[minji-go] Week 2

3sum/minji-go.java

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+/*
+    Problem: https://leetcode.com/problems/3sum/
+    Description: return all the triplets (i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0)
+    Concept: Array, Two Pointers, Sorting
+    Time Complexity: O(N²), Runtime 70ms
+    Space Complexity: O(N), Memory 51.63MB
+*/
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        Map<Integer, Integer> number = new HashMap<>();
+        for(int i=0; i<nums.length; i++) {
+            number.put(nums[i], number.getOrDefault(nums[i], 0)+1);
+        }
+
+        Arrays.sort(nums);
+        Set<List<Integer>> set = new HashSet<>();
+        List<List<Integer>> triplets = new ArrayList<>();
+        List<Integer> lastTriplet = null;
+        for(int i=0; i<nums.length-1; i++) {
+            if(i>0 && nums[i]==nums[i-1]) continue;
+
+            for(int j=i+1; j<nums.length; j++){
+                if(j>i+1 && nums[j]==nums[j-1]) continue;
+
+                int target = -(nums[i]+nums[j]);
+                if(nums[j]>target) continue;
+
+                int count = number.getOrDefault(target,0);
+                if(nums[i]==target) count--;
+                if(nums[j]==target) count--;
+                if(count<=0) continue;
+
+                List<Integer> triplet = List.of(nums[i], nums[j], target);
+                if(triplet.equals(lastTriplet)) continue;
+                lastTriplet = triplet;
+                triplets.add(triplet);
+            }
+        }
+        return triplets;
+    }
+}

climbing-stairs/minji-go.java

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+/*
+    Problem: https://leetcode.com/problems/climbing-stairs/
+    Description: how many distinct ways can you climb to the top, if you can either climb 1 or 2 steps
+    Concept: Dynamic Programming, Memoization, Recursion, Math, Array, Iterator, Combinatorics ...
+    Time Complexity: O(n), Runtime: 0ms
+    Space Complexity: O(1), Memory: 40.63MB
+*/
+class Solution {
+    public int climbStairs(int n) {
+        if(n==1) return 1;
+        if(n==2) return 2;
+
+        int prev=1, cur=2;
+        for(int i=3; i<=n; i++){
+            int next=prev+cur;
+            prev=cur;
+            cur=next;
+        }
+        return cur;
+    }
+}

construct-binary-tree-from-preorder-and-inorder-traversal/minji-go.java

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+/*
+    Problem: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
+    Description: Given two integer arrays preorder and inorder, construct and return the binary tree.
+    Concept: Array, Hash Table, Divide and Conquer, Tree, Binary Tree
+    Time Complexity: O(NM), Runtime 2ms
+    Space Complexity: O(N), Memory 45.02MB
+*/
+import java.util.HashMap;
+import java.util.Map;
+
+class Solution {
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+        boolean isLeft = true;
+        Map<Integer, TreeNode> parents = new HashMap<>();
+        TreeNode rootNode = null, parentNode = null;
+
+        for (int pidx=0, iidx=0; pidx<preorder.length; pidx++) {
+            int pval = preorder[pidx];
+            int ival = inorder[iidx];
+
+            if(pidx==0) {
+                rootNode = parentNode = new TreeNode(pval);
+            } else if (isLeft) {
+                parents.put(pval, parentNode);
+                parentNode = parentNode.left = new TreeNode(pval);
+            } else {
+                isLeft = true;
+                parents.put(pval, parentNode);
+                parentNode = parentNode.right = new TreeNode(pval);
+            }
+
+            if(pval==ival) {
+                isLeft = false;
+                TreeNode targetNode = parentNode;
+                while (iidx<inorder.length-1 && parents.get(parentNode.val)!=null) {
+                    if(parentNode.val == inorder[iidx+1]){
+                        iidx++;
+                        targetNode = parentNode;
+                    }
+                    parentNode = parents.get(parentNode.val);
+                }
+                if(iidx<inorder.length-1 && parentNode.val != inorder[iidx+1]){
+                    iidx++;
+                    parentNode = targetNode;
+                } else iidx = iidx+2;
+            }
+        }
+
+        return rootNode;
+    }
+}
+
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */

decode-ways/minji-go.java

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+/*
+    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
+*/
+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];
+
+        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];
+        }
+        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;
+    }
+}

valid-anagram/minji-go.java

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+/*
+    Problem: https://leetcode.com/problems/valid-anagram/
+    Description: return true if one string is an anagram of the other, one formed by rearranging the letters of the other
+    Concept:String, Hash Table, Sorting, Array, Counting, String Matching, Ordered Map, Ordered Set, Hash Function ...
+    Time Complexity: O(n), Runtime: 27ms
+    Space Complexity: O(n), Memory: 43.05MB
+*/
+import java.util.HashMap;
+import java.util.Map;
+
+class Solution {
+    public boolean isAnagram(String s, String t) {
+        if(s.length() != t.length()) return false;
+
+        Map<Character, Integer> charCount = new HashMap<>();
+        for(int i=0; i<s.length(); i++){
+            charCount.put(s.charAt(i), charCount.getOrDefault(s.charAt(i), 0)+1);
+            charCount.put(t.charAt(i), charCount.getOrDefault(t.charAt(i), 0)-1);
+        }
+        for(Integer count : charCount.values()){
+            if(count !=0) return false;
+        }
+        return true;
+    }
+}