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Updated 383-ransom-note.md.
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‎en/1-1000/383-ransom-note.md‎

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3. Then, traverse `ransomNote` and perform reverse operations on the data in `Map`. If the count of a character is less than 0, return `false`.
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## Steps
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1. First count the characters in `magazine`, and store the results in `Map`.
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```javascript
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charToCount = new Map()
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for (character in magazine) {
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charToCount[character] += 1
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}
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```
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```javascript
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charToCount = new Map()
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for (character in magazine) {
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charToCount[character] += 1
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}
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```
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2. Then, traverse `ransomNote` and perform reverse operations on the data in `Map`. If the count of a character is less than 0, return `false`.
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```javascript
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charToCount = new Map()
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for (character in magazine) {
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charToCount[character] += 1
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}
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for (character in ransomNote) {
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charToCount[character] -= 1
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if (charToCount[character] < 0) {
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return false
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}
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}
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return true
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```
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```javascript
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charToCount = new Map()
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for (character in magazine) {
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charToCount[character] += 1
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}
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for (character in ransomNote) {
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charToCount[character] -= 1
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if (charToCount[character] < 0) {
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return false
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}
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}
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return true
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```
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## Complexity
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* Time: `O(n)`.

‎en/3001-4000/unorganized.md‎

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- https://leetcode.com/problems/find-special-substring-of-length-k/
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- https://leetcode.cn/problems/eat-pizzas/
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- https://leetcode.cn/problems/merge-intervals/
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## Other algorithm
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* 线段树 https://leetcode.cn/problems/fruits-into-baskets-iii
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*

‎zh/1-1000/383-ransom-note.md‎

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# 383. Ransom Note - Best Practices of LeetCode Solutions
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LeetCode link: [383. Ransom Note](https://leetcode.com/problems/ransom-note),
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[383. 赎金信](https://leetcode.cn/problems/ransom-note)
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原文链接：[coding5.com - 力扣题解最佳实践](https://coding5.com/zh/leetcode/383-ransom-note)
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[中文题解](#中文题解)
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# 383. 赎金信 - 力扣题解最佳实践
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## LeetCode problem description
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Given two strings `ransomNote` and `magazine`, return `true` if `ransomNote` can be constructed by using the letters from `magazine` and `false` otherwise.
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力扣链接：[383. 赎金信](https://leetcode.cn/problems/ransom-note), 难度：**简单**。
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Each letter in `magazine` can only be used once in `ransomNote`.
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## 力扣“383. 赎金信”问题描述
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Difficulty: **Easy**
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给你两个字符串：`ransomNote` 和 `magazine` ，判断 `ransomNote` 能不能由 `magazine` 里面的字符构成。
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### [Example 1]
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**Input**: `ransomNote = "a", magazine = "b"`
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如果可以，返回 `true` ；否则返回 `false` 。
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**Output**: `false`
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`magazine` 中的每个字符只能在 `ransomNote` 中使用一次。
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### [Example 2]
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**Input**: `ransomNote = "aa", magazine = "ab"`
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### [示例 1]
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**Output**: `false`
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**输入**: `ransomNote = "a", magazine = "b"`
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### [Example 3]
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**Input**: `ransomNote = "aa", magazine = "aab"`
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**输出**: `false`
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### [示例 2]
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**Output**: `true`
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**输入**: `ransomNote = "aa", magazine = "ab"`
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### [Constraints]
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- `1 <= ransomNote.length, magazine.length <= 100000`
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- `ransomNote` and `magazine` consist of lowercase English letters.
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**输出**: `false`
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### [示例 3]
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## Intuition
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[中文题解](#中文题解)
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**输入**: `ransomNote = "aa", magazine = "aab"`
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1. This question is equivalent to asking whether `magazine` can contain all the characters in `ransomNote`.
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2. First count the characters in `magazine`, and store the results in `Map`.
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3. Then, traverse `ransomNote` and perform reverse operations on the data in `Map`. If the count of a character is less than 0, return `false`.
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**输出**: `true`
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### [约束]
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## Steps
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1. First count the characters in `magazine`, and store the results in `Map`.
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```javascript
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charToCount = new Map()
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- `1 <= ransomNote.length, magazine.length <= 10^5`
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- `ransomNote` 和 `magazine` 由小写英文字母组成
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for (character in magazine) {
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charToCount[character] += 1
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}
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```
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## 思路
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2. Then, traverse `ransomNote` and perform reverse operations on the data in `Map`. If the count of a character is less than 0, return `false`.
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```javascript
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charToCount = new Map()
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1. 本题等同于求`magazine`是否能包含`ransomNote`中的所有字符。
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2. 先对`magazine`进行统计，得出每个字符对应的字数，结果存储在`Map`中。每一次都是一个加一的操作。
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3. 下一步做什么？
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<details><summary>点击查看答案</summary><p>遍历`ransomNote`，对当前字符对应的数量进行减一操作（反向操作）。如果某个字符的数量小于0，则返回`false`。</p></details>
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for (character in magazine) {
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charToCount[character] += 1
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}
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## 步骤
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for (character in ransomNote) {
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charToCount[character] -= 1
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1. 先对`magazine`进行字符和字数统计，结果存储在`Map`中。
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if (charToCount[character] < 0) {
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return false
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}
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}
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```javascript
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charToCount = new Map()
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return true
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```
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for (character in magazine) {
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charToCount[character] += 1
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}
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```
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2. 然后，遍历`ransomNote`，并对`Map`中的数据进行反向操作。如果某个字符的字数小于0，则返回`false`。
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## Complexity
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* Time: `O(n)`.
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* Space: `O(n)`.
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```javascript
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charToCount = new Map()
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for (character in magazine) {
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charToCount[character] += 1
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}
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for (character in ransomNote) {
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charToCount[character] -= 1
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if (charToCount[character] < 0) {
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return false
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}
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}
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return true
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```
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## 复杂度
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- 时间复杂度: `O(N)`.
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- 空间复杂度: `O(N)`.
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## Java
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```java
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class Solution {
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public boolean canConstruct(String ransomNote, String magazine) {
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```
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## Python
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```python
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# from collections import defaultdict
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return True
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```
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## C++
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```cpp
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## JavaScript
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```javascript
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var canConstruct = function (ransomNote, magazine) {
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const charToCount = new Map()
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```
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## C#
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```c#
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public class Solution
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{
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}
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```
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## Go
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```go
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Ruby
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```ruby
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# Welcome to create a PR to complete the code of this language, thanks!
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```
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## C
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```c
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Kotlin
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```kotlin
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Swift
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```swift
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Rust
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```rust
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Other languages
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## 力扣问题描述
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[383. 赎金信](https://leetcode.cn/problems/ransom-note) ，难度：**简单**。
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给你两个字符串：`ransomNote` 和 `magazine` ，判断 `ransomNote` 能不能由 `magazine` 里面的字符构成。
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如果可以，返回 `true` ；否则返回 `false` 。
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`magazine` 中的每个字符只能在 `ransomNote` 中使用一次。
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### [示例 2]
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**输入**: `ransomNote = "aa", magazine = "ab"`
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**输出**: `false`
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### [示例 3]
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**输入**: `ransomNote = "aa", magazine = "aab"`
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**输出**: `true`
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# 中文题解
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## 思路
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1. 本题等同于求`magazine`是否能包含`ransomNote`中的所有字符。
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2. 先对`magazine`进行字符和字数统计，结果存储在`Map`中。
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3. 然后，遍历`ransomNote`，并对`Map`中的数据进行反向操作。如果某个字符的字数小于0，则返回`false`。
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## 步骤
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1. 先对`magazine`进行字符和字数统计，结果存储在`Map`中。
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```javascript
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charToCount = new Map()
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for (character in magazine) {
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charToCount[character] += 1
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}
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```
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2. 然后，遍历`ransomNote`，并对`Map`中的数据进行反向操作。如果某个字符的字数小于0，则返回`false`。
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```javascript
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charToCount = new Map()
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for (character in magazine) {
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charToCount[character] += 1
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}
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for (character in ransomNote) {
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charToCount[character] -= 1
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if (charToCount[character] < 0) {
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return false
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}
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}
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return true
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```

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