diff --git a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README.md b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README.md index c7c9ffab5de7c..51a1d50ee6932 100644 --- a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README.md +++ b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README.md @@ -79,7 +79,13 @@ tags: -### 方法一 +### 方法一：枚举子串左端点 + +如果我们固定子字符串的左端点，那么子字符串越长，字典序越大。假设子字符串左端点为 $i$，剩余子字符串的最小长度为 $\text{numFriends} - 1$，那么子字符串的右端点可以取到 $\min(n, i + n - (\text{numFriends} - 1))$，其中 $n$ 为字符串的长度。注意我们说的是左开右闭。 + +我们枚举所有可能的左端点，取出对应的子字符串，比较字典序，最终得到字典序最大的子字符串。 + +时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。 @@ -91,11 +97,7 @@ class Solution: if numFriends == 1: return word n = len(word) - ans = "" - for i in range(n): - k = min(n - i, n - numFriends + 1) - ans = max(ans, word[i : i + k]) - return ans + return max(word[i : i + n - (numFriends - 1)] for i in range(n)) ``` #### Java @@ -109,8 +111,7 @@ class Solution { int n = word.length(); String ans = ""; for (int i = 0; i < n; ++i) { - int k = Math.min(n - i, n - numFriends + 1); - String t = word.substring(i, i + k); + String t = word.substring(i, Math.min(n, i + n - (numFriends - 1))); if (ans.compareTo(t) < 0) { ans = t; } @@ -129,12 +130,13 @@ public: if (numFriends == 1) { return word; } - int n = word.size(); - string ans; + int n = word.length(); + string ans = ""; for (int i = 0; i < n; ++i) { - int k = min(n - i, n - numFriends + 1); - string t = word.substr(i, k); - ans = max(ans, t); + string t = word.substr(i, min(n - i, n - (numFriends - 1))); + if (ans < t) { + ans = t; + } } return ans; } @@ -149,9 +151,8 @@ func answerString(word string, numFriends int) (ans string) { return word } n := len(word) - for i := range word { - k := min(n-i, n-numFriends+1) - t := word[i : i+k] + for i := 0; i < n; i++ { + t := word[i:min(n, i+n-(numFriends-1))] ans = max(ans, t) } return @@ -165,14 +166,11 @@ function answerString(word: string, numFriends: number): string { if (numFriends === 1) { return word; } - let ans: string = ''; const n = word.length; - for (let i = 0; i < n; ++i) { - const k = Math.min(n - i, n - numFriends + 1); - const t = word.slice(i, i + k); - if (ans < t) { - ans = t; - } + let ans = ''; + for (let i = 0; i < n; i++) { + const t = word.slice(i, Math.min(n, i + n - (numFriends - 1))); + ans = t > ans ? t : ans; } return ans; } diff --git a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README_EN.md b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README_EN.md index 97348781ffe9f..a642be1a8a20c 100644 --- a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README_EN.md +++ b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README_EN.md @@ -77,7 +77,13 @@ tags: -### Solution 1 +### Solution 1: Enumerate Substring Left Endpoints + +If we fix the left endpoint of the substring, the longer the substring, the larger its lexicographical order. Suppose the left endpoint of the substring is $i$, and the minimum length of the remaining substrings is $\text{numFriends} - 1$, then the right endpoint of the substring can be up to $\min(n, i + n - (\text{numFriends} - 1))$, where $n$ is the length of the string. Note that we are talking about left-closed, right-open intervals. + +We enumerate all possible left endpoints, extract the corresponding substrings, compare their lexicographical order, and finally obtain the lexicographically largest substring. + +The time complexity is $O(n^2)$, and the space complexity is $O(n)$, where $n$ is the length of the string. @@ -89,11 +95,7 @@ class Solution: if numFriends == 1: return word n = len(word) - ans = "" - for i in range(n): - k = min(n - i, n - numFriends + 1) - ans = max(ans, word[i : i + k]) - return ans + return max(word[i : i + n - (numFriends - 1)] for i in range(n)) ``` #### Java @@ -107,8 +109,7 @@ class Solution { int n = word.length(); String ans = ""; for (int i = 0; i < n; ++i) { - int k = Math.min(n - i, n - numFriends + 1); - String t = word.substring(i, i + k); + String t = word.substring(i, Math.min(n, i + n - (numFriends - 1))); if (ans.compareTo(t) < 0) { ans = t; } @@ -127,12 +128,13 @@ public: if (numFriends == 1) { return word; } - int n = word.size(); - string ans; + int n = word.length(); + string ans = ""; for (int i = 0; i < n; ++i) { - int k = min(n - i, n - numFriends + 1); - string t = word.substr(i, k); - ans = max(ans, t); + string t = word.substr(i, min(n - i, n - (numFriends - 1))); + if (ans < t) { + ans = t; + } } return ans; } @@ -147,9 +149,8 @@ func answerString(word string, numFriends int) (ans string) { return word } n := len(word) - for i := range word { - k := min(n-i, n-numFriends+1) - t := word[i : i+k] + for i := 0; i < n; i++ { + t := word[i:min(n, i+n-(numFriends-1))] ans = max(ans, t) } return @@ -163,14 +164,11 @@ function answerString(word: string, numFriends: number): string { if (numFriends === 1) { return word; } - let ans: string = ''; const n = word.length; - for (let i = 0; i < n; ++i) { - const k = Math.min(n - i, n - numFriends + 1); - const t = word.slice(i, i + k); - if (ans < t) { - ans = t; - } + let ans = ''; + for (let i = 0; i < n; i++) { + const t = word.slice(i, Math.min(n, i + n - (numFriends - 1))); + ans = t > ans ? t : ans; } return ans; } diff --git a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.cpp b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.cpp index 9732f5751b30f..6bfd212edf630 100644 --- a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.cpp +++ b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.cpp @@ -4,12 +4,13 @@ class Solution { if (numFriends == 1) { return word; } - int n = word.size(); - string ans; + int n = word.length(); + string ans = ""; for (int i = 0; i < n; ++i) { - int k = min(n - i, n - numFriends + 1); - string t = word.substr(i, k); - ans = max(ans, t); + string t = word.substr(i, min(n - i, n - (numFriends - 1))); + if (ans < t) { + ans = t; + } } return ans; } diff --git a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.go b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.go index 9da1acb21b408..f64abf6f13986 100644 --- a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.go +++ b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.go @@ -3,10 +3,9 @@ func answerString(word string, numFriends int) (ans string) { return word } n := len(word) - for i := range word { - k := min(n-i, n-numFriends+1) - t := word[i : i+k] + for i := 0; i < n; i++ { + t := word[i:min(n, i+n-(numFriends-1))] ans = max(ans, t) } return -} +} \ No newline at end of file diff --git a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.java b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.java index 552745c67fefe..695176f37ddee 100644 --- a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.java +++ b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.java @@ -6,12 +6,11 @@ public String answerString(String word, int numFriends) { int n = word.length(); String ans = ""; for (int i = 0; i < n; ++i) { - int k = Math.min(n - i, n - numFriends + 1); - String t = word.substring(i, i + k); + String t = word.substring(i, Math.min(n, i + n - (numFriends - 1))); if (ans.compareTo(t) < 0) { ans = t; } } return ans; } -} +} \ No newline at end of file diff --git a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.py b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.py index e3cc6c7fec6a9..cc427322d54f1 100644 --- a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.py +++ b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.py @@ -3,8 +3,4 @@ def answerString(self, word: str, numFriends: int) -> str: if numFriends == 1: return word n = len(word) - ans = "" - for i in range(n): - k = min(n - i, n - numFriends + 1) - ans = max(ans, word[i : i + k]) - return ans + return max(word[i : i + n - (numFriends - 1)] for i in range(n)) diff --git a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.ts b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.ts index 467d1db75b33a..b978782caf2d8 100644 --- a/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.ts +++ b/solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.ts @@ -2,14 +2,11 @@ function answerString(word: string, numFriends: number): string { if (numFriends === 1) { return word; } - let ans: string = ''; const n = word.length; - for (let i = 0; i < n; ++i) { - const k = Math.min(n - i, n - numFriends + 1); - const t = word.slice(i, i + k); - if (ans < t) { - ans = t; - } + let ans = ''; + for (let i = 0; i < n; i++) { + const t = word.slice(i, Math.min(n, i + n - (numFriends - 1))); + ans = t > ans ? t : ans; } return ans; }