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feat: add solutions to lc problems: No.1961~1964 #2145

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Dec 23, 2023
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feat: add solutions to lc problems: No.1961~1964 #2145

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34dde01
feat: add solutions to lc problems: No.1961~1964
yanglbme Dec 23, 2023
67c7853
fix: readme en
yanglbme Dec 23, 2023
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66 changes: 53 additions & 13 deletions solution/1900-1999/1961.Check If String Is a Prefix of Array/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -46,6 +46,14 @@ s 可以由 "i"、"love" 和 "leetcode" 相连得到。

<!-- 这里可写通用的实现逻辑 -->

**方法一：遍历**

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 $s$。

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 $false$。

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。

<!-- tabs:start -->

### **Python3**
Expand All @@ -55,11 +63,11 @@ s 可以由 "i"、"love" 和 "leetcode" 相连得到。
```python
class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
t = 0
n, m = len(s), 0
for i, w in enumerate(words):
t += len(w)
if len(s) == t:
return ''.join(words[: i + 1]) == s
m += len(w)
if m == n:
return "".join(words[: i + 1]) == s
return False
```

Expand All @@ -71,9 +79,12 @@ class Solution:
class Solution {
public boolean isPrefixString(String s, String[] words) {
StringBuilder t = new StringBuilder();
for (String w : words) {
for (var w : words) {
t.append(w);
if (s.length() == t.length()) {
if (t.length() > s.length()) {
return false;
}
if (t.length() == s.length()) {
return s.equals(t.toString());
}
}
Expand All @@ -88,10 +99,15 @@ class Solution {
class Solution {
public:
bool isPrefixString(string s, vector<string>& words) {
string t = "";
for (string& w : words) {
string t;
for (auto& w : words) {
t += w;
if (t.size() == s.size()) return t == s;
if (t.size() > s.size()) {
return false;
}
if (t.size() == s.size()) {
return t == s;
}
}
return false;
}
Expand All @@ -102,17 +118,41 @@ public:

```go
func isPrefixString(s string, words []string) bool {
t := ""
t := strings.Builder{}
for _, w := range words {
t += w
if t == s {
return true
t.WriteString(w)
if t.Len() > len(s) {
return false
}
if t.Len() == len(s) {
return t.String() == s
}
}
return false
}
```

### **TypeScript**

```ts
function isPrefixString(s: string, words: string[]): boolean {
const t: string[] = [];
const n = s.length;
let m = 0;
for (const w of words) {
m += w.length;
if (m > n) {
return false;
}
t.push(w);
if (m === n) {
return s === t.join('');
}
}
return false;
}
```

### **...**

```
Expand Down
66 changes: 53 additions & 13 deletions solution/1900-1999/1961.Check If String Is a Prefix of Array/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -40,18 +40,26 @@ It is impossible to make s using a prefix of arr.</pre>

## Solutions

**Solution 1: Traversal**

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 $s$.

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 $false$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

<!-- tabs:start -->

### **Python3**

```python
class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
t = 0
n, m = len(s), 0
for i, w in enumerate(words):
t += len(w)
if len(s) == t:
return ''.join(words[: i + 1]) == s
m += len(w)
if m == n:
return "".join(words[: i + 1]) == s
return False
```

Expand All @@ -61,9 +69,12 @@ class Solution:
class Solution {
public boolean isPrefixString(String s, String[] words) {
StringBuilder t = new StringBuilder();
for (String w : words) {
for (var w : words) {
t.append(w);
if (s.length() == t.length()) {
if (t.length() > s.length()) {
return false;
}
if (t.length() == s.length()) {
return s.equals(t.toString());
}
}
Expand All @@ -78,10 +89,15 @@ class Solution {
class Solution {
public:
bool isPrefixString(string s, vector<string>& words) {
string t = "";
for (string& w : words) {
string t;
for (auto& w : words) {
t += w;
if (t.size() == s.size()) return t == s;
if (t.size() > s.size()) {
return false;
}
if (t.size() == s.size()) {
return t == s;
}
}
return false;
}
Expand All @@ -92,17 +108,41 @@ public:

```go
func isPrefixString(s string, words []string) bool {
t := ""
t := strings.Builder{}
for _, w := range words {
t += w
if t == s {
return true
t.WriteString(w)
if t.Len() > len(s) {
return false
}
if t.Len() == len(s) {
return t.String() == s
}
}
return false
}
```

### **TypeScript**

```ts
function isPrefixString(s: string, words: string[]): boolean {
const t: string[] = [];
const n = s.length;
let m = 0;
for (const w of words) {
m += w.length;
if (m > n) {
return false;
}
t.push(w);
if (m === n) {
return s === t.join('');
}
}
return false;
}
```

### **...**

```
Expand Down
25 changes: 15 additions & 10 deletions solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,11 +1,16 @@
class Solution {
public:
bool isPrefixString(string s, vector<string>& words) {
string t = "";
for (string& w : words) {
t += w;
if (t.size() == s.size()) return t == s;
}
return false;
}
class Solution {
public:
bool isPrefixString(string s, vector<string>& words) {
string t;
for (auto& w : words) {
t += w;
if (t.size() > s.size()) {
return false;
}
if (t.size() == s.size()) {
return t == s;
}
}
return false;
}
};
11 changes: 7 additions & 4 deletions solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,9 +1,12 @@
func isPrefixString(s string, words []string) bool {
t := ""
t := strings.Builder{}
for _, w := range words {
t += w
if t == s {
return true
t.WriteString(w)
if t.Len() > len(s) {
return false
}
if t.Len() == len(s) {
return t.String() == s
}
}
return false
Expand Down
25 changes: 14 additions & 11 deletions solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,12 +1,15 @@
class Solution {
public boolean isPrefixString(String s, String[] words) {
StringBuilder t = new StringBuilder();
for (String w : words) {
t.append(w);
if (s.length() == t.length()) {
return s.equals(t.toString());
}
}
return false;
}
class Solution {
public boolean isPrefixString(String s, String[] words) {
StringBuilder t = new StringBuilder();
for (var w : words) {
t.append(w);
if (t.length() > s.length()) {
return false;
}
if (t.length() == s.length()) {
return s.equals(t.toString());
}
}
return false;
}
}
16 changes: 8 additions & 8 deletions solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,8 +1,8 @@
class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
t = 0
for i, w in enumerate(words):
t += len(w)
if len(s) == t:
return ''.join(words[: i + 1]) == s
return False
class Solution:
def isPrefixString(self, s: str, words: List[str]) -> bool:
n, m = len(s), 0
for i, w in enumerate(words):
m += len(w)
if m == n:
return "".join(words[: i + 1]) == s
return False
16 changes: 16 additions & 0 deletions solution/1900-1999/1961.Check If String Is a Prefix of Array/Solution.ts
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,16 @@
function isPrefixString(s: string, words: string[]): boolean {
const t: string[] = [];
const n = s.length;
let m = 0;
for (const w of words) {
m += w.length;
if (m > n) {
return false;
}
t.push(w);
if (m === n) {
return s === t.join('');
}
}
return false;
}
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