[minji-go] week 06 solutions

combination-sum/minji-go.java

@@ -1,31 +1,31 @@
-/*
-    Problem: https://leetcode.com/problems/combination-sum/
-    Description: return a list of all unique combinations of candidates where the chosen numbers sum to target
-    Concept: Array, Backtracking
-    Time Complexity: O(Nᵀ), Runtime 2ms 
-    Space Complexity: O(T), Memory 44.88MB
-*/
-class Solution {
-    public List<List<Integer>> answer = new ArrayList<>();
+/**
+ * <a href="https://leetcode.com/problems/combination-sum/">week03-3.combination-sum</a>
+ * <li>Description: return a list of all unique combinations of candidates where the chosen numbers sum to target </li>
+ * <li>Topics: Array, Backtracking              </li>
+ * <li>Time Complexity: O(K^T), Runtime 2ms     </li>
+ * <li>Space Complexity: O(T), Memory 44.9MB    </li>
+ */
 
+class Solution {
     public List<List<Integer>> combinationSum(int[] candidates, int target) {
+        List<List<Integer>> combinations = new ArrayList<>();
         Arrays.sort(candidates);
-        findCombination(candidates, target, new ArrayList<>(), 0);
-        return answer;
+        dfs(candidates, target, 0, new ArrayList<>(), combinations);
+        return combinations;
     }
 
-    public void findCombination(int[] candidates, int target, List<Integer> combination, int idx) {
-        if(target == 0) {
-            answer.add(new ArrayList<>(combination));
+    public void dfs(int[] candidates, int target, int index, List<Integer> combination, List<List<Integer>> combinations) {
+        if (target == 0) {
+            combinations.add(new ArrayList<>(combination));
             return;
         }
 
-        for(int i=idx; i<candidates.length; i++) {
-            if(candidates[i] > target) break;
+        for (int i = index; i < candidates.length; i++) {
+            if (target - candidates[i] < 0) break;
 
             combination.add(candidates[i]);
-            findCombination(candidates, target-candidates[i], combination, i);
-            combination.remove(combination.size()-1);
+            dfs(candidates, target - candidates[i], i, combination, combinations);
+            combination.remove(combination.size() - 1);
         }
     }
 }

container-with-most-water/minji-go.java

@@ -0,0 +1,28 @@
+/**
+ * <a href="https://leetcode.com/problems/container-with-most-water/">week06-2.container-with-most-water</a>
+ * <li>Description: Return the maximum amount of water a container can store</li>
+ * <li>Topics: Array, Two Pointers, Greedy      </li>
+ * <li>Time Complexity: O(N), Runtime 5ms       </li>
+ * <li>Space Complexity: O(1), Memory 57.42MB   </li>
+ */
+class Solution {
+    public int maxArea(int[] height) {
+        int maxArea = 0;
+        int left = 0;
+        int right = height.length - 1;
+
+        while (left < right) {
+            int width = right - left;
+            int minHeight = Math.min(height[left], height[right]);
+            maxArea = Math.max(maxArea, width * minHeight);
+
+            if (height[left] < height[right]) {
+                left++;
+            } else {
+                right--;
+            }
+        }
+
+        return maxArea;
+    }
+}

decode-ways/minji-go.java

@@ -1,28 +1,34 @@
-/*
-    Problem: https://leetcode.com/problems/decode-ways/
-    Description: Given a string s containing only digits, return the number of ways to decode it
-    Concept: String, Dynamic Programming
-    Time Complexity: O(N), Runtime 1ms
-    Space Complexity: O(N), Memory 42.12MB
-*/
+/**
+ * <a href="https://leetcode.com/problems/decode-ways/">week03-5.decode-ways</a>
+ * <li>Description: return the number of ways to decode it </li>
+ * <li>Topics: String, Dynamic Programming  </li>
+ * <li>Time Complexity: O(N), Runtime 1ms   </li>
+ * <li>Space Complexity: O(1), Memory 41.8MB</li>
+ */
 class Solution {
     public int numDecodings(String s) {
-        int[] dp = new int[s.length()];
-        if(decode(s.substring(0, 1))) dp[0]=1;
-        if(s.length()>1 && decode(s.substring(1, 2))) dp[1]+=dp[0];
-        if(s.length()>1 && decode(s.substring(0, 2))) dp[1]+=dp[0];
+        if (s.charAt(0) == '0') {
+            return 0;
+        }
+
+        int last2 = 1;
+        int last1 = 1;
+
+        for (int i = 1; i < s.length(); i++) {
+            int curr = 0;
+            if (s.charAt(i) != '0') {
+                curr += last1;
+            }
 
-        for(int i=2; i<s.length(); i++){
-            if(decode(s.substring(i,i+1))) dp[i]+=dp[i-1];
-            if(decode(s.substring(i-1,i+1))) dp[i]+=dp[i-2];
+            int num = Integer.parseInt(s.substring(i - 1, i + 1));
+            if (num >= 10 && num <= 26) {
+                curr += last2;
+            }
+
+            last2 = last1;
+            last1 = curr;
         }
-        return dp[s.length()-1];
-    }
 
-    public boolean decode(String s){
-        int num = Integer.parseInt(s);
-        int numLength = (int) Math.log10(num) + 1;
-        if(num<0 || num>26 || numLength != s.length()) return false;
-        return true;
+        return last1;
     }
 }

design-add-and-search-words-data-structure/minji-go.java

@@ -0,0 +1,69 @@
+/**
+ * <a href="https://leetcode.com/problems/design-add-and-search-words-data-structure/">week06-3.design-add-and-search-words-data-structure</a>
+ * <li>Description: Design a data structure that supports adding new words and finding if a string matches any previously added string</li>
+ * <li>Topics: String, Depth-First Search, Design, Trie </li>
+ * <li>Time Complexity: O(N), Runtime 274ms             </li>
+ * <li>Space Complexity: O(N), Memory 118.44MB          </li>
+ */
+class WordDictionary {
+    private Trie dictionary;
+
+    public WordDictionary() {
+        dictionary = new Trie();
+    }
+
+    public void addWord(String word) {
+        dictionary.add(word);
+    }
+
+    public boolean search(String word) {
+        return dictionary.contains(word);
+    }
+
+}
+
+public class Trie {
+    private boolean isEnd;
+    private Map<Character, Trie> next;
+
+    Trie() {
+        next = new HashMap<>();
+    }
+
+    public void add(String word) {
+        Trie trie = this;
+        for(char c : word.toCharArray()){
+            trie = trie.next.computeIfAbsent(c, k -> new Trie());
+        }
+        trie.isEnd = true;
+    }
+
+    public boolean contains(String word) {
+        Trie trie = this;
+        for(int i=0; i<word.length(); i++){
+            char c = word.charAt(i);
+            if(c == '.') {
+                for(Trie newTrie : trie.next.values()) {
+                    if(newTrie.contains(word.substring(i+1))){
+                        return true;
+                    }
+                }
+                return false;
+            } else {
+                trie = trie.next.get(c);
+                if(trie == null) {
+                    return false;
+                }
+            }
+        }
+
+        return trie.isEnd;
+    }
+}
+
+/**
+ * Your WordDictionary object will be instantiated and called as such:
+ * WordDictionary obj = new WordDictionary();
+ * obj.addWord(word);
+ * boolean param_2 = obj.search(word);
+ */

longest-increasing-subsequence/minji-go.java

@@ -0,0 +1,28 @@
+/**
+ * <a href="https://leetcode.com/problems/longest-increasing-subsequence/">week06-4.longest-increasing-subsequence</a>
+ * <li>Description: return the length of the longest strictly increasing subsequence</li>
+ * <li>Topics: Array, Binary Search, Dynamic Programming </li>
+ * <li>Time Complexity: O(NLogN), Runtime 6ms            </li>
+ * <li>Space Complexity: O(N), Memory 44.3MB            </li>
+ */
+class Solution {
+    public int lengthOfLIS(int[] nums) {
+        List<Integer> dp = new ArrayList<>();
+
+        for(int num : nums){
+            int idx = Collections.binarySearch(dp, num);
+
+            if(idx < 0) {
+                idx = -idx -1;
+            }
+
+            if(idx == dp.size()) {
+                dp.add(num);
+            } else {
+                dp.set(idx, num);
+            }
+        }
+
+        return dp.size();
+    }
+}

number-of-1-bits/minji-go.java

@@ -0,0 +1,17 @@
+/**
+ * <a href="https://leetcode.com/problems/number-of-1-bits/">week03-2.number-of-1-bits</a>
+ * <li>Description: returns the number of set bits in its binary representation</li>
+ * <li>Topics: Divide and Conquer, Bit Manipulation </li>
+ * <li>Time Complexity: O(logN), Runtime 0ms        </li>
+ * <li>Space Complexity: O(1), Memory 41.95MB       </li>
+ */
+class Solution {
+    public int hammingWeight(int n) {
+        int count = 0;
+        while(n != 0) {
+            n &= (n-1);
+            count++;
+        }
+        return count;
+    }
+}

spiral-matrix/minji-go.java

@@ -0,0 +1,39 @@
+/**
+ * <a href="https://leetcode.com/problems/spiral-matrix/">week06-5.spiral-matrix</a>
+ * <li>Description: return all elements of the matrix in spiral order</li>
+ * <li>Topics: Array, Matrix, Simulation        </li>
+ * <li>Time Complexity: O(N*M), Runtime 0ms     </li>
+ * <li>Space Complexity: O(1), Memory 41.95MB   </li>
+ */
+class Solution {
+    public List<Integer> spiralOrder(int[][] matrix) {
+        List<Integer> answer = new ArrayList<>();
+        int lr = 0;
+        int hr = matrix.length-1;
+        int lc = 0;
+        int hc = matrix[0].length-1;
+
+        while (lr<=hr && lc<=hc) {
+            for(int c=lc; c<=hc && lr<=hr; c++) {
+                answer.add(matrix[lr][c]);
+            }
+            lr++;
+
+            for(int r=lr; r<=hr && lc<=hc; r++) {
+                answer.add(matrix[r][hc]);
+            }
+            hc--;
+
+            for(int c=hc; c>=lc && lr<=hr; c--){
+                answer.add(matrix[hr][c]);
+            }
+            hr--;
+
+            for(int r=hr; r>=lr && lc<=hc; r--){
+                answer.add(matrix[r][lc]);
+            }
+            lc++;
+        }
+        return answer;
+    }
+}

valid-palindrome/minji-go.java

@@ -1,22 +1,19 @@
-/*
-    Problem: https://leetcode.com/problems/valid-palindrome/
-    Description: return true if it is a palindrome, alphanumeric characters(letters and numbers) reads the same forward and backward
-    Concept: Two Pointers, String
-    Time Complexity: O(n), Runtime: 10ms
-    Space Complexity: O(n), Memory: 58.6MB
-*/
+/**
+ * <a href="https://leetcode.com/problems/valid-palindrome/">week03-1.valid-palindrome</a>
+ * <li>Description: return true if it is a palindrome </li>
+ * <li>Topics: Two Pointers, String             </li>
+ * <li>Time Complexity: O(N), Runtime 13ms      </li>
+ * <li>Space Complexity: O(N), Memory 45.55MB   </li>
+ */
 class Solution {
     public boolean isPalindrome(String s) {
-        String regex ="[^A-Za-z0-9]";
-        String palindrome = s.replaceAll(regex,"").toLowerCase(); //replaceAll(), toLowerCase(): O(n)
+        String str = s.toLowerCase().replaceAll("[^0-9a-z]", "");
 
-        boolean answer = true;
-        for(int i=0; i<palindrome.length()/2; i++){
-            if(palindrome.charAt(i) != palindrome.charAt(palindrome.length()-1-i)) {
-                answer = false;
-                break;
+        for (int i = 0; i < str.length() / 2; i++) {
+            if (str.charAt(i) != str.charAt(str.length() - 1 - i)) {
+                return false;
             }
         }
-        return answer;
+        return true;
     }
 }

valid-parentheses/minji-go.java

@@ -0,0 +1,36 @@
+/**
+ * <a href="https://leetcode.com/problems/valid-parentheses/">week06-1.valid-parentheses</a>
+ * <li>Description: determine if the input string is valid </li>
+ * <li>Topics: String, Stack                    </li>
+ * <li>Time Complexity: O(N), Runtime 2ms       </li>
+ * <li>Space Complexity: O(N), Memory 41.98MB   </li>
+ */
+
+class Solution {
+    public boolean isValid(String s) {
+        Deque<Character> stack = new ArrayDeque<>();
+        for(char c : s.toCharArray()) {
+            if(isOpenBracket(c)){
+                stack.push(c);
+                continue;
+            }
+
+            if(stack.isEmpty() || isNoneMatchBracket(stack.pop(), c)) {
+                return false;
+            }
+        }
+
+        return stack.isEmpty();
+    }
+
+    private boolean isOpenBracket(char open) {
+        return open == '(' || open == '{' || open == '[';
+    }
+
+    private boolean isNoneMatchBracket(char open, char close) {
+        if(open == '(') return close != ')';
+        if(open == '{') return close != '}';
+        if(open == '[') return close != ']';
+        return true;
+    }
+}