Merge pull request #857 from ekgns33/main

[byteho0n] Week5

Author: ekgns33

best-time-to-buy-and-sell-stock/ekgns33.java

@@ -0,0 +1,11 @@
+class Solution {
+  public int maxProfit(int[] prices) {
+    int minPrice = Integer.MAX_VALUE;
+    int maxProfit = 0;
+    for(int i = 0; i < prices.length; i++) {
+      minPrice = Math.min(minPrice, prices[i]);
+      maxProfit = Math.max(maxProfit, prices[i] - minPrice);
+    }
+    return maxProfit;
+  }
+}

encode-and-decode-strings/ekgns33.java

@@ -0,0 +1,36 @@
+class Solution {
+  public String encode(List<String> strs) {
+    StringBuilder res = new StringBuilder();
+    for (String s : strs) {
+      res.append(s.length()).append('|').append(s);
+    }
+    return res.toString();
+  }
+
+
+  /*
+  * number + | + string
+  * read number until |
+  * move pointer, read substring
+  *
+  * tc : O(n) when n is the length of encoded string
+  * sc : O(1)
+  * */
+  public List<String> decode(String str) {
+    List<String> res = new ArrayList<>();
+    int start = 0;
+    while (start < str.length()) {
+      int cur = start;
+      //read until |
+      while (str.charAt(cur) != '|') {
+        cur++;
+      }
+      int length = Integer.parseInt(str.substring(start, cur));
+      start = cur + 1;
+      cur = start + length;
+      res.add(str.substring(start, cur));
+      start = cur;
+    }
+    return res;
+  }
+}

group-anagrams/ekgns33.java

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+/**
+
+ input : array of strings
+ output : grouped anagrams
+
+ ex) eat ate tea > same group
+
+ solution 1) brute force
+ ds : hashmap
+ algo : sort
+ generate Key by sorting string O(nlogn)
+
+ save to hashmap
+
+ tc : O(m* nlogn) when m is length of array, n is max length of string
+ sc : O(m)
+
+ solutino 2) better?
+ cannot optimize m times read
+ how about generating key
+
+ sort = nlogn
+ read = n << use frequency
+ 26 * n
+ tc : O(m * (26 * n)) ~= O(mn)
+ sc : O(m * 26) ~= O(m)
+ */
+
+class Solution {
+  public List<List<String>> groupAnagrams(String[] strs) {
+    Map<String, List<String>> map = new HashMap<>();
+    int n = strs.length;
+    for(String str : strs) {
+      int l = str.length();
+      char[] freq = new char[26];
+      for(int i = 0; i < l; i++) {
+        freq[str.charAt(i) - 'a']++;
+      }
+      String key = new String(freq);
+      map.putIfAbsent(key, new ArrayList<>());
+      map.get(key).add(str);
+    }
+    return List.copyOf(map.values());
+  }
+}

implement-trie-prefix-tree/ekgns33.java

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+class Trie {
+
+  class Node {
+    boolean isEnd;
+    Node[] childs = new Node[26];
+    Node() {}
+  }
+
+  private Node root;
+
+  public Trie() {
+    this.root = new Node();
+  }
+
+  // O(m) when m is the length of word
+  public void insert(String word) {
+    Node curN = this.root;
+    for(char c : word.toCharArray()) {
+      if(curN.childs[c-'a'] == null) {
+        curN.childs[c-'a'] = new Node();
+      }
+      curN = curN.childs[c-'a'];
+    }
+    //end
+    curN.isEnd = true;
+  }
+  // O(k) when k is the length of searching word
+  public boolean search(String word) {
+    Node curN = this.root;
+    for(char c : word.toCharArray()) {
+      if(curN.childs[c-'a'] == null) return false;
+      curN = curN.childs[c-'a'];
+    }
+    return curN.isEnd;
+  }
+
+  // O(k) when k is the length of prefix
+  public boolean startsWith(String prefix) {
+    Node curN = this.root;
+    for(char c : prefix.toCharArray()) {
+      if(curN.childs[c-'a'] == null) return false;
+      curN = curN.childs[c-'a'];
+    }
+    return true;
+  }
+}
+
+/**
+ * Your Trie object will be instantiated and called as such:
+ * Trie obj = new Trie();
+ * obj.insert(word);
+ * boolean param_2 = obj.search(word);
+ * boolean param_3 = obj.startsWith(prefix);
+ */

word-break/ekgns33.java

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+/**
+ input : string s and array of strings
+ output : return true if s can be segmented into one or more words in dictionary
+ constraints :
+ 1) same word can be used multiple times
+ 2) every word in dictionary is unique
+ 3) s.length [1, 300]
+ 4) length of tokens in dictionary [1, 20]
+
+ ex)
+ leetcode >> leet code
+ applepenapple >> apple pen
+ applepenapple >> apple applepen
+
+
+ solution 1) bruteforce
+ for each character c, check if word exists in dictionary
+ that the first letter is c
+
+ match string
+ move to next
+ unmatch
+ next word
+
+ >> get all the combinations
+ O(n * m * k)
+ n is the length of string s,
+ m is number of tokens,
+ k is length of token
+
+ tc : O (nmk) (300 * 1000 * 20) ~= 6 * 10^6 < 1sec
+ sc : O (mk)
+
+ solution 2) optimize? heuristic?
+
+ using trie will reduce tc
+ for each character:
+ get all substring substr(i, j)
+ search trie
+
+ tc : O(n^2 + m*k)
+
+ solution 3) dp
+
+ let dp[i] true if possible to build s which length is i
+ with segemented tokens
+
+ build HashSet for tokens
+ iterate i from 1 to n
+ iterate j from 0 to i-1
+ dp[i] = dp[j] & substr(j,i) presents // substr O(n)
+ if true break;
+
+ tc : O(n^3 + m*k)
+ */
+class Solution {
+  public boolean wordBreak(String s, List<String> wordDict) {
+    Set<String> dict = new HashSet<>(wordDict);
+    boolean[] dp = new boolean[s.length() + 1];
+    dp[0] = true;
+    for(int i = 1; i < dp.length; i++) {
+      for(int j = 0; j < i; j++) {
+        if(dp[j] && dict.contains(s.substring(j, i))){
+          dp[i] = true;
+          break;
+        }
+      }
+    }
+    return dp[dp.length - 1];
+  }
+}
+
+
+// class Solution1 {
+//     public boolean wordBreak(String s, List<String> wordDict) {
+//         Map<Character, List<String>> tokens = new HashMap<>();
+//         for(String word : wordDict) {
+//             tokens.putIfAbsent(word.charAt(word.length()-1), new ArrayList<>());
+//             tokens.get(word.charAt(word.length()-1)).add(word);
+//         }
+//         boolean[] dp = new boolean[s.length()+1];
+//         dp[0] = true;
+//         for(int i = 1; i < dp.length; i++) {
+//             List<String> token = tokens.get(s.charAt(i-1));
+//             if(token == null) {dp[i] = false; continue;}
+//             for(String word : token) {
+//                 if(i - word.length() < 0) continue;
+//                 dp[i] = dp[i-word.length()] && s.substring(i-word.length(), i).equals(word);
+//                 if(dp[i]) break;
+//             }
+//         }
+//         return dp[dp.length-1];
+//     }
+// }