prices[indexes[i]] - indexes[i] == prices[indexes[j]] - indexes[j].",
+ "So now we define a new array named group and is constructed as group[i] = prices[i] - i.",
+ "A subarray of prices is linear if they belong to the same group.",
+ "Since all elements are positive, if we choose some index i, the optimum way is to choose all elements from group[i].",
+ "So for each group, we calculate the sum of its prices and the answer would be the maximum sum over all groups."
+ ],
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diff --git a/leetcode-cn/originData/[no content]minimizing-array-after-replacing-pairs-with-their-product.json b/leetcode-cn/originData/[no content]minimizing-array-after-replacing-pairs-with-their-product.json
new file mode 100644
index 00000000..7c1db1de
--- /dev/null
+++ b/leetcode-cn/originData/[no content]minimizing-array-after-replacing-pairs-with-their-product.json
@@ -0,0 +1,73 @@
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+ "questionId": "3177",
+ "questionFrontendId": "2892",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2469582,
+ "title": "Minimizing Array After Replacing Pairs With Their Product",
+ "titleSlug": "minimizing-array-after-replacing-pairs-with-their-product",
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+ "name": "Array",
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diff --git a/leetcode-cn/originData/apply-operations-on-array-to-maximize-sum-of-squares.json b/leetcode-cn/originData/apply-operations-on-array-to-maximize-sum-of-squares.json
new file mode 100644
index 00000000..7cb1130a
--- /dev/null
+++ b/leetcode-cn/originData/apply-operations-on-array-to-maximize-sum-of-squares.json
@@ -0,0 +1,189 @@
+{
+ "data": {
+ "question": {
+ "questionId": "3153",
+ "questionFrontendId": "2897",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2469480,
+ "title": "Apply Operations on Array to Maximize Sum of Squares",
+ "titleSlug": "apply-operations-on-array-to-maximize-sum-of-squares",
+ "content": " You are given a 0-indexed integer array You can do the following operation on the array any number of times: You have to choose Return the maximum sum of squares you can achieve. Since the answer can be very large, return it modulo Example 1: Example 2: Constraints: 给你一个下标从 0 开始的整数数组 你可以对数组执行以下操作 任意次 ： 你需要从最终的数组里选择 请你返回你可以得到的 最大 平方和。 由于答案可能会很大，将答案对 示例 1： 示例 2： 提示： \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 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diff --git a/leetcode-cn/originData/apply-operations-to-make-two-strings-equal.json b/leetcode-cn/originData/apply-operations-to-make-two-strings-equal.json
new file mode 100644
index 00000000..c9ab9cf5
--- /dev/null
+++ b/leetcode-cn/originData/apply-operations-to-make-two-strings-equal.json
@@ -0,0 +1,177 @@
+{
+ "data": {
+ "question": {
+ "questionId": "3033",
+ "questionFrontendId": "2896",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2470809,
+ "title": "Apply Operations to Make Two Strings Equal",
+ "titleSlug": "apply-operations-to-make-two-strings-equal",
+ "content": " You are given two 0-indexed binary strings You can perform any of the following operations on the string Return the minimum cost needed to make the strings Note that flipping a character means changing it from Example 1: Example 2: Constraints: 给你两个下标从 0 开始的二进制字符串 你可以对字符串 请你返回使字符串 注意 ，反转字符的意思是将 示例 1： 示例 2： 提示： \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\nclang 11<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 20\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/code>\\u7ea7\\u4f18\\u5316\\u3002AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\nOpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\nPython 2.7.12<\\/code><\\/p>\\r\\n\\r\\nGCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU11\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O1<\\/code>\\u7ea7\\u4f18\\u5316\\u3002 AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\nSwift 5.5.2<\\/code><\\/p>\\r\\n\\r\\nGo 1.18<\\/code><\\/p>\\r\\n\\r\\nPython 3.10<\\/code><\\/p>\\r\\n\\r\\nScala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\nrust 1.58.1<\\/code><\\/p>\\r\\n\\r\\nPHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n1.",
+ "Merge all adjacent ones (since 1 * 1 = 1 and k >= 1).",
+ "Let dp[i] be the answer to the problem for the first i elements.",
+ "To calculate dp[i], try to brute-force all indices j such that elements from j to i are merged together to create a new element.",
+ "For a fixed i, you could go backward from ith elements and multiply them together until the product is at most k. Now if you are currently on index j and you've merged all elements from jth element to ith element, dp[i] = min(dp[i], dp[j - 1] + 1).",
+ "The above backward moving can be done at most 2 * log2(k) times. Since we've merged adjacent ones, every two adjacent elements have a product of at least 2.",
+ "So the total time complexity would be n * 2 * log2(k)."
+ ],
+ "solution": null,
+ "status": null,
+ "sampleTestCase": "[2,3,3,7,3,5]\n20",
+ "metaData": "{\n \"name\": \"minArrayLength\",\n \"params\": [\n {\n \"name\": \"nums\",\n \"type\": \"integer[]\"\n },\n {\n \"type\": \"integer\",\n \"name\": \"k\"\n }\n ],\n \"return\": {\n \"type\": \"integer\"\n }\n}",
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+ "envInfo": "{\"cpp\":[\"C++\",\"clang 11<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 20\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/code>\\u7ea7\\u4f18\\u5316\\u3002AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\nOpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\nPython 2.7.12<\\/code><\\/p>\\r\\n\\r\\nGCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU11\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O1<\\/code>\\u7ea7\\u4f18\\u5316\\u3002 AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\nSwift 5.5.2<\\/code><\\/p>\\r\\n\\r\\nGo 1.18<\\/code><\\/p>\\r\\n\\r\\nPython 3.10<\\/code><\\/p>\\r\\n\\r\\nScala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\nrust 1.58.1<\\/code><\\/p>\\r\\n\\r\\nPHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\nnums and a positive integer k.\n\t
\n\ni and j and simultaneously update the values of nums[i] to (nums[i] AND nums[j]) and nums[j] to (nums[i] OR nums[j]). Here, OR denotes the bitwise OR operation, and AND denotes the bitwise AND operation.k elements from the final array and calculate the sum of their squares.109 + 7.\nInput: nums = [2,6,5,8], k = 2\nOutput: 261\nExplanation: We can do the following operations on the array:\n- Choose i = 0 and j = 3, then change nums[0] to (2 AND 8) = 0 and nums[3] to (2 OR 8) = 10. The resulting array is nums = [0,6,5,10].\n- Choose i = 2 and j = 3, then change nums[2] to (5 AND 10) = 0 and nums[3] to (5 OR 10) = 15. The resulting array is nums = [0,6,0,15].\nWe can choose the elements 15 and 6 from the final array. The sum of squares is 152 + 62 = 261.\nIt can be shown that this is the maximum value we can get.\n
\n\n\nInput: nums = [4,5,4,7], k = 3\nOutput: 90\nExplanation: We do not need to apply any operations.\nWe can choose the elements 7, 5, and 4 with a sum of squares: 72 + 52 + 42 = 90.\nIt can be shown that this is the maximum value we can get.\n
\n\n\n\t
\n",
+ "translatedTitle": "对数组执行操作使平方和最大",
+ "translatedContent": "1 <= k <= nums.length <= 1051 <= nums[i] <= 109nums 和一个 正 整数 k 。\n\t
\n\ni 和 j ，同时 将 nums[i] 更新为 (nums[i] AND nums[j]) 且将 nums[j] 更新为 (nums[i] OR nums[j]) ，OR 表示按位 或 运算，AND 表示按位 与 运算。k 个元素，并计算它们的 平方 之和。109 + 7 取余 后返回。\n输入：nums = [2,6,5,8], k = 2\n输出：261\n解释：我们可以对数组执行以下操作：\n- 选择 i = 0 和 j = 3 ，同时将 nums[0] 变为 (2 AND 8) = 0 且 nums[3] 变为 (2 OR 8) = 10 ，结果数组为 nums = [0,6,5,10] 。\n- 选择 i = 2 和 j = 3 ，同时将 nums[2] 变为 (5 AND 10) = 0 且 nums[3] 变为 (5 OR 10) = 15 ，结果数组为 nums = [0,6,0,15] 。\n从最终数组里选择元素 15 和 6 ，平方和为 152 + 62 = 261 。\n261 是可以得到的最大结果。\n
\n\n\n输入：nums = [4,5,4,7], k = 3\n输出：90\n解释：不需要执行任何操作。\n选择元素 7 ，5 和 4 ，平方和为 72 + 52 + 42 = 90 。\n90 是可以得到的最大结果。\n
\n\n\n\t
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\n\ni and j, and flip both s1[i] and s1[j]. The cost of this operation is x.i such that i < n - 1 and flip both s1[i] and s1[i + 1]. The cost of this operation is 1.s1 and s2 equal, or return -1 if it is impossible.0 to 1 or vice-versa.\nInput: s1 = "1100011000", s2 = "0101001010", x = 2\nOutput: 4\nExplanation: We can do the following operations:\n- Choose i = 3 and apply the second operation. The resulting string is s1 = "1101111000".\n- Choose i = 4 and apply the second operation. The resulting string is s1 = "1101001000".\n- Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = "0101001010" = s2.\nThe total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.\n
\n\n\nInput: s1 = "10110", s2 = "00011", x = 4\nOutput: -1\nExplanation: It is not possible to make the two strings equal.\n
\n\n\n\t
\n",
+ "translatedTitle": "执行操作使两个字符串相等",
+ "translatedContent": "n == s1.length == s2.length1 <= n, x <= 500s1 and s2 consist only of the characters '0' and '1'.s1 和 s2 ，两个字符串的长度都是 n ，再给你一个正整数 x 。s1 执行以下操作 任意次 ：\n\t
\n\ni 和 j ，将 s1[i] 和 s1[j] 都反转，操作的代价为 x 。i < n - 1 的下标 i ，反转 s1[i] 和 s1[i + 1] ，操作的代价为 1 。s1 和 s2 相等的 最小 操作代价之和，如果无法让二者相等，返回 -1 。0 变成 1 ，或者 1 变成 0 。\n输入：s1 = \"1100011000\", s2 = \"0101001010\", x = 2\n输出：4\n解释：我们可以执行以下操作：\n- 选择 i = 3 执行第二个操作。结果字符串是 s1 = \"1101111000\" 。\n- 选择 i = 4 执行第二个操作。结果字符串是 s1 = \"1101001000\" 。\n- 选择 i = 0 和 j = 8 ，执行第一个操作。结果字符串是 s1 = \"0101001010\" = s2 。\n总代价是 1 + 1 + 2 = 4 。这是最小代价和。\n
\n\n\n输入：s1 = \"10110\", s2 = \"00011\", x = 4\n输出：-1\n解释：无法使两个字符串相等。\n
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