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Commit 40f3d8b

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feat: add solutions to lc problem: No.3403 (doocs#4436)
No.3403.Find the Lexicographically Largest String From the Box I
1 parent 1ebe39f commit 40f3d8b

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7 files changed

+58
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‎solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README.md‎

Lines changed: 21 additions & 23 deletions
Original file line numberDiff line numberDiff line change
@@ -79,7 +79,13 @@ tags:
7979

8080
<!-- solution:start -->
8181

82-
### 方法一
82+
### 方法一:枚举子串左端点
83+
84+
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,ドル剩余子字符串的最小长度为 $\text{numFriends} - 1,ドル那么子字符串的右端点可以取到 $\min(n, i + n - (\text{numFriends} - 1)),ドル其中 $n$ 为字符串的长度。注意我们说的是左开右闭。
85+
86+
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。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。
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8490
<!-- tabs:start -->
8591

@@ -91,11 +97,7 @@ class Solution:
9197
if numFriends == 1:
9298
return word
9399
n = len(word)
94-
ans = ""
95-
for i in range(n):
96-
k = min(n - i, n - numFriends + 1)
97-
ans = max(ans, word[i : i + k])
98-
return ans
100+
return max(word[i : i + n - (numFriends - 1)] for i in range(n))
99101
```
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101103
#### Java
@@ -109,8 +111,7 @@ class Solution {
109111
int n = word.length();
110112
String ans = "";
111113
for (int i = 0; i < n; ++i) {
112-
int k = Math.min(n - i, n - numFriends + 1);
113-
String t = word.substring(i, i + k);
114+
String t = word.substring(i, Math.min(n, i + n - (numFriends - 1)));
114115
if (ans.compareTo(t) < 0) {
115116
ans = t;
116117
}
@@ -129,12 +130,13 @@ public:
129130
if (numFriends == 1) {
130131
return word;
131132
}
132-
int n = word.size();
133-
string ans;
133+
int n = word.length();
134+
string ans = "";
134135
for (int i = 0; i < n; ++i) {
135-
int k = min(n - i, n - numFriends + 1);
136-
string t = word.substr(i, k);
137-
ans = max(ans, t);
136+
string t = word.substr(i, min(n - i, n - (numFriends - 1)));
137+
if (ans < t) {
138+
ans = t;
139+
}
138140
}
139141
return ans;
140142
}
@@ -149,9 +151,8 @@ func answerString(word string, numFriends int) (ans string) {
149151
return word
150152
}
151153
n := len(word)
152-
for i := range word {
153-
k := min(n-i, n-numFriends+1)
154-
t := word[i : i+k]
154+
for i := 0; i < n; i++ {
155+
t := word[i:min(n, i+n-(numFriends-1))]
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ans = max(ans, t)
156157
}
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return
@@ -165,14 +166,11 @@ function answerString(word: string, numFriends: number): string {
165166
if (numFriends === 1) {
166167
return word;
167168
}
168-
let ans: string = '';
169169
const n = word.length;
170-
for (let i = 0; i < n; ++i) {
171-
const k = Math.min(n - i, n - numFriends + 1);
172-
const t = word.slice(i, i + k);
173-
if (ans < t) {
174-
ans = t;
175-
}
170+
let ans = '';
171+
for (let i = 0; i < n; i++) {
172+
const t = word.slice(i, Math.min(n, i + n - (numFriends - 1)));
173+
ans = t > ans ? t : ans;
176174
}
177175
return ans;
178176
}

‎solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/README_EN.md‎

Lines changed: 21 additions & 23 deletions
Original file line numberDiff line numberDiff line change
@@ -77,7 +77,13 @@ tags:
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Enumerate Substring Left Endpoints
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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,ドル and the minimum length of the remaining substrings is $\text{numFriends} - 1,ドル then the right endpoint of the substring can be up to $\min(n, i + n - (\text{numFriends} - 1)),ドル where $n$ is the length of the string. Note that we are talking about left-closed, right-open intervals.
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We enumerate all possible left endpoints, extract the corresponding substrings, compare their lexicographical order, and finally obtain the lexicographically largest substring.
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The time complexity is $O(n^2),ドル and the space complexity is $O(n),ドル where $n$ is the length of the string.
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<!-- tabs:start -->
8389

@@ -89,11 +95,7 @@ class Solution:
8995
if numFriends == 1:
9096
return word
9197
n = len(word)
92-
ans = ""
93-
for i in range(n):
94-
k = min(n - i, n - numFriends + 1)
95-
ans = max(ans, word[i : i + k])
96-
return ans
98+
return max(word[i : i + n - (numFriends - 1)] for i in range(n))
9799
```
98100

99101
#### Java
@@ -107,8 +109,7 @@ class Solution {
107109
int n = word.length();
108110
String ans = "";
109111
for (int i = 0; i < n; ++i) {
110-
int k = Math.min(n - i, n - numFriends + 1);
111-
String t = word.substring(i, i + k);
112+
String t = word.substring(i, Math.min(n, i + n - (numFriends - 1)));
112113
if (ans.compareTo(t) < 0) {
113114
ans = t;
114115
}
@@ -127,12 +128,13 @@ public:
127128
if (numFriends == 1) {
128129
return word;
129130
}
130-
int n = word.size();
131-
string ans;
131+
int n = word.length();
132+
string ans = "";
132133
for (int i = 0; i < n; ++i) {
133-
int k = min(n - i, n - numFriends + 1);
134-
string t = word.substr(i, k);
135-
ans = max(ans, t);
134+
string t = word.substr(i, min(n - i, n - (numFriends - 1)));
135+
if (ans < t) {
136+
ans = t;
137+
}
136138
}
137139
return ans;
138140
}
@@ -147,9 +149,8 @@ func answerString(word string, numFriends int) (ans string) {
147149
return word
148150
}
149151
n := len(word)
150-
for i := range word {
151-
k := min(n-i, n-numFriends+1)
152-
t := word[i : i+k]
152+
for i := 0; i < n; i++ {
153+
t := word[i:min(n, i+n-(numFriends-1))]
153154
ans = max(ans, t)
154155
}
155156
return
@@ -163,14 +164,11 @@ function answerString(word: string, numFriends: number): string {
163164
if (numFriends === 1) {
164165
return word;
165166
}
166-
let ans: string = '';
167167
const n = word.length;
168-
for (let i = 0; i < n; ++i) {
169-
const k = Math.min(n - i, n - numFriends + 1);
170-
const t = word.slice(i, i + k);
171-
if (ans < t) {
172-
ans = t;
173-
}
168+
let ans = '';
169+
for (let i = 0; i < n; i++) {
170+
const t = word.slice(i, Math.min(n, i + n - (numFriends - 1)));
171+
ans = t > ans ? t : ans;
174172
}
175173
return ans;
176174
}

‎solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.cpp‎

Lines changed: 6 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -4,12 +4,13 @@ class Solution {
44
if (numFriends == 1) {
55
return word;
66
}
7-
int n = word.size();
8-
string ans;
7+
int n = word.length();
8+
string ans = "";
99
for (int i = 0; i < n; ++i) {
10-
int k = min(n - i, n - numFriends + 1);
11-
string t = word.substr(i, k);
12-
ans = max(ans, t);
10+
string t = word.substr(i, min(n - i, n - (numFriends - 1)));
11+
if (ans < t) {
12+
ans = t;
13+
}
1314
}
1415
return ans;
1516
}

‎solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.go‎

Lines changed: 3 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -3,10 +3,9 @@ func answerString(word string, numFriends int) (ans string) {
33
return word
44
}
55
n := len(word)
6-
for i := range word {
7-
k := min(n-i, n-numFriends+1)
8-
t := word[i : i+k]
6+
for i := 0; i < n; i++ {
7+
t := word[i:min(n, i+n-(numFriends-1))]
98
ans = max(ans, t)
109
}
1110
return
12-
}
11+
}

‎solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.java‎

Lines changed: 2 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -6,12 +6,11 @@ public String answerString(String word, int numFriends) {
66
int n = word.length();
77
String ans = "";
88
for (int i = 0; i < n; ++i) {
9-
int k = Math.min(n - i, n - numFriends + 1);
10-
String t = word.substring(i, i + k);
9+
String t = word.substring(i, Math.min(n, i + n - (numFriends - 1)));
1110
if (ans.compareTo(t) < 0) {
1211
ans = t;
1312
}
1413
}
1514
return ans;
1615
}
17-
}
16+
}

‎solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.py‎

Lines changed: 1 addition & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -3,8 +3,4 @@ def answerString(self, word: str, numFriends: int) -> str:
33
if numFriends == 1:
44
return word
55
n = len(word)
6-
ans = ""
7-
for i in range(n):
8-
k = min(n - i, n - numFriends + 1)
9-
ans = max(ans, word[i : i + k])
10-
return ans
6+
return max(word[i : i + n - (numFriends - 1)] for i in range(n))

‎solution/3400-3499/3403.Find the Lexicographically Largest String From the Box I/Solution.ts‎

Lines changed: 4 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -2,14 +2,11 @@ function answerString(word: string, numFriends: number): string {
22
if (numFriends === 1) {
33
return word;
44
}
5-
let ans: string = '';
65
const n = word.length;
7-
for (let i = 0; i < n; ++i) {
8-
const k = Math.min(n - i, n - numFriends + 1);
9-
const t = word.slice(i, i + k);
10-
if (ans < t) {
11-
ans = t;
12-
}
6+
let ans = '';
7+
for (let i = 0; i < n; i++) {
8+
const t = word.slice(i, Math.min(n, i + n - (numFriends - 1)));
9+
ans = t > ans ? t : ans;
1310
}
1411
return ans;
1512
}

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