You are given two integer arrays, nums and cost, of the same size, and an integer k.
You can divide nums into subarrays. The cost of the ith subarray consisting of elements nums[l..r] is:
You can divide nums into subarrays. The cost of the ith subarray consisting of elements nums[l..r] is:
(nums[0] + nums[1] + ... + nums[r] + k * i) * (cost[l] + cost[l + 1] + ... + cost[r]).Return the minimum total cost possible from any valid division.
-A subarray is a contiguous non-empty sequence of elements within an array.
-
Example 1:
diff --git a/solution/3500-3599/3501.Maximize Active Section with Trade II/README_EN.md b/solution/3500-3599/3501.Maximize Active Section with Trade II/README_EN.md index 527245c83a435..f9820eafe32ea 100644 --- a/solution/3500-3599/3501.Maximize Active Section with Trade II/README_EN.md +++ b/solution/3500-3599/3501.Maximize Active Section with Trade II/README_EN.md @@ -20,7 +20,6 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3501.Ma'1' represents an active section.'0' represents an inactive section.You can perform at most one trade to maximize the number of active sections in s. In a trade, you:
'0's that is surrounded by '1's to all '1's.Additionally, you are given a 2D array queries, where queries[i] = [li, ri] represents a substring s[li...ri].
Additionally, you are given a 2D array queries, where queries[i] = [li, ri] represents a substring s[li...ri].
For each query, determine the maximum possible number of active sections in s after making the optimal trade on the substring s[li...ri].
Return an array answer, where answer[i] is the result for queries[i].
A substring is a contiguous non-empty sequence of characters within a string.
-Note
You are given two strings, s and t.
You can create a new string by selecting a substring from s (possibly empty) and a substring from t (possibly empty), then concatenating them in order.
You can create a new string by selecting a substring from s (possibly empty) and a substring from t (possibly empty), then concatenating them in order.
Return the length of the longest palindrome that can be formed this way.
- -A substring is a contiguous sequence of characters within a string.
- -A palindrome is a string that reads the same forward and backward.
+Return the length of the longest palindrome that can be formed this way.
Example 1:
@@ -87,7 +83,26 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3503.Lo -### Solution 1 +### Solution 1: Enumerate Palindrome Centers + Dynamic Programming + +According to the problem description, the concatenated palindrome string can be composed entirely of string $s$, entirely of string $t$, or a combination of both strings $s$ and $t$. Additionally, there may be extra palindromic substrings in either string $s$ or $t$. + +Therefore, we first reverse string $t$ and preprocess arrays $\textit{g1}$ and $\textit{g2}$, where $\textit{g1}[i]$ represents the length of the longest palindromic substring starting at index $i$ in string $s$, and $\textit{g2}[i]$ represents the length of the longest palindromic substring starting at index $i$ in string $t$. + +We can initialize the answer $\textit{ans}$ as the maximum value in $\textit{g1}$ and $\textit{g2}$. + +Next, we define $\textit{f}[i][j]$ as the length of the palindromic substring ending at the $i$-th character of string $s$ and the $j$-th character of string $t$. + +For $\textit{f}[i][j]$, if $s[i - 1]$ equals $t[j - 1]$, then $\textit{f}[i][j] = \textit{f}[i - 1][j - 1] + 1$. We then update the answer: + +$$ +\textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } i \geq m \text{ else } \textit{g1}[i])) \\ +\textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } j \geq n \text{ else } \textit{g2}[j])) +$$ + +Finally, we return the answer $\textit{ans}$. + +The time complexity is $O(m \times (m + n))$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of strings $s$ and $t$, respectively. diff --git a/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README.md b/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README.md index 17ee66466df32..269de847bbde6 100644 --- a/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README.md +++ b/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README.md @@ -90,7 +90,27 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3504.Lo -### 方法一 +### 方法一：枚举回文中点 + 动态规划 + +根据题目描述，连接后的回文串，可以只由字符串 $s$ 组成，也可以只由字符串 $t$ 组成，也可以由字符串 $s$ 和字符串 $t$ 组成，并且还可能在字符串 $s$ 或 $t$ 中多出一部分回文子串。 + +因此，我们先将字符串 $t$ 反转，然后预处理出数组 $\textit{g1}$ 和 $\textit{g2}$，其中 $\textit{g1}[i]$ 表示在字符串 $s$ 中以下标 $i$ 开始的最长回文子串长度，而 $\textit{g2}[i]$ 表示在字符串 $t$ 中以下标 $i$ 开始的最长回文子串长度。 + +那么我们可以初始化答案 $\textit{ans}$ 为 $\textit{g1}$ 和 $\textit{g2}$ 中的最大值。 + +接下来，我们定义 $\textit{f}[i][j]$ 表示以字符串 $s$ 的第 $i$ 个字符结尾，以字符串 $t$ 的第 $j$ 个字符结尾的回文子串的长度。 + +对于 $\textit{f}[i][j]$，如果 $s[i - 1]$ 等于 $t[j - 1]$，那么有 $\textit{f}[i][j] = \textit{f}[i - 1][j - 1] + 1$。然后，我们更新答案： + +$$ +\textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } i \geq m \text{ else } \textit{g1}[i])) \\ + +\textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } j \geq n \text{ else } \textit{g2}[j])) \ +$$ + +最后，我们返回答案 $\textit{ans}$ 即可。 + +时间复杂度 $O(m \times (m + n))$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是字符串 $s$ 和 $t$ 的长度。 diff --git a/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README_EN.md b/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README_EN.md index 3a2e00ef7e1cc..95221be84beba 100644 --- a/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README_EN.md +++ b/solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README_EN.md @@ -15,15 +15,10 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3504.LoYou are given two strings, s and t.
You can create a new string by selecting a substring from s (possibly empty) and a substring from t (possibly empty), then concatenating them in order.
You can create a new string by selecting a substring from s (possibly empty) and a substring from t (possibly empty), then concatenating them in order.
Return the length of the longest palindrome that can be formed this way.
- -A substring is a contiguous sequence of characters within a string.
- -A palindrome is a string that reads the same forward and backward.
+Return the length of the longest palindrome that can be formed this way.
Example 1:
@@ -88,7 +83,26 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3504.Lo -### Solution 1 +### Solution 1: Enumerate Palindrome Centers + Dynamic Programming + +According to the problem description, the concatenated palindrome string can be composed entirely of string $s$, entirely of string $t$, or a combination of both strings $s$ and $t$. Additionally, there may be extra palindromic substrings in either string $s$ or $t$. + +Therefore, we first reverse string $t$ and preprocess arrays $\textit{g1}$ and $\textit{g2}$, where $\textit{g1}[i]$ represents the length of the longest palindromic substring starting at index $i$ in string $s$, and $\textit{g2}[i]$ represents the length of the longest palindromic substring starting at index $i$ in string $t$. + +We can initialize the answer $\textit{ans}$ as the maximum value in $\textit{g1}$ and $\textit{g2}$. + +Next, we define $\textit{f}[i][j]$ as the length of the palindromic substring ending at the $i$-th character of string $s$ and the $j$-th character of string $t$. + +For $\textit{f}[i][j]$, if $s[i - 1]$ equals $t[j - 1]$, then $\textit{f}[i][j] = \textit{f}[i - 1][j - 1] + 1$. We then update the answer: + +$$ +\textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } i \geq m \text{ else } \textit{g1}[i])) \\ +\textit{ans} = \max(\textit{ans}, \textit{f}[i][j] \times 2 + (0 \text{ if } j \geq n \text{ else } \textit{g2}[j])) +$$ + +Finally, we return the answer $\textit{ans}$. + +The time complexity is $O(m \times (m + n))$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of strings $s$ and $t$, respectively. diff --git a/solution/3500-3599/3505.Minimum Operations to Make Elements Within K Subarrays Equal/README_EN.md b/solution/3500-3599/3505.Minimum Operations to Make Elements Within K Subarrays Equal/README_EN.md index 4ad8f5bb9a43a..c6bf631bd3eec 100644 --- a/solution/3500-3599/3505.Minimum Operations to Make Elements Within K Subarrays Equal/README_EN.md +++ b/solution/3500-3599/3505.Minimum Operations to Make Elements Within K Subarrays Equal/README_EN.md @@ -15,14 +15,13 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3505.MiYou are given an integer array nums and two integers, x and k. You can perform the following operation any number of times (including zero):
nums by 1.Return the minimum number of operations needed to have at least k non-overlapping subarrays of size exactly x in nums, where all elements within each subarray are equal.
Return the minimum number of operations needed to have at least k non-overlapping subarrays of size exactly x in nums, where all elements within each subarray are equal.
Example 1: