[minji-go] Week 2

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: 40.51MB
+*/
+class Solution {
+    public int climbStairs(int n) {
+        int step1=1, step2=2;
+        for(int i=3; i<n; i++){
+            int step3=step1+step2;
+            step1=step2;
+            step2=step3;
+        }
+        return n==1?step1:step2;
+    }
+}

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: 43.11MB
+*/
+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> count = new HashMap<>();
+        for(int i=0; i<s.length(); i++){
+            count.put(s.charAt(i), count.getOrDefault(s.charAt(i), 0)+1);
+            count.put(t.charAt(i), count.getOrDefault(t.charAt(i), 0)-1);
+        }
+        for(Character key : count.keySet()){
+            if(count.get(key)!=0) return false;
+        }
+        return true;
+    }
+}