Merge pull request #2 from ParkJangSu/green_study

Added java solution for 14 / python solution for 2420

java/014_Longest_Common_Prefix.java

@@ -1,16 +1,17 @@
+//014_Longest_Common_Prefix.java
 class Solution {
     public String longestCommonPrefix(String[] strs) {
         String result ="";
         String temp = "";
-        int c = 0;
+        int c = 0; //move first point
         boolean check = true;
         while(true){
-            for(int i = 0; i<strs.length; i++){
+            for(int i = 0; i<strs.length; i++){ //move second point
                 if(c>=strs[i].length()){
                     check = false;
                     break;
                 }
-                if(i==0){
+                if(i==0){ //temp -> check same Character
                     temp = Character.toString(strs[0].charAt(c));
                 }
                 if(!temp.equals(Character.toString(strs[i].charAt(c)))){

python/2420_Find_All_Good_Indices.py

@@ -0,0 +1,38 @@
+#2420_Find_All_Good_Indices.py
+class Solution:
+    def goodIndices(self, nums: List[int], k: int) -> List[int]:
+        # posi : count the increasing idxes
+        # nega : count the decreasing idxes
+        posi, nega = [0], [0]
+
+        for i in range(1, len(nums)):
+            diff = nums[i] - nums[i - 1]
+
+            posi.append(posi[i - 1])
+            nega.append(nega[i - 1])
+
+            # if diff show positive or negative
+            # then the value will updated
+            if diff > 0:
+                posi[i] += 1
+            elif diff < 0:
+                nega[i] += 1
+
+        # ans : count the idxes that
+        # before k element is non increasing
+        # after k element is non decreasing
+        ans = []
+        for i in range(k, len(nums) - k):
+            if i + k >= len(nums):
+                break
+
+            # check the condition with
+            # for after, nega[i + 1], nega[i + k] is the two to check
+            # for brfore, posi[i - 1], posi[i - k] is the two to check
+            if nega[i + k] - nega[i + 1] > 0:
+                continue
+            if posi[i - 1] - posi[i - k] > 0:
+                continue
+
+            ans.append(i)
+        return ans