> For the complete documentation index, see [llms.txt](https://heunnajo.gitbook.io/algorithms-problem-solving-skills/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://heunnajo.gitbook.io/algorithms-problem-solving-skills/algorithm-problems/longest-substring-with-at-most-two-distinct.md).

# Longest Substring With At Most Two Distinct

Given a string s, find the length of the longest substring *t* that contains at most 2 distinct characters.\
최대 2개의 구분되 철자를 포함하는 가장 긴 substring의 길이를 구해라.

> Example 1 :\
> Input : "eceba"\
> Output : 3\
> Explanation : t is "ece" which its length is 3.
>
> Example 2 :\
> Input : "ccaabbb"\
> Output : 5\
> Explanation : t is "aabbb" which its length is 5.

**자료구조** : **정수형 변수 4개, Map(Character, Integer)**\
1\. start : 시작 인덱스\
2\. end : 끝 인덱스\
3\. count : 문자열의 갯수\
4\. leng : 문자열의 길이(end-start 중 큰 값) 

**알고리즘**\
1\. Map에 저장한다. - c 2개, a 2개, b 4개 => (c,2) (a,2) (b,4)\
2\. 문자의 갯수는 최대 2개이다. - 문자 갯수(count)가 2개 넘는다면 start(시작 인덱스)로 이를 제어한다.

알고리즘을 Java로 구현

```java
package java_basic;
import java.util.*;

public class LongestSubMostTwo {

	public static void main(String[] args) {
		String s = "ccaabbb";
		System.out.println(solve(s));
	}

	private static int solve(String s) {
		//1. 그릇 생성 
		Map<Character, Integer> map = new HashMap<>();//(c,2) (a,2) (b,3)
		int start = 0, end = 0, count = 0, leng = 0;
		
		//2. 반복문 돌린다.
		while(end < s.length()) {
			char c = s.charAt(end);
			map.put(c, map.getOrDefault(c, 0)+1);//(c,2) (a,2)
			if(map.get(c) == 1) count++;//new char, count는 2가 된다. 
			end++;
			
			while(count > 2) {//문자 갯수가 2보다 많으면! 
				char tmp = s.charAt(start);
				map.put(tmp,map.get(tmp)-1);
				if(map.get(tmp) == 0) { 
					count--;
				}
				start++;
			}
			//3. leng 리턴
			leng = Math.max(leng, end-start);
		}
		return leng;
		
	}

}

```

**배운 내용 정리**(Java 문법 및 메소드)\
1\. Map.getOrDefault() : 찾는 키가 존재한다면 찾는 키의 값을 반환하고 없다면 기본 값을 반환하는 메서드

```java
getOrDefault(Object key, V DefaultValue)
```

```java
char c = s.charAt(end);
map.put(c, map.getOrDefault(c, 0)+1);
//처음에는 c가 map에 없으니 0을 리턴.
//=>map.put(c,1)
//map에 (c,1)이 존재하니까 해당 값인 1을 리턴!
//=>map.put(c,1+1)=>map.put(c,2)
```

\ <br>


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