Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [1.00000,-1.00000,-1.00000,3.00000,5.00000,6.00000]
-Explanation:
+Explanation:
Window position Median
--------------- -----
[1 3 -1] -3 5 3 6 7 1
diff --git a/solution/0500-0599/0588.Design In-Memory File System/README.md b/solution/0500-0599/0588.Design In-Memory File System/README.md
index b543fc76a3ce8..357f38fcca065 100644
--- a/solution/0500-0599/0588.Design In-Memory File System/README.md
+++ b/solution/0500-0599/0588.Design In-Memory File System/README.md
@@ -77,6 +77,7 @@ fileSystem.readContentFromFile("/a/b/c/d"); // 返回 "hello"
path 和 filePath 都是绝对路径，除非是根目录 ‘/’ 自身，其他路径都是以 ‘/’ 开头且 不 以 ‘/’ 结束。addContentToFile 中的文件的父目录都存在。1 <= content.length <= 50ls, mkdir, addContentToFile, and readContentFromFile 最多被调用 300 次如果对 words2 中的每一个单词 b，b 都是 a 的子集，那么我们称 words1 中的单词 a 是 通用单词 。
以数组形式返回 words1 中所有的通用单词。你可以按 任意顺序 返回答案。
以数组形式返回 words1 中所有的 通用 单词。你可以按 任意顺序 返回答案。
示例 1：
+示例 1：
--输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"] -输出：["facebook","google","leetcode"] -+
输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]
-示例 2：
+输出：["facebook","google","leetcode"]
+-输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["l","e"] -输出：["apple","google","leetcode"] -+
示例 2：
-示例 3：
+输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lc","eo"]
--输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","oo"] -输出：["facebook","google"] -+
输出：["leetcode"]
+示例 4：
+示例 3：
--输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lo","eo"] -输出：["google","leetcode"] -+
输入：words1 = ["acaac","cccbb","aacbb","caacc","bcbbb"], words2 = ["c","cc","b"]
-示例 5：
- --输入：words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["ec","oc","ceo"] -输出：["facebook","leetcode"] -+
输出：["cccbb"]
+diff --git a/solution/0900-0999/0975.Odd Even Jump/README_EN.md b/solution/0900-0999/0975.Odd Even Jump/README_EN.md index f3d8f0f75c976..99f76a4a016e2 100644 --- a/solution/0900-0999/0975.Odd Even Jump/README_EN.md +++ b/solution/0900-0999/0975.Odd Even Jump/README_EN.md @@ -40,7 +40,7 @@ tags:
Input: arr = [10,13,12,14,15]
Output: 2
-Explanation:
+Explanation:
From starting index i = 0, we can make our 1st jump to i = 2 (since arr[2] is the smallest among arr[1], arr[2], arr[3], arr[4] that is greater or equal to arr[0]), then we cannot jump any more.
From starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more.
From starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end.
@@ -54,7 +54,7 @@ jumps.
Input: arr = [2,3,1,1,4]
Output: 3
-Explanation:
+Explanation:
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd-numbered), we first jump to i = 1 because arr[1] is the smallest value in [arr[1], arr[2], arr[3], arr[4]] that is greater than or equal to arr[0].
During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr[2] is the largest value in [arr[2], arr[3], arr[4]] that is less than or equal to arr[1]. arr[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3
diff --git a/solution/1100-1199/1172.Dinner Plate Stacks/README.md b/solution/1100-1199/1172.Dinner Plate Stacks/README.md
index ced8ea8849c43..dcc64da0f7884 100644
--- a/solution/1100-1199/1172.Dinner Plate Stacks/README.md
+++ b/solution/1100-1199/1172.Dinner Plate Stacks/README.md
@@ -65,14 +65,14 @@ D.popAtStack(0); // 返回 20。栈的现状为： 4 21
﹈ ﹈ ﹈
D.popAtStack(2); // 返回 21。栈的现状为： 4
1 3 5
- ﹈ ﹈ ﹈
+ ﹈ ﹈ ﹈
D.pop() // 返回 5。栈的现状为： 4
- 1 3
- ﹈ ﹈
-D.pop() // 返回 4。栈的现状为： 1 3
- ﹈ ﹈
-D.pop() // 返回 3。栈的现状为： 1
- ﹈
+ 1 3
+ ﹈ ﹈
+D.pop() // 返回 4。栈的现状为： 1 3
+ ﹈ ﹈
+D.pop() // 返回 3。栈的现状为： 1
+ ﹈
D.pop() // 返回 1。现在没有栈。
D.pop() // 返回 -1。仍然没有栈。
diff --git a/solution/1100-1199/1172.Dinner Plate Stacks/README_EN.md b/solution/1100-1199/1172.Dinner Plate Stacks/README_EN.md
index a1b5f39c39692..7f1837953bb7a 100644
--- a/solution/1100-1199/1172.Dinner Plate Stacks/README_EN.md
+++ b/solution/1100-1199/1172.Dinner Plate Stacks/README_EN.md
@@ -42,7 +42,7 @@ tags:
Output
[null, null, null, null, null, null, 2, null, null, 20, 21, 5, 4, 3, 1, -1]
-Explanation:
+Explanation:
DinnerPlates D = DinnerPlates(2); // Initialize with capacity = 2
D.push(1);
D.push(2);
@@ -65,14 +65,14 @@ D.popAtStack(0); // Returns 20. The stacks are now: 4 21
﹈ ﹈ ﹈
D.popAtStack(2); // Returns 21. The stacks are now: 4
1 3 5
- ﹈ ﹈ ﹈
+ ﹈ ﹈ ﹈
D.pop() // Returns 5. The stacks are now: 4
- 1 3
- ﹈ ﹈
-D.pop() // Returns 4. The stacks are now: 1 3
- ﹈ ﹈
-D.pop() // Returns 3. The stacks are now: 1
- ﹈
+ 1 3
+ ﹈ ﹈
+D.pop() // Returns 4. The stacks are now: 1 3
+ ﹈ ﹈
+D.pop() // Returns 3. The stacks are now: 1
+ ﹈
D.pop() // Returns 1. There are no stacks.
D.pop() // Returns -1. There are still no stacks.
diff --git a/solution/1300-1399/1348.Tweet Counts Per Frequency/README.md b/solution/1300-1399/1348.Tweet Counts Per Frequency/README.md
index 83dbe80cffbf8..d3cf52c0d5cfe 100644
--- a/solution/1300-1399/1348.Tweet Counts Per Frequency/README.md
+++ b/solution/1300-1399/1348.Tweet Counts Per Frequency/README.md
@@ -68,7 +68,7 @@ tweetCounts.recordTweet("tweet3", 0);
tweetCounts.recordTweet("tweet3", 60);
tweetCounts.recordTweet("tweet3", 10); // "tweet3" 发布推文的时间分别是 0, 10 和 60 。
tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 59); // 返回 [2]。统计频率是每分钟（60 秒），因此只有一个有效时间间隔 [0,60> - > 2 条推文。
-tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 60); // 返回 [2,1]。统计频率是每分钟（60 秒），因此有两个有效时间间隔 1) [0,60> - > 2 条推文，和 2) [60,61> - > 1 条推文。
+tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 60); // 返回 [2,1]。统计频率是每分钟（60 秒），因此有两个有效时间间隔 1) [0,60> - > 2 条推文，和 2) [60,61> - > 1 条推文。
tweetCounts.recordTweet("tweet3", 120); // "tweet3" 发布推文的时间分别是 0, 10, 60 和 120 。
tweetCounts.getTweetCountsPerFrequency("hour", "tweet3", 0, 210); // 返回 [4]。统计频率是每小时（3600 秒），因此只有一个有效时间间隔 [0,211> - > 4 条推文。
diff --git a/solution/1400-1499/1400.Construct K Palindrome Strings/README.md b/solution/1400-1499/1400.Construct K Palindrome Strings/README.md
index b8ec712c5656a..fe8d098302ddd 100644
--- a/solution/1400-1499/1400.Construct K Palindrome Strings/README.md
+++ b/solution/1400-1499/1400.Construct K Palindrome Strings/README.md
@@ -21,7 +21,7 @@ tags:
-给你一个字符串 s 和一个整数 k 。请你用 s 字符串中 所有字符 构造 k 个非空 回文串 。
给你一个字符串 s 和一个整数 k 。请你用 s 字符串中 所有字符 构造 k 个非空 回文串 。
如果你可以用 s 中所有字符构造 k 个回文字符串，那么请你返回 True ，否则返回 False 。
示例 1：
-输入：s = "annabelle", k = 2 +输入：s = "annabelle", k = 2 输出：true 解释：可以用 s 中所有字符构造 2 个回文字符串。 -一些可行的构造方案包括："anna" + "elble"，"anbna" + "elle"，"anellena" + "b" +一些可行的构造方案包括："anna" + "elble"，"anbna" + "elle"，"anellena" + "b"
示例 2：
-输入：s = "leetcode", k = 3 +输入：s = "leetcode", k = 3 输出：false 解释：无法用 s 中所有字符构造 3 个回文串。@@ -47,35 +47,19 @@ tags:
示例 3：
-输入：s = "true", k = 4 +输入：s = "true", k = 4 输出：true 解释：唯一可行的方案是让 s 中每个字符单独构成一个字符串。-
示例 4：
- --输入：s = "yzyzyzyzyzyzyzy", k = 2 -输出：true -解释：你只需要将所有的 z 放在一个字符串中，所有的 y 放在另一个字符串中。那么两个字符串都是回文串。 -- -
示例 5：
- --输入：s = "cr", k = 7 -输出：false -解释：我们没有足够的字符去构造 7 个回文串。 --
提示：
1 <= s.length <= 10^51 <= s.length <= 105s 中所有字符都是小写英文字母。1 <= k <= 10^51 <= k <= 105示例 1：
输入：nums = [8,2,4,7], limit = 4 -输出：2 +输出：2 解释：所有子数组如下： [8] 最大绝对差 |8-8| = 0 <= 4. -[8,2] 最大绝对差 |8-2| = 6 > 4. +[8,2] 最大绝对差 |8-2| = 6 > 4. [8,2,4] 最大绝对差 |8-2| = 6 > 4. [8,2,4,7] 最大绝对差 |8-2| = 6 > 4. [2] 最大绝对差 |2-2| = 0 <= 4. @@ -43,14 +43,14 @@ tags: [2,4,7] 最大绝对差 |2-7| = 5 > 4. [4] 最大绝对差 |4-4| = 0 <= 4. [4,7] 最大绝对差 |4-7| = 3 <= 4. -[7] 最大绝对差 |7-7| = 0 <= 4. +[7] 最大绝对差 |7-7| = 0 <= 4. 因此，满足题意的最长子数组的长度为 2 。
示例 2：
输入：nums = [10,1,2,4,7,2], limit = 5 -输出：4 +输出：4 解释：满足题意的最长子数组是 [2,4,7,2]，其最大绝对差 |2-7| = 5 <= 5 。diff --git a/solution/1400-1499/1438.Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README_EN.md b/solution/1400-1499/1438.Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README_EN.md index 07f167125aab5..838e7dae2e4b2 100644 --- a/solution/1400-1499/1438.Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README_EN.md +++ b/solution/1400-1499/1438.Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README_EN.md @@ -30,10 +30,10 @@ tags:
Input: nums = [8,2,4,7], limit = 4 -Output: 2 -Explanation: All subarrays are: +Output: 2 +Explanation: All subarrays are: [8] with maximum absolute diff |8-8| = 0 <= 4. -[8,2] with maximum absolute diff |8-2| = 6 > 4. +[8,2] with maximum absolute diff |8-2| = 6 > 4. [8,2,4] with maximum absolute diff |8-2| = 6 > 4. [8,2,4,7] with maximum absolute diff |8-2| = 6 > 4. [2] with maximum absolute diff |2-2| = 0 <= 4. @@ -41,7 +41,7 @@ tags: [2,4,7] with maximum absolute diff |2-7| = 5 > 4. [4] with maximum absolute diff |4-4| = 0 <= 4. [4,7] with maximum absolute diff |4-7| = 3 <= 4. -[7] with maximum absolute diff |7-7| = 0 <= 4. +[7] with maximum absolute diff |7-7| = 0 <= 4. Therefore, the size of the longest subarray is 2.@@ -49,7 +49,7 @@ Therefore, the size of the longest subarray is 2.
Input: nums = [10,1,2,4,7,2], limit = 5 -Output: 4 +Output: 4 Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.diff --git a/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README.md b/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README.md index 8c534cf5a9efe..a78b2580463c1 100644 --- a/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README.md +++ b/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README.md @@ -41,8 +41,8 @@ tags:
-输入：k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] -输出：[1] +输入：k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] +输出：[1] 解释： 所有服务器一开始都是空闲的。 前 3 个请求分别由前 3 台服务器依次处理。 diff --git a/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README_EN.md b/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README_EN.md index 6c8ae0631143e..11f80e8f7c508 100644 --- a/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README_EN.md +++ b/solution/1600-1699/1606.Find Servers That Handled Most Number of Requests/README_EN.md @@ -38,9 +38,9 @@ tags:diff --git a/solution/2700-2799/2762.Continuous Subarrays/README_EN.md b/solution/2700-2799/2762.Continuous Subarrays/README_EN.md index 258846bd742cb..0b3ca5f943041 100644 --- a/solution/2700-2799/2762.Continuous Subarrays/README_EN.md +++ b/solution/2700-2799/2762.Continuous Subarrays/README_EN.md @@ -39,7 +39,7 @@ tags:Example 1:
-Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] -Output: [1] -Explanation: +Input: k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] +Output: [1] +Explanation: All of the servers start out available. The first 3 requests are handled by the first 3 servers in order. Request 3 comes in. Server 0 is busy, so it's assigned to the next available server, which is 1. @@ -53,7 +53,7 @@ Servers 0 and 2 handled one request each, while server 1 handled two requests. Hdiff --git a/solution/2600-2699/2653.Sliding Subarray Beauty/README_EN.md b/solution/2600-2699/2653.Sliding Subarray Beauty/README_EN.md index 0de9ec290eff7..0ff195c6f0e26 100644 --- a/solution/2600-2699/2653.Sliding Subarray Beauty/README_EN.md +++ b/solution/2600-2699/2653.Sliding Subarray Beauty/README_EN.md @@ -38,7 +38,7 @@ tags:Input: k = 3, arrival = [1,2,3,4], load = [1,2,1,2] Output: [0] -Explanation: +Explanation: The first 3 requests are handled by first 3 servers. Request 3 comes in. It is handled by server 0 since the server is available. Server 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server. diff --git a/solution/1800-1899/1825.Finding MK Average/README.md b/solution/1800-1899/1825.Finding MK Average/README.md index f7b67495227d5..fa970e64153d0 100644 --- a/solution/1800-1899/1825.Finding MK Average/README.md +++ b/solution/1800-1899/1825.Finding MK Average/README.md @@ -52,7 +52,7 @@ tags: [null, null, null, -1, null, 3, null, null, null, 5] 解释： -MKAverage obj = new MKAverage(3, 1); +MKAverage obj = new MKAverage(3, 1); obj.addElement(3); // 当前元素为 [3] obj.addElement(1); // 当前元素为 [3,1] obj.calculateMKAverage(); // 返回 -1 ，因为 m = 3 ，但数据流中只有 2 个元素 diff --git a/solution/1800-1899/1825.Finding MK Average/README_EN.md b/solution/1800-1899/1825.Finding MK Average/README_EN.md index 12b3147398684..47e210d29904b 100644 --- a/solution/1800-1899/1825.Finding MK Average/README_EN.md +++ b/solution/1800-1899/1825.Finding MK Average/README_EN.md @@ -51,7 +51,7 @@ tags: [null, null, null, -1, null, 3, null, null, null, 5] Explanation -diff --git a/solution/2500-2599/2590.Design a Todo List/README.md b/solution/2500-2599/2590.Design a Todo List/README.md index c66941f981134..8a3a3cb1e1ff0 100644 --- a/solution/2500-2599/2590.Design a Todo List/README.md +++ b/solution/2500-2599/2590.Design a Todo List/README.md @@ -44,17 +44,17 @@ tags: [null, 1, 2, ["Task1", "Task2"], [], 3, ["Task3", "Task2"], null, null, ["Task3"], ["Task3", "Task1"]] 解释 -TodoList todoList = new TodoList(); -todoList.addTask(1, "Task1", 50, []); // 返回1。为ID为1的用户添加一个新任务。 -todoList.addTask(1, "Task2", 100, ["P1"]); // 返回2。为ID为1的用户添加另一个任务，并给它添加标签“P1”。 -todoList.getAllTasks(1); // 返回["Task1", "Task2"]。用户1目前有两个未完成的任务。 -todoList.getAllTasks(5); // 返回[]。用户5目前没有任务。 -todoList.addTask(1, "Task3", 30, ["P1"]); // 返回3。为ID为1的用户添加另一个任务，并给它添加标签“P1”。 -todoList.getTasksForTag(1, "P1"); // 返回["Task3", "Task2"]。返回ID为1的用户未完成的带有“P1”标签的任务。 -todoList.completeTask(5, 1); // 不做任何操作，因为任务1不属于用户5。 -todoList.completeTask(1, 2); // 将任务2标记为已完成。 -todoList.getTasksForTag(1, "P1"); // 返回["Task3"]。返回ID为1的用户未完成的带有“P1”标签的任务。 - // 注意，现在不包括 “Task2” ，因为它已经被标记为已完成。 +TodoList todoList = new TodoList(); +todoList.addTask(1, "Task1", 50, []); // 返回1。为ID为1的用户添加一个新任务。 +todoList.addTask(1, "Task2", 100, ["P1"]); // 返回2。为ID为1的用户添加另一个任务，并给它添加标签“P1”。 +todoList.getAllTasks(1); // 返回["Task1", "Task2"]。用户1目前有两个未完成的任务。 +todoList.getAllTasks(5); // 返回[]。用户5目前没有任务。 +todoList.addTask(1, "Task3", 30, ["P1"]); // 返回3。为ID为1的用户添加另一个任务，并给它添加标签“P1”。 +todoList.getTasksForTag(1, "P1"); // 返回["Task3", "Task2"]。返回ID为1的用户未完成的带有“P1”标签的任务。 +todoList.completeTask(5, 1); // 不做任何操作，因为任务1不属于用户5。 +todoList.completeTask(1, 2); // 将任务2标记为已完成。 +todoList.getTasksForTag(1, "P1"); // 返回["Task3"]。返回ID为1的用户未完成的带有“P1”标签的任务。 + // 注意，现在不包括 “Task2” ，因为它已经被标记为已完成。 todoList.getAllTasks(1); // 返回["Task3", "Task1"]。用户1现在有两个未完成的任务。MKAverage obj = new MKAverage(3, 1); +MKAverage obj = new MKAverage(3, 1); obj.addElement(3); // current elements are [3] obj.addElement(1); // current elements are [3,1] obj.calculateMKAverage(); // return -1, because m = 3 and only 2 elements exist. diff --git a/solution/2300-2399/2349.Design a Number Container System/README_EN.md b/solution/2300-2399/2349.Design a Number Container System/README_EN.md index 8480f11edb6db..b55a94b3c5cfd 100644 --- a/solution/2300-2399/2349.Design a Number Container System/README_EN.md +++ b/solution/2300-2399/2349.Design a Number Container System/README_EN.md @@ -54,7 +54,7 @@ nc.change(1, 10); // Your container at index 1 will be filled with number 10. nc.change(3, 10); // Your container at index 3 will be filled with number 10. nc.change(5, 10); // Your container at index 5 will be filled with number 10. nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1. -nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. +nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.Input: nums = [1,-1,-3,-2,3], k = 3, x = 2 Output: [-1,-2,-2] -Explanation: There are 3 subarrays with size k = 3. +Explanation: There are 3 subarrays with size k = 3. The first subarray isdiff --git a/solution/2700-2799/2749.Minimum Operations to Make the Integer Zero/README_EN.md b/solution/2700-2799/2749.Minimum Operations to Make the Integer Zero/README_EN.md index a017d3e80647a..45530dc18abf5 100644 --- a/solution/2700-2799/2749.Minimum Operations to Make the Integer Zero/README_EN.md +++ b/solution/2700-2799/2749.Minimum Operations to Make the Integer Zero/README_EN.md @@ -34,9 +34,9 @@ tags: Input: num1 = 3, num2 = -2 Output: 3 Explanation: We can make 3 equal to 0 with the following operations: -- We choose i = 2 and substract 22 + (-2) from 3, 3 - (4 + (-2)) = 1. -- We choose i = 2 and substract 22 + (-2) from 1, 1 - (4 + (-2)) = -1. -- We choose i = 0 and substract 20 + (-2) from -1, (-1) - (1 + (-2)) = 0. +- We choose i = 2 and subtract 22 + (-2) from 3, 3 - (4 + (-2)) = 1. +- We choose i = 2 and subtract 22 + (-2) from 1, 1 - (4 + (-2)) = -1. +- We choose i = 0 and subtract 20 + (-2) from -1, (-1) - (1 + (-2)) = 0. It can be proven, that 3 is the minimum number of operations that we need to perform.[1, -1, -3]and the 2nd smallest negative integer is -1. The second subarray is[-1, -3, -2]and the 2nd smallest negative integer is -2. The third subarray is[-3, -2, 3]and the 2nd smallest negative integer is -2.
Input: nums = [5,4,2,4] Output: 8 -Explanation: +Explanation: Continuous subarray of size 1: [5], [4], [2], [4]. Continuous subarray of size 2: [5,4], [4,2], [2,4]. Continuous subarray of size 3: [4,2,4]. @@ -55,7 +55,7 @@ It can be shown that there are no more continuous subarrays.diff --git a/solution/2800-2899/2817.Minimum Absolute Difference Between Elements With Constraint/README_EN.md b/solution/2800-2899/2817.Minimum Absolute Difference Between Elements With Constraint/README_EN.md index fb1c71f35d61f..a770f63c0f35c 100644 --- a/solution/2800-2899/2817.Minimum Absolute Difference Between Elements With Constraint/README_EN.md +++ b/solution/2800-2899/2817.Minimum Absolute Difference Between Elements With Constraint/README_EN.md @@ -34,8 +34,8 @@ tags:Input: nums = [1,2,3] Output: 6 -Explanation: +Explanation: Continuous subarray of size 1: [1], [2], [3]. Continuous subarray of size 2: [1,2], [2,3]. Continuous subarray of size 3: [1,2,3]. diff --git a/solution/2700-2799/2763.Sum of Imbalance Numbers of All Subarrays/README_EN.md b/solution/2700-2799/2763.Sum of Imbalance Numbers of All Subarrays/README_EN.md index 0243dafa27604..ca4df4150344f 100644 --- a/solution/2700-2799/2763.Sum of Imbalance Numbers of All Subarrays/README_EN.md +++ b/solution/2700-2799/2763.Sum of Imbalance Numbers of All Subarrays/README_EN.md @@ -43,7 +43,7 @@ tags: - Subarray [3, 1] with an imbalance number of 1. - Subarray [3, 1, 4] with an imbalance number of 1. - Subarray [1, 4] with an imbalance number of 1. -The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 3. +The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 3.Example 2:
@@ -55,8 +55,8 @@ The imbalance number of all other subarrays is 0. Hence, the sum of imbalance nu - Subarray [1, 3] with an imbalance number of 1. - Subarray [1, 3, 3] with an imbalance number of 1. - Subarray [1, 3, 3, 3] with an imbalance number of 1. -- Subarray [1, 3, 3, 3, 5] with an imbalance number of 2. -- Subarray [3, 3, 3, 5] with an imbalance number of 1. +- Subarray [1, 3, 3, 3, 5] with an imbalance number of 2. +- Subarray [3, 3, 3, 5] with an imbalance number of 1. - Subarray [3, 3, 5] with an imbalance number of 1. - Subarray [3, 5] with an imbalance number of 1. The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 8.
Input: nums = [4,3,2,4], x = 2 Output: 0 -Explanation: We can select nums[0] = 4 and nums[3] = 4. -They are at least 2 indices apart, and their absolute difference is the minimum, 0. +Explanation: We can select nums[0] = 4 and nums[3] = 4. +They are at least 2 indices apart, and their absolute difference is the minimum, 0. It can be shown that 0 is the optimal answer.diff --git a/solution/3200-3299/3278.Find Candidates for Data Scientist Position II/README.md b/solution/3200-3299/3278.Find Candidates for Data Scientist Position II/README.md index ba80832e5d3d5..f48d4d484d420 100644 --- a/solution/3200-3299/3278.Find Candidates for Data Scientist Position II/README.md +++ b/solution/3200-3299/3278.Find Candidates for Data Scientist Position II/README.md @@ -53,6 +53,7 @@ tags:
100 分 开始。10 分。5 分。100 points10 points for each skill where proficiency > importance5 points for each skill where proficiency < importance结果表以 student_id，subject 升序排序。
diff --git a/solution/3400-3499/3422.Minimum Operations to Make Subarray Elements Equal/README.md b/solution/3400-3499/3422.Minimum Operations to Make Subarray Elements Equal/README.md index 1be8c45c55821..ad4685f9375f9 100644 --- a/solution/3400-3499/3422.Minimum Operations to Make Subarray Elements Equal/README.md +++ b/solution/3400-3499/3422.Minimum Operations to Make Subarray Elements Equal/README.md @@ -6,7 +6,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3422.Mi -# [3422. Minimum Operations to Make Subarray Elements Equal 🔒](https://leetcode.cn/problems/minimum-operations-to-make-subarray-elements-equal) +# [3422. 将子数组元素变为相等所需的最小操作数 🔒](https://leetcode.cn/problems/minimum-operations-to-make-subarray-elements-equal) [English Version](/solution/3400-3499/3422.Minimum%20Operations%20to%20Make%20Subarray%20Elements%20Equal/README_EN.md) @@ -14,49 +14,51 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3422.Mi -You are given an integer array nums and an integer k. You can perform the following operation any number of times:
给定一个整数数组 nums 和一个整数 k。你可以进行任意次以下操作：
nums by 1.nums 的任何元素增加或减少 1。Return the minimum number of operations required to ensure that at least one subarray of size k in nums has all elements equal.
返回确保 至少 有一个大小为 k 的 nums 中的 子数组 的所有元素都相等的所需的 最小 操作数。
-
Example 1:
+ +示例 1：
Input: nums = [4,-3,2,1,-4,6], k = 3
+输入：nums = [4,-3,2,1,-4,6], k = 3
-Output: 5
+输出：5
-Explanation:
+解释：
nums[1]. The resulting array is [4, 1, 2, 1, -4, 6].nums[2]. The resulting array is [4, 1, 1, 1, -4, 6].[1, 1, 1] of size k = 3 with all elements equal. Hence, the answer is 5.nums[1] 增加 4。结果数组为 [4, 1, 2, 1, -4, 6]。nums[2] 减少 1。结果数组为 [4, 1, 1, 1, -4, 6]。k = 3 的子数组 [1, 1, 1]，所有元素都想等。因此，答案为 5。Example 2:
+示例 2：
Input: nums = [-2,-2,3,1,4], k = 2
+输入：nums = [-2,-2,3,1,4], k = 2
-Output: 0
+输出：0
-Explanation:
+解释：
The subarray [-2, -2] of size k = 2 already contains all equal elements, so no operations are needed. Hence, the answer is 0.
大小为 k = 2 的子数组 [-2, -2] 已经包含了所有相等的元素，所以不需要操作。因此答案为 0。
-
Constraints:
+ +提示：
2 <= nums.length <= 105