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feat: add solutions to lc problems: No.1961~1964 (#2145)
* No.1961.Check If String Is a Prefix of Array * No.1963.Minimum Number of Swaps to Make the String Balanced * No.1964.Find the Longest Valid Obstacle Course at Each Position
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‎solution/1900-1999/1961.Check If String Is a Prefix of Array/README.md

Lines changed: 53 additions & 13 deletions
Original file line numberDiff line numberDiff line change
@@ -46,6 +46,14 @@ s 可以由 "i"、"love" 和 "leetcode" 相连得到。
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：遍历**
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我们遍历数组 $words$，用一个变量 $t$ 记录当前已经拼接的字符串，如果 $t$ 的长度大于 $s$ 的长度，说明 $s$ 不是 $words$ 的前缀字符串，返回 $false$；如果 $t$ 的长度等于 $s$ 的长度，返回 $t$ 是否等于 $s$。
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遍历结束后，如果 $t$ 的长度小于 $s$ 的长度，说明 $s$ 不是 $words$ 的前缀字符串，返回 $false$。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是字符串 $s$ 的长度。
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<!-- tabs:start -->
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### **Python3**
@@ -55,11 +63,11 @@ s 可以由 "i"、"love" 和 "leetcode" 相连得到。
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```python
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class Solution:
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def isPrefixString(self, s: str, words: List[str]) -> bool:
58-
t = 0
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n, m = len(s), 0
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for i, w in enumerate(words):
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t += len(w)
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if len(s) == t:
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return ''.join(words[: i + 1]) == s
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m += len(w)
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if m == n:
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return "".join(words[: i + 1]) == s
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return False
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```
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@@ -71,9 +79,12 @@ class Solution:
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class Solution {
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public boolean isPrefixString(String s, String[] words) {
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StringBuilder t = new StringBuilder();
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for (String w : words) {
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for (var w : words) {
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t.append(w);
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if (s.length() == t.length()) {
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if (t.length() > s.length()) {
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return false;
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}
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if (t.length() == s.length()) {
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return s.equals(t.toString());
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}
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}
@@ -88,10 +99,15 @@ class Solution {
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class Solution {
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public:
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bool isPrefixString(string s, vector<string>& words) {
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string t = "";
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for (string& w : words) {
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string t;
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for (auto& w : words) {
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t += w;
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if (t.size() == s.size()) return t == s;
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if (t.size() > s.size()) {
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return false;
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}
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if (t.size() == s.size()) {
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return t == s;
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}
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}
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return false;
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}
@@ -102,17 +118,41 @@ public:
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```go
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func isPrefixString(s string, words []string) bool {
105-
t := ""
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t := strings.Builder{}
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for _, w := range words {
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t += w
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if t == s {
109-
return true
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t.WriteString(w)
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if t.Len() > len(s) {
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return false
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}
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if t.Len() == len(s) {
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return t.String() == s
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}
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}
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return false
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}
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```
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### **TypeScript**
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```ts
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function isPrefixString(s: string, words: string[]): boolean {
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const t: string[] = [];
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const n = s.length;
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let m = 0;
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for (const w of words) {
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m += w.length;
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if (m > n) {
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return false;
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}
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t.push(w);
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if (m === n) {
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return s === t.join('');
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}
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}
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return false;
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}
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```
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### **...**
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```

‎solution/1900-1999/1961.Check If String Is a Prefix of Array/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -40,18 +40,26 @@ It is impossible to make s using a prefix of arr.</pre>
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## Solutions
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**Solution 1: Traversal**
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We traverse the array $words$, using a variable $t$ to record the currently concatenated string. If the length of $t$ is greater than the length of $s$, it means that $s$ is not a prefix string of $words$, so we return $false$; if the length of $t$ is equal to the length of $s$, we return whether $t$ is equal to $s$.
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At the end of the traversal, if the length of $t$ is less than the length of $s$, it means that $s$ is not a prefix string of $words$, so we return $false$.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def isPrefixString(self, s: str, words: List[str]) -> bool:
50-
t = 0
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n, m = len(s), 0
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for i, w in enumerate(words):
52-
t += len(w)
53-
if len(s) == t:
54-
return ''.join(words[: i + 1]) == s
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m += len(w)
61+
if m == n:
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return "".join(words[: i + 1]) == s
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return False
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```
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@@ -61,9 +69,12 @@ class Solution:
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class Solution {
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public boolean isPrefixString(String s, String[] words) {
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StringBuilder t = new StringBuilder();
64-
for (String w : words) {
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for (var w : words) {
6573
t.append(w);
66-
if (s.length() == t.length()) {
74+
if (t.length() > s.length()) {
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return false;
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}
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if (t.length() == s.length()) {
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return s.equals(t.toString());
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}
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}
@@ -78,10 +89,15 @@ class Solution {
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class Solution {
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public:
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bool isPrefixString(string s, vector<string>& words) {
81-
string t = "";
82-
for (string& w : words) {
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string t;
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for (auto& w : words) {
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t += w;
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if (t.size() == s.size()) return t == s;
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if (t.size() > s.size()) {
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return false;
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}
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if (t.size() == s.size()) {
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return t == s;
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}
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}
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return false;
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}
@@ -92,17 +108,41 @@ public:
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```go
94110
func isPrefixString(s string, words []string) bool {
95-
t := ""
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t := strings.Builder{}
96112
for _, w := range words {
97-
t += w
98-
if t == s {
99-
return true
113+
t.WriteString(w)
114+
if t.Len() > len(s) {
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return false
116+
}
117+
if t.Len() == len(s) {
118+
return t.String() == s
100119
}
101120
}
102121
return false
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}
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```
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### **TypeScript**
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```ts
128+
function isPrefixString(s: string, words: string[]): boolean {
129+
const t: string[] = [];
130+
const n = s.length;
131+
let m = 0;
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for (const w of words) {
133+
m += w.length;
134+
if (m > n) {
135+
return false;
136+
}
137+
t.push(w);
138+
if (m === n) {
139+
return s === t.join('');
140+
}
141+
}
142+
return false;
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}
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```
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### **...**
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```

‎solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.cpp

Lines changed: 15 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,16 @@
1-
class Solution {
2-
public:
3-
bool isPrefixString(string s, vector<string>& words) {
4-
string t = "";
5-
for (string& w : words) {
6-
t += w;
7-
if (t.size() == s.size()) return t == s;
8-
}
9-
return false;
10-
}
1+
class Solution {
2+
public:
3+
bool isPrefixString(string s, vector<string>& words) {
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string t;
5+
for (auto& w : words) {
6+
t += w;
7+
if (t.size() > s.size()) {
8+
return false;
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}
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if (t.size() == s.size()) {
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return t == s;
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}
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}
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return false;
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}
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};

‎solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.go

Lines changed: 7 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,9 +1,12 @@
11
func isPrefixString(s string, words []string) bool {
2-
t := ""
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t := strings.Builder{}
33
for _, w := range words {
4-
t += w
5-
if t == s {
6-
return true
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t.WriteString(w)
5+
if t.Len() > len(s) {
6+
return false
7+
}
8+
if t.Len() == len(s) {
9+
return t.String() == s
710
}
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}
912
return false

‎solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.java

Lines changed: 14 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,15 @@
1-
class Solution {
2-
public boolean isPrefixString(String s, String[] words) {
3-
StringBuilder t = new StringBuilder();
4-
for (String w : words) {
5-
t.append(w);
6-
if (s.length() == t.length()) {
7-
return s.equals(t.toString());
8-
}
9-
}
10-
return false;
11-
}
1+
class Solution {
2+
public boolean isPrefixString(String s, String[] words) {
3+
StringBuilder t = new StringBuilder();
4+
for (var w : words) {
5+
t.append(w);
6+
if (t.length() > s.length()) {
7+
return false;
8+
}
9+
if (t.length() == s.length()) {
10+
return s.equals(t.toString());
11+
}
12+
}
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return false;
14+
}
1215
}

‎solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.py

Lines changed: 8 additions & 8 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,8 @@
1-
class Solution:
2-
def isPrefixString(self, s: str, words: List[str]) -> bool:
3-
t = 0
4-
for i, w in enumerate(words):
5-
t += len(w)
6-
if len(s) == t:
7-
return ''.join(words[: i + 1]) == s
8-
return False
1+
class Solution:
2+
def isPrefixString(self, s: str, words: List[str]) -> bool:
3+
n, m = len(s), 0
4+
for i, w in enumerate(words):
5+
m += len(w)
6+
if m == n:
7+
return "".join(words[: i + 1]) == s
8+
return False

‎solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.ts

Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
1+
function isPrefixString(s: string, words: string[]): boolean {
2+
const t: string[] = [];
3+
const n = s.length;
4+
let m = 0;
5+
for (const w of words) {
6+
m += w.length;
7+
if (m > n) {
8+
return false;
9+
}
10+
t.push(w);
11+
if (m === n) {
12+
return s === t.join('');
13+
}
14+
}
15+
return false;
16+
}

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