 @@ class Solution { * @return Integer */ function isPrefixOfWord($sentence, $searchWord) { - $arr = explode(' ', $sentence); - for ($i = 0; $i < count($arr); $i++) { - if (strpos($arr[$i], $searchWord) === 0) { + $words = explode(' ', $sentence); + for ($i = 0; $i < count($words); ++$i) { + if (strpos($words[$i], $searchWord) === 0) { return $i + 1; } } diff --git a/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/Solution.php b/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/Solution.php index 2025bc4f30248..2e5d79c9f4e00 100644 --- a/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/Solution.php +++ b/solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/Solution.php @@ -5,9 +5,9 @@ class Solution { * @return Integer */ function isPrefixOfWord($sentence, $searchWord) { - $arr = explode(' ', $sentence); - for ($i = 0; $i < count($arr); $i++) { - if (strpos($arr[$i], $searchWord) === 0) { + $words = explode(' ', $sentence); + for ($i = 0; $i < count($words); ++$i) { + if (strpos($words[$i], $searchWord) === 0) { return $i + 1; } }