From 6ed8fbd4be7a2e45bf9c7a9e6b0d36233e8567d6 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Wed, 23 Oct 2024 11:43:06 +0800 Subject: [PATCH] feat: update solutions to lc problem: No.0555 --- .../0555.Split Concatenated Strings/README_EN.md | 12 ++++++++++-- 1 file changed, 10 insertions(+), 2 deletions(-) diff --git a/solution/0500-0599/0555.Split Concatenated Strings/README_EN.md b/solution/0500-0599/0555.Split Concatenated Strings/README_EN.md index 1045aaa0f3373..8565aae1cc864 100644 --- a/solution/0500-0599/0555.Split Concatenated Strings/README_EN.md +++ b/solution/0500-0599/0555.Split Concatenated Strings/README_EN.md @@ -37,7 +37,7 @@ tags: 
 Input: strs = ["abc","xyz"]
 Output: "zyxcba"
-Explanation: You can get the looped string "-abcxyz-", "-abczyx-", "-cbaxyz-", "-cbazyx-", where '-' represents the looped status. 
+Explanation: You can get the looped string "-abcxyz-", "-abczyx-", "-cbaxyz-", "-cbazyx-", where '-' represents the looped status.
 The answer string came from the fourth looped one, where you could cut from the middle character 'a' and get "zyxcba".
@@ -64,7 +64,15 @@ The answer string came from the fourth looped one, where you could cut from the -### Solution 1 +### Solution 1: Greedy + +We first traverse the string array `strs`. For each string $s$, if the reversed string $t$ is greater than $s$, we replace $s$ with $t$. + +Then we enumerate each position $i$ in the string array `strs` as a split point, dividing the string array `strs` into two parts: $strs[i + 1:]$ and $strs[:i]$. We then concatenate these two parts to get a new string $t$. Next, we enumerate each position $j$ in the current string $strs[i]$. The suffix part is $a = strs[i][j:]$, and the prefix part is $b = strs[i][:j]$. We can concatenate $a$, $t$, and $b$ to get a new string $cur$. If $cur$ is greater than the current answer, we update the answer. This considers the case where $strs[i]$ is reversed. We also need to consider the case where $strs[i]$ is not reversed, i.e., concatenate $a$, $t$, and $b$ in reverse order to get a new string $cur$. If $cur$ is greater than the current answer, we update the answer. + +Finally, we return the answer. + +The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string array `strs`.