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YoungSeok-ChoiYoungSeok-Choi
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Merge pull request #1409 from YoungSeok-Choi/feature/week-6
[YoungSeok-Choi] week 6 solutions
2 parents 2560589 + ae12971 commit 58386e4

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6 files changed

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‎container-with-most-water/YoungSeok-Choi.java

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// tc: O(n)
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// 투 포인터라는 문제풀이 방법으로 간단히 (답지보고) 해결...
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class Solution {
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public int maxArea(int[] h) {
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int start = 0;
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int end = h.length - 1;
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int mx = -987654321;
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while(start < end) {
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mx = Math.max(mx, (end - start) * Math.min(h[start], h[end]));
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if(h[start] < h[end]) {
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start++;
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} else {
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end--;
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}
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}
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return mx;
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}
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}
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// 예외처리가 덕지덕지 붙기 시작할 때. (잘못됨을 깨닫고)다른 방법으로 풀어햐 하는건 아닌지 생각하는 습관 필요함..ㅠ
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class WrongSolution {
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public int maxArea(int[] h) {
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int mx = Math.min(h[0], h[1]);
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int idx = 0;
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for(int i = 1; i < h.length; i++) {
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int offset = i - idx;
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int prevCalc = Math.min(h[i - 1], h[i]);
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int calc = Math.min(h[idx], h[i]);
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int newMx = calc * offset;
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// 새롭게 인덱스를 바꿔버리는게 더 나을때.
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if(prevCalc > newMx) {
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idx = i - 1;
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mx = Math.max(mx, prevCalc);
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continue;
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}
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// 물을 더 많이 담을 수 있을 때.
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if(h[idx] < h[i] && newMx > mx) {
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if(i == 2) {
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int exc = Math.min(h[1], h[i]) * offset;
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if(exc > newMx) {
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idx = 1;
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mx = Math.max(exc, mx);
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continue;
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}
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}
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idx = i;
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}
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mx = Math.max(newMx, mx);
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}
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return mx;
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}
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}

‎design-add-and-search-words-data-structure/YoungSeok-Choi.java

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import java.util.ArrayList;
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import java.util.HashMap;
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import java.util.List;
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import java.util.Map;
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// NOTE: dot에 대한 wildcard 검색이 핵심이었던 문제.
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// NOTE: tc -> 문자 입력 O(n), 패턴검색 O(26^n), 패턴X검색 O(1)
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/**
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검색어 | 의미
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"." | 길이가 1인 아무 문자와 매치
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".a" | 길이 2: 첫 글자는 아무 문자, 두 번째는 a
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"..d" | 길이 3: 처음 두 글자는 아무 문자, 마지막은 d
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"..." | 길이 3인 모든 단어와 매치
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*/
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class WordDictionary {
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private WordDictionary[] child;
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private Character val;
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private Map<String, Boolean> cMap;
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public WordDictionary() {
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this.cMap = new HashMap<>();
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this.val = null;
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}
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public void addWord(String word) {
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if(this.cMap.containsKey(word)) return;
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this.cMap.put(word, true);
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this.innerInsert(word);
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}
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public void innerInsert(String word) {
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if(word.length() == 0) return;
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if(this.child == null) {
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this.child = new WordDictionary[26];
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}
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char c = word.charAt(0);
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int idx = c - 97;
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if(this.child[idx] == null) {
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this.child[idx] = new WordDictionary();
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this.child[idx].val = c;
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}
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this.child[idx].innerInsert(word.substring(1));
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}
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public boolean search(String word) {
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if(!word.contains(".")) return this.cMap.containsKey(word);
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return this.patternSearch(word);
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}
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private boolean patternSearch(String wildcard) {
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if(wildcard.length() == 0) return true;
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char c = wildcard.charAt(0);
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if(c == '.') { // NOTE: wildcard를 만났을 때, 해당 depth는 검사하지 않고, 다음 번(자식) 문자들에 대해서 검색 이어나가기.
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List<Boolean> res = new ArrayList<>();
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for(WordDictionary children : this.child) {
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if(children != null) {
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res.add(children.patternSearch(wildcard.substring(1)));
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}
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}
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for(boolean b : res) {
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if(b) return true;
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}
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return false;
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}
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int idx = c - 97;
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if(this.child == null || this.child[idx] == null || this.child[idx].val == null) return false;
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return this.child[idx].patternSearch(wildcard.substring(1));
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}
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}
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/**
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* Your WordDictionary object will be instantiated and called as such:
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* WordDictionary obj = new WordDictionary();
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* obj.addWord(word);
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* boolean param_2 = obj.search(word);
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*/
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// 이전 TRIE 자료구조를 모른 채로 풀이했던 문제..
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class PrevWordDictionary {
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private String word;
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private PrevWordDictionary leftChild;
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private PrevWordDictionary rightChild;
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public PrevWordDictionary() {
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this.leftChild = null;
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this.rightChild = null;
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this.word = null;
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}
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public void addWord(String word) {
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if(this.word == null) {
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this.word = word;
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return;
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}
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// NOTE: 사전 순으로 크고 작음을 두어 이진처리.
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if(this.word.compareTo(word) == 1) {
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this.rightChild = new PrevWordDictionary();
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this.rightChild.addWord(word);
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} else if (this.word.compareTo(word) == -1) {
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this.leftChild = new PrevWordDictionary();
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this.leftChild.addWord(word);
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} else {
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// already exists. just return
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}
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}
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public boolean search(String word) {
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if(word.contains(".")) {
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// String replaced = word.replace(".", "");
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// return word.startsWith(".") ? this.startsWith(replaced) : this.endsWith(replaced);
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return this.patternSearch(word.replace(".", ""));
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}
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if(this.word == null) {
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return false;
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}
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if(this.word.equals(word)) {
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return true;
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}
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if(this.rightChild == null && this.leftChild == null) {
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return false;
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}
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if(this.word.compareTo(word) == 1) {
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return this.rightChild == null ? false : this.rightChild.search(word);
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} else {
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return this.leftChild == null ? false : this.leftChild.search(word);
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}
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}
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private boolean patternSearch(String wildcard) {
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if(this.word.contains(wildcard)) return true;
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boolean left = false;
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boolean right = false;
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if(this.leftChild != null) {
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left = this.leftChild.patternSearch(wildcard);
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}
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if(this.rightChild != null) {
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right = this.rightChild.patternSearch(wildcard);
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}
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return left || right;
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}
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}

‎implement-trie-prefix-tree/YoungSeok-Choi.java

Lines changed: 72 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -1,30 +1,63 @@
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import java.util.HashMap;
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import java.util.Map;
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// Map으로 풀려버려서 당황..
5-
// 이진트리? 어떤식으로 풀어야 할지 자료구조 정하고 다시 풀어보기..
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// 한 글자씩 잘라서 하위 자식 노드들로 관리
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// search 동작은 기존 그대로 Map자료구조에서 찾고, 이후 prefix연산에서 속도를 개선 (230ms -> 30ms)
66
class Trie {
77

8-
Map<String, Boolean> tMap;
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private Trie[] child;
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private Character val;
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private Map<String, Boolean> cMap;
911

1012
public Trie() {
11-
this.tMap = new HashMap<>();
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this.cMap = new HashMap<>();
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this.val = null;
1215
}
1316

1417
public void insert(String word) {
15-
this.tMap.put(word, true);
18+
if(this.cMap.containsKey(word)) return;
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this.cMap.put(word, true);
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this.innerInsert(word);
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}
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public void innerInsert(String word) {
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if(word.length() == 0) return;
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if(this.child == null) {
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this.child = new Trie[26];
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}
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char c = word.charAt(0);
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int idx = c - 97;
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if(this.child[idx] == null) {
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this.child[idx] = new Trie();
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this.child[idx].val = c;
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}
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this.child[idx].innerInsert(word.substring(1));
1640
}
1741

1842
public boolean search(String word) {
19-
return this.tMap.containsKey(word);
43+
return this.cMap.containsKey(word);
2044
}
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46+
// public boolean search(String word) {
47+
48+
// }
2149

22-
public boolean startsWith(String prefix) {
23-
for(String key : this.tMap.keySet()) {
24-
if(key.startsWith(prefix)) return true;
50+
public boolean startsWith(String word) {
51+
if(word.length() == 0) {
52+
return true;
2553
}
2654

27-
return false;
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char c = word.charAt(0);
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int idx = c - 97;
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if(this.child == null || this.child[idx] == null || this.child[idx].val == null) return false;
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60+
return this.child[idx].startsWith(word.substring(1));
2861
}
2962
}
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@@ -35,3 +68,32 @@ public boolean startsWith(String prefix) {
3568
* boolean param_2 = obj.search(word);
3669
* boolean param_3 = obj.startsWith(prefix);
3770
*/
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// Map으로 풀려버려서 당황..
74+
// 이진트리? 어떤식으로 풀어야 할지 자료구조 정하고 다시 풀어보기..
75+
class BeforeTrie {
76+
77+
Map<String, Boolean> tMap;
78+
79+
public BeforeTrie() {
80+
this.tMap = new HashMap<>();
81+
}
82+
83+
public void insert(String word) {
84+
this.tMap.put(word, true);
85+
}
86+
87+
public boolean search(String word) {
88+
return this.tMap.containsKey(word);
89+
}
90+
91+
public boolean startsWith(String prefix) {
92+
for(String key : this.tMap.keySet()) {
93+
if(key.startsWith(prefix)) return true;
94+
}
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return false;
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}
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}
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