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Merged
merged 1 commit into from
May 26, 2025
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feat: add solutions to lc problem: No.3403 #4436

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44 changes: 21 additions & 23 deletions ...00-3499/3403.Find the Lexicographically Largest String From the Box I/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -79,7 +79,13 @@ tags:

<!-- solution:start -->

### 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。

<!-- tabs:start -->

Expand All @@ -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
Expand All @@ -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;
}
Expand All @@ -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;
}
Expand All @@ -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
Expand All @@ -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;
}
Expand Down
44 changes: 21 additions & 23 deletions ...3499/3403.Find the Lexicographically Largest String From the Box I/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -77,7 +77,13 @@ tags:

<!-- solution:start -->

### 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.

<!-- tabs:start -->

Expand All @@ -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
Expand All @@ -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;
}
Expand All @@ -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;
}
Expand All @@ -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
Expand All @@ -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;
}
Expand Down
11 changes: 6 additions & 5 deletions ...tion/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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;
}
Expand Down
7 changes: 3 additions & 4 deletions solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.go
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Original file line number Diff line number Diff line change
Expand Up @@ -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
}
}
5 changes: 2 additions & 3 deletions ...ion/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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;
}
}
}
6 changes: 1 addition & 5 deletions solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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))
11 changes: 4 additions & 7 deletions solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.ts
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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;
}