diff --git a/README.md b/README.md
index 2d866fc..c34ae3b 100644
--- a/README.md
+++ b/README.md
@@ -166,6 +166,7 @@ I'm currently working on [Analytics-Zoo](https://github.com/intel-analytics/anal
 | 482 | [License Key Formatting](https://leetcode.com/problems/license-key-formatting/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/482_License_Key_Formatting.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/482_License_Key_Formatting.java) | String processing, lower and len % K, O(n) and O(n) |
 | 485 | [Max Consecutive Ones](https://leetcode.com/problems/max-consecutive-ones/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/485_Max_Consecutive_Ones.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/485_Max_Consecutive_Ones.java) | Add one when encounter 1, set to 0 when encounter 0, O(n) and O(1) |
 | 509 | [Fibonacci Number](https://leetcode.com/problems/fibonacci-number/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/509_Fibonacci_Number.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/509_Fibonacci_Number.java) | 1. Recursive, O(n) <br>2. DP with memo, O(n). Note that N<=30, which means that we can keep a memo from 0 to 30. |
+| 523 | [Continuous Subarray Sum](https://leetcode.com/problems/continuous-subarray-sum/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/523_Continuous_Subarray_Sum.py) | O(n) solution using dict() |
 | 538 | [Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/538_Convert_BST_to_Greater_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/538_Convert_BST_to_Greater_Tree.java) | Right first DFS with a variable recording sum of node.val and right.val. 1. Recursive.<br>2. Stack 3. Reverse Morris In-order Traversal |
 | 541 | [Reverse String II](https://leetcode.com/problems/reverse-string-ii/solution/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/541_Reverse_String_II.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/541_Reverse_String_II.java) | Handle each 2k until reaching end, On(n) and O(n) |
 | 543 | [Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/543_Diameter_of_Binary_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/543_Diameter_of_Binary_Tree.java) | DFS with O(1) for max answer |
@@ -242,7 +243,11 @@ I'm currently working on [Analytics-Zoo](https://github.com/intel-analytics/anal
 | 1365 | [How Many Numbers Are Smaller Than the Current Number](https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1365_How_Many_Numbers_Are_Smaller_Than_the_Current_Number.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1365_How_Many_Numbers_Are_Smaller_Than_the_Current_Number.java) | 1. Sort and get position in sorted nums, O(nlogn) and O(n)<br>2. Fill count into 0-100, O(n) and O(1) |
 | 1480 | [Running Sum of 1d Array](https://leetcode.com/problems/running-sum-of-1d-array/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1480_Running_Sum_of_1d_Array.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1480_Running_Sum_of_1d_Array.java) | 1. Go through the array, O(n) and O(1)<br>2. Accumulate API |
 | 1539 | [Kth Missing Positive Number](https://leetcode.com/problems/kth-missing-positive-number/) | [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1539_Kth_Missing_Positive_Number.java) | Binary search, num of missing = arr[i]-i-1 |
+| 1909 | [Remove One Element to Make the Array Strictly Increasing](https://leetcode.com/problems/remove-one-element-to-make-the-array-strictly-increasing/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1909_Remove_One_Element_to_Make_the_Array_Strictly_Increasing.py )| Use brute-force. O( (nums.length)<sup>2</sup>) |
 | 1981 | [Minimize the Difference Between Target and Chosen Elements](https://leetcode.com/problems/minimize-the-difference-between-target-and-chosen-elements/) | [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1981_Minimize_the_Difference_Between_Target_and_Chosen_Elements.java) | DP memo[row][sum] to avoid recomputing |
+| 2409 | [Count Days Spent Together](https://leetcode.com/problems/count-days-spent-together/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/2409_Count_Days_Spent_Together.py)| Use month as a day |
+| 2413 | [Smallest Even Multiple](https://leetcode.com/problems/smallest-even-multiple/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/2413_Smallest_Even_Multiple.py)| Check the n is multiply by 2 |
+| 2429 | [Minimize XOR](https://leetcode.com/problems/minimize-xor/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/2429_Minimize_XOR.py.py) | check the num1, num2 with length and replace "0" compare with num1. |
 
 | # | To Understand |
 |---| ----- |
diff --git a/cpp/2553_Separate_the_Digits_in_an_Array.cpp b/cpp/2553_Separate_the_Digits_in_an_Array.cpp
new file mode 100644
index 0000000..e88e8c6
--- /dev/null
+++ b/cpp/2553_Separate_the_Digits_in_an_Array.cpp
@@ -0,0 +1,19 @@
+class Solution {
+public:
+    vector<int> separateDigits(vector<int>& nums) {
+        vector<int> answer;
+        int n;
+        for( int i=0; i < nums.size(); i++){
+            n=nums[i];
+            vector<int> temp;
+            while( n>0 ){
+                temp.push_back(n%10);
+                n = n / 10;
+            }
+            for(int j= temp.size()-1; j>=0; j--){
+                answer.push_back(temp[j]);
+            }
+        }
+        return answer;
+    }
+};
diff --git a/java/006_ZigZag_Conversion.java b/java/006_ZigZag_Conversion.java
index ea64188..97e4b1a 100644
--- a/java/006_ZigZag_Conversion.java
+++ b/java/006_ZigZag_Conversion.java
@@ -1 +1,33 @@
-//TODO
\ No newline at end of file
+//006_ZigZag_Conversion.java
+class Solution {
+    public String convert(String s, int numRows) {
+        if(numRows==1) {
+            return s;
+        }
+
+        String answer = "";
+        String[] str_array = new String[numRows];
+
+        for(int i=0;i<numRows;i++){
+            str_array[i]="";
+        }
+
+        int mod = 2*numRows-2;
+
+        for(int i=0;i<s.length();i++){
+            char c = s.charAt(i);
+            int index = i%mod;
+            if(index >= numRows) {
+                index = 2*(numRows-1) - index;
+            }
+            str_array[index]+=c;
+        }
+
+        for(int i=0;i<numRows;i++){
+            answer += str_array[i];
+        }
+
+        return answer;
+    }
+}
+
diff --git a/java/014_Longest_Common_Prefix.java b/java/014_Longest_Common_Prefix.java
new file mode 100644
index 0000000..4ff7134
--- /dev/null
+++ b/java/014_Longest_Common_Prefix.java
@@ -0,0 +1,33 @@
+//014_Longest_Common_Prefix.java
+class Solution {
+    public String longestCommonPrefix(String[] strs) {
+        String result ="";
+        String temp = "";
+        int c = 0; //move first point
+        boolean check = true;
+        while(true){
+            for(int i = 0; i<strs.length; i++){ //move second point
+                if(c>=strs[i].length()){
+                    check = false;
+                    break;
+                }
+                if(i==0){ //temp -> check same Character
+                    temp = Character.toString(strs[0].charAt(c));
+                }
+                if(!temp.equals(Character.toString(strs[i].charAt(c)))){
+                    check = false;
+                    break;
+                }
+                if(i==strs.length-1){
+                    result += temp;
+                }
+            }
+            if(!check){
+                break;
+            }
+            c++;
+        }
+        return result;
+        
+    }
+}
\ No newline at end of file
diff --git a/java/015_3Sum.java b/java/015_3Sum.java
new file mode 100644
index 0000000..cc0a069
--- /dev/null
+++ b/java/015_3Sum.java
@@ -0,0 +1,47 @@
+//015. 3Sum
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        //create result list to store i,j,k
+        List<List<Integer>> result = new LinkedList<List<Integer>>();
+        
+        //sorting nums
+        Arrays.sort(nums);
+        
+        for (int i = 0; i < nums.length - 2; i++) {
+
+            int left = i + 1;
+            int right = nums.length - 1;
+
+            if (i > 0 && nums[i] == nums[i-1]) {
+                continue; //if nums have same numbers, just check one time.
+            } 
+            
+            while (left < right) {
+                int sum = nums[left] + nums[right] + nums[i];
+                
+                if (sum == 0) {
+                    //if sum == 0, store i,j,k
+                    result.add(Arrays.asList(nums[i], nums[left], nums[right]));
+                    left++; //check anoter case
+                    right--;
+                    //if next number == now number
+                    while (nums[left] == nums[left - 1] && left < right) {
+                        left++;
+                    }
+                    while (nums[right] == nums[right + 1] && left < right) {
+                        right--;
+                    } 
+                } else if (sum > 0) {
+                    //if sum > 0, right--;
+                    right--;
+                } else {
+                    //if sum < 0, left++;
+                    left++;
+                }
+            }
+        }
+        
+        return result; //return result list
+    }
+}
+ 
diff --git a/java/026_Remove_Duplicates_from_Sorted_Array.java b/java/026_Remove_Duplicates_from_Sorted_Array.java
new file mode 100644
index 0000000..13f03b1
--- /dev/null
+++ b/java/026_Remove_Duplicates_from_Sorted_Array.java
@@ -0,0 +1,16 @@
+//026. Remove Duplicates from Sorted Array
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        int index = 1;
+
+        for (int i = 0; i < nums.length - 1; i++) {
+            if (nums[i] != nums[i + 1]) {
+                nums[index] = nums[i + 1];
+                index++;
+            }
+        }
+
+        return index;
+    }
+}
+
diff --git a/java/034_Find_First_and_Last_Position_of_Element_in_Sorted_Array.java b/java/034_Find_First_and_Last_Position_of_Element_in_Sorted_Array.java
new file mode 100644
index 0000000..01a70a7
--- /dev/null
+++ b/java/034_Find_First_and_Last_Position_of_Element_in_Sorted_Array.java
@@ -0,0 +1,59 @@
+class Solution {
+    /*
+        Rule 1
+        - Given array is sorted in non-decreasing order
+        
+        Rule 2
+        - Limit time complexity O(log n).
+        
+        So I can use Binary Search Algorithm.
+    */
+    public int[] searchRange(int[] nums, int target) {
+        // initialize arr[0] -> maximum integer, arr[1] -> minimum integer
+        int[] arr = {100001, -10};
+        int s = 0;
+        int e = nums.length - 1;
+        
+        // find minimum index in nums with Binary Search alogorithm
+        while (s <= e) {
+            
+            int mid = (s + e) / 2;
+            
+            if(nums[mid] > target) {
+                e = mid - 1;
+            }
+            else if(nums[mid] <= target) {
+                if(nums[mid] == target) {
+                    arr[0] = Math.min(arr[0], mid);
+                    arr[1] = Math.max(arr[1], mid);
+                }
+                s = mid + 1;
+            }
+        }
+        
+        s = 0;
+        e = nums.length - 1;
+        
+        // find maximum index in nums with Binary Search alogorithm
+        while(s <= e) {
+            int mid = (s + e) / 2;
+            
+            if(nums[mid] >= target) {
+                if(nums[mid] == target) {
+                    arr[0] = Math.min(arr[0], mid);
+                    arr[1] = Math.max(arr[1], mid);
+                }
+                e = mid - 1;
+            }
+            else if(nums[mid] < target) {
+                s = mid + 1;
+            }
+        }
+        
+        // if arr data is initial data, set -1.
+        if(arr[0] == 100001) arr[0] = -1;
+        if(arr[1] == -10) arr[1] = -1;
+        
+        return arr;
+    }
+}
diff --git a/java/039_Combination_Sum.java b/java/039_Combination_Sum.java
new file mode 100644
index 0000000..7a54dd3
--- /dev/null
+++ b/java/039_Combination_Sum.java
@@ -0,0 +1,29 @@
+//039_Combination_Sum
+class Solution {
+    List<List<Integer>> answer = new ArrayList<List<Integer>>();
+
+    public List<List<Integer>> combinationSum(int[] candidates, int target) {
+        int clen =candidates.length;
+        for (int i = 0; i < clen; i++) {
+            List<Integer> tlist = new ArrayList<Integer>();
+            tlist.add(candidates[i]);
+            backtracking(candidates, i, 1, (target - candidates[i]), tlist);
+        }
+        return answer;
+    }
+    private void backtracking(int[] candidates, int index, int tsize, int target, List<Integer> temp) {
+        if (target == 0) {
+            answer.add(new ArrayList(temp));
+            return;
+        }
+        
+        for (int i = index, len = candidates.length; i < len; i++) {
+            if (candidates[i] <= target) {
+                temp.add(candidates[i]);
+                backtracking(candidates, i, (tsize + 1), (target - candidates[i]), temp);
+                temp.remove(tsize);
+            }
+        }
+    }
+}
+
diff --git a/java/049_Group_Anagrams.java b/java/049_Group_Anagrams.java
new file mode 100644
index 0000000..a787625
--- /dev/null
+++ b/java/049_Group_Anagrams.java
@@ -0,0 +1,42 @@
+class Solution {
+    public List<List<String>> groupAnagrams(String[] strs) {
+        int[][] alphabets = new int[strs.length]['z' - 'a' + 1];
+        
+        for(int i = 0; i < strs.length; i++) {
+            String str = strs[i];
+            
+            for(int j = 0; j < str.length(); j++) 
+                alphabets[i][str.charAt(j) - 'a']++;
+        }
+        
+        boolean[] visited = new boolean[strs.length];
+        
+        List<List<String>> answer = new ArrayList<>();
+        
+        for(int i = 0; i < strs.length; i++) {
+            if(visited[i]) continue;
+            
+            List<String> list = new ArrayList<>();
+            
+            for(int j = i; j < strs.length; j++) {
+                if(visited[j]) continue;
+                if(isAnagram(alphabets[i], alphabets[j])) {
+                    list.add(strs[j]);
+                    visited[j] = true;
+                }
+            }
+            
+            answer.add(list);
+        }
+        
+        return answer;
+    }
+    
+    public boolean isAnagram(int[] arr1, int[] arr2) {
+        for(int i = 0; i < arr1.length; i++) {
+            if(arr1[i] != arr2[i])
+                return false;
+        }
+        return true;
+    }
+}
diff --git a/java/055_Jump_Game.java b/java/055_Jump_Game.java
new file mode 100644
index 0000000..50971d4
--- /dev/null
+++ b/java/055_Jump_Game.java
@@ -0,0 +1,49 @@
+import java.util.*;
+
+class Solution {
+    public boolean canJump(int[] nums) {
+        /*
+         * Constraints
+         * 1 <= nums.length <= 10^4
+         * 0 <= nums[i] <= 10^5
+         * 
+         * Solution
+         * 1. Use BFS Algorithm.
+         *   - reason 1 : have to ignore array which is not visited.
+         *   - reason 2 : we have to visit all possible array from array[start].
+         */
+
+        int N = nums.length;
+        ArrayDeque<Integer> q = new ArrayDeque<>();
+        boolean[] visited = new boolean[N];
+        
+        // First, add first array index.
+        // And, set visited[first_index] to true.
+        q.add(0);
+        visited[0] = true;
+        
+        // Axiom : if N is 1, result is true.
+        if(N == 1) return true;
+        
+        // BFS algorithm
+        while(!q.isEmpty()) {
+            int cur = q.poll();
+            
+            // find cur + 1 to cur + nums[cur]
+            for(int i = 1; i <= nums[cur]; i++) {
+                if(cur + i >= N - 1) return true;
+                int next = Math.min(cur + i, N - 1);
+                
+                // set visited[next] to true and add index into queue.
+                // because of time limit(No overlap steps.)
+                if(!visited[next]) {
+                    visited[next] = true;
+                    q.add(next);
+                }
+            }
+        }
+        
+        return false;
+        
+    }
+}
\ No newline at end of file
diff --git a/java/088_Merge_Sorted_Array.java b/java/088_Merge_Sorted_Array.java
new file mode 100644
index 0000000..95c7c8e
--- /dev/null
+++ b/java/088_Merge_Sorted_Array.java
@@ -0,0 +1,29 @@
+class Solution {
+    //we are being given two int arrays and two int variables that state the number of elements in each array, respectively
+    public void merge(int[] nums1, int m, int[] nums2, int n) {
+        //I am using a counter so that I can iterate through the second array inside the for loop
+        int counter = 0;
+        //We know that nums1 is of the size n + m, so we can add all the elements to the array and sort them later
+        //For loop adds all values of nums2 to the end of nums1
+        for(int i=m; i<nums1.length; i++){
+            nums1[i] = nums2[counter++];
+        }
+        //Now that all of the elements are in the array, we can begin the sort process
+        //Though we could use the Arrays.sort function to quickly sort the array, I wanted to implement a sorting algorithm myself
+        //I am implementing what's called a "Bubble Sorting Algorithm" to sort this array
+        //This should work best in our scenario as we don't have to work with a large list of numbers
+        for(int i=0; i<nums1.length; i++){
+            for(int j=0; j<nums1.length-i-1; j++){
+                if(nums1[j] > nums1[j+1]){
+                    int temp = nums1[j+1];
+                    nums1[j+1] = nums1[j];
+                    nums1[j] = temp;
+                }
+            }
+        }
+        //The following function simply prints out everything that is contained within our num1 array
+        for(int i=0; i<nums1.length; i++){
+            System.out.println(nums1[i]);
+        }
+    }
+}
\ No newline at end of file
diff --git a/java/17_Letter_Combinations_of_a_Phone_Number.java b/java/17_Letter_Combinations_of_a_Phone_Number.java
new file mode 100644
index 0000000..7588d2c
--- /dev/null
+++ b/java/17_Letter_Combinations_of_a_Phone_Number.java
@@ -0,0 +1,40 @@
+class Solution {
+  	// make list for return.
+    public ArrayList<String> list = new ArrayList<>();
+	// make array for get phone number's characters.
+    public char[][] arr = {
+            {'0', '0', '0', '-'},
+            {'0', '0', '0', '-'},
+            {'a', 'b', 'c', '-'},
+            {'d', 'e', 'f', '-'},
+            {'g', 'h', 'i', '-'},
+            {'j', 'k', 'l', '-'},
+            {'m', 'n', 'o', '-'},
+            {'p', 'q', 'r', 's'},
+            {'t', 'u', 'v', '-'},
+            {'w', 'x', 'y', 'z'},
+        };
+    
+	// main function
+    public List<String> letterCombinations(String digits) {
+        addString(digits, 0, "");
+        return list;
+    }
+    
+	// axiom : if input == "", return []
+	// if index == digits.length(), add str in list
+	// else do loop number's character, and function recursion.
+    public void addString(String digits, int index, String str) {
+        if(digits.equals("")) return;
+        if(index == digits.length()) {
+            list.add(str);
+        }
+        else {
+            for(int i = 0; i < 4; i++) {
+                int number = Integer.parseInt(digits.charAt(index) + "");
+                if(arr[number][i] == '-') continue;
+                addString(digits, index + 1, str + arr[number][i]);
+            }
+        }
+    }
+}
diff --git a/java/219_Contains_Duplicate_II.java b/java/219_Contains_Duplicate_II.java
new file mode 100644
index 0000000..f88a1a9
--- /dev/null
+++ b/java/219_Contains_Duplicate_II.java
@@ -0,0 +1,44 @@
+import java.util.*;
+
+class Solution {
+  /*
+   * I have to save indice for each index.
+   * So I use the HashMap<Integer, List<Integer>>
+   */
+
+    public boolean containsNearbyDuplicate(int[] nums, int k) {
+        HashMap<Integer, List<Integer>> map = new HashMap<>();
+        
+        for(int i = 0; i < nums.length; i++) {
+            if(!map.containsKey(nums[i])) {
+                map.put(nums[i], new ArrayList<>());
+            }
+            map.get(nums[i]).add(i);
+        }
+        
+		// use Iterator to find appropriate two indice.
+		// Each list guarantee ascending.
+		// So list.get(i) and list.get(i + 1) is minimum.
+        Iterator<Integer> keys = map.keySet().iterator();
+        boolean answer = false;
+        
+        while(keys.hasNext()) {
+            int key = keys.next();
+            List<Integer> list = map.get(key);
+            
+            if(list.size() < 2) continue;
+            
+            for(int i = 1; i < list.size(); i++) {
+                int a = list.get(i - 1);
+                int b = list.get(i);
+                
+                if(b - a <= k) {
+                    answer = true;
+                    break;
+                }
+            }
+            if(answer) break;
+        }
+        return answer;
+    }
+}
diff --git a/java/22_Generate_Parentheses.java b/java/22_Generate_Parentheses.java
new file mode 100644
index 0000000..1667c87
--- /dev/null
+++ b/java/22_Generate_Parentheses.java
@@ -0,0 +1,26 @@
+class Solution {
+  	// main function
+    public List<String> generateParenthesis(int n) {
+        ArrayList<String> list = new ArrayList<>();
+        rec(list, "(", n - 1, n);
+        return list;
+    }
+    
+	// axiom : if start == end == 0, add str in list.
+	// IDEA :
+	// In well-formed parentheses
+	// close character(")") has to be bigger than open character("(")
+	// So, we can solve this problem with recursion.
+    public void rec(List<String> list, String str, int start, int end) {
+        if(start == 0 && end == 0) {
+            list.add(str);
+        }
+        
+        if(start > 0) {
+            rec(list, str + "(", start - 1, end);
+        }
+        if(end > start) {
+            rec(list, str + ")", start, end - 1);
+        }
+    }
+}
diff --git a/java/442_Find_All_Duplicates_in_an_Array.java b/java/442_Find_All_Duplicates_in_an_Array.java
new file mode 100644
index 0000000..b12d6f5
--- /dev/null
+++ b/java/442_Find_All_Duplicates_in_an_Array.java
@@ -0,0 +1,17 @@
+class Solution {
+    public List<Integer> findDuplicates(int[] nums) {
+        int n = nums.length;
+
+        List<Integer> ans = new ArrayList<>();
+
+        for(int i=0;i<n;i++) {
+            int possibleVal = Math.abs(nums[i]);
+
+            if(nums[possibleVal - 1] < 0) {
+                ans.add(possibleVal);
+            }
+            nums[possibleVal - 1] = -1 * nums[possibleVal - 1];
+        }
+        return ans;
+    }
+}
diff --git a/java/787_Cheapest_Flight_Within_K_Stops.java b/java/787_Cheapest_Flight_Within_K_Stops.java
new file mode 100644
index 0000000..9355cf9
--- /dev/null
+++ b/java/787_Cheapest_Flight_Within_K_Stops.java
@@ -0,0 +1,41 @@
+class Solution {
+    //using bellman ford
+    
+    public void computePrice(int[][]flights, int[] prices, int [] temp){
+        for(int[] flight: flights){
+            int u = flight[0];
+            int v = flight[1];
+            int price = flight[2];
+            
+            if(prices[u] != Integer.MAX_VALUE){
+                if(prices[u] + price < temp[v]){
+                    temp[v] = prices[u] + price;
+                }
+            }
+        }
+    }
+    
+    public void copyTempToPrice(int[] prices, int[] temp){
+        for(int i=0; i<prices.length; i++){
+            prices[i] = temp[i];
+        }
+    }
+    
+    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
+        int[] prices = new int[n];
+        int[] temp = new int[n];
+        
+        Arrays.fill(prices, Integer.MAX_VALUE);
+        Arrays.fill(temp, Integer.MAX_VALUE);
+        
+        prices[src] = 0;
+        temp[src] = 0;
+        
+        for(int i=0; i<=k; i++){
+            computePrice(flights, prices, temp);
+            copyTempToPrice(prices, temp);
+        }
+        
+        return prices[dst] == Integer.MAX_VALUE ? -1 : prices[dst];
+    }
+}
\ No newline at end of file
diff --git a/python/001_Two_Sum.py b/python/001_Two_Sum.py
index e4bd58a..52399a9 100644
--- a/python/001_Two_Sum.py
+++ b/python/001_Two_Sum.py
@@ -58,9 +58,4 @@ def twoSum(self, nums, target):
                 begin += 1
             else:
                 end -= 1
-
-
-if __name__ == '__main__':
-    # begin
-    s = Solution()
-    print s.twoSum([3, 2, 4], 6)
+                
diff --git a/python/007_Reverse_Integer.py b/python/007_Reverse_Integer.py
index a125d47..19aeba8 100644
--- a/python/007_Reverse_Integer.py
+++ b/python/007_Reverse_Integer.py
@@ -1,39 +1,34 @@
 class Solution:
-    # @return an integer
+    def reverse(self, x):
+        # https://leetcode.com/problems/reverse-integer/
+#        flag = True if x < 0 else False
+#       if flag:
+#           x = -x
+#       x = str(x)[::-1]
 
-    # def reverse(self, x):
-    #     max_int = 2147483647
-    #     if x == 0:
-    #         return 0
-    #     isPos = True
-    #     if x < 0:
-    #         x *= (-1)
-    #         isPos = False
-    #     ltemp = []
-    #     while x != 0:
-    #         temp = x % 10
-    #         ltemp.append(temp)
-    #         x /= 10
-    #     result = 0
-    #     # the main solution
-    #     for t in ltemp:
-    #         result = result * 10 + t
-    #     if result > max_int:
-    #         result = 0
-    #     if isPos:
-    #         return result
-    #     else:
-    #         return -1 * result
+#       if flag:
+#           x = "-" + x
 
-    def reverse(self, x):
-        # Note that in Python -1 / 10 = -1
-        res, isPos = 0, 1
+#       value = 2 ** 31
+#       x = int(x)
+#       if -value <= x < value:
+#           return x
+#       return 0
+        
+        is_neg = False
         if x < 0:
-            isPos = -1
-            x = -1 * x
-        while x != 0:
-            res = res * 10 + x % 10
-            if res > 2147483647:
-                return 0
-            x /= 10
-        return res * isPos
\ No newline at end of file
+            x = -x
+            is_neg = True
+
+        res = 0
+        while x > 0:
+            res *= 10
+            res += x % 10
+            x //= 10
+        if is_neg:
+            res = -res
+
+        if res < -2**31 or res > 2**31-1:
+            return 0
+        return res
+    
\ No newline at end of file
diff --git a/python/1909_Remove_One_Element_to_Make_the_Array_Strictly_Increasing.py b/python/1909_Remove_One_Element_to_Make_the_Array_Strictly_Increasing.py
new file mode 100644
index 0000000..5ae5c6f
--- /dev/null
+++ b/python/1909_Remove_One_Element_to_Make_the_Array_Strictly_Increasing.py
@@ -0,0 +1,37 @@
+class Solution:
+    def canBeIncreasing(self, nums: List[int]) -> bool:
+        # bruteforcing the whole idxes.
+        canBe = [0] * len(nums)
+
+        # choosing the idx that will be removed.
+        for bannedIdx in range(len(nums)):
+            Flag = 1
+
+            # if the bannedIdx is 0 than the startIdx will be 2.
+            # when bannedIdx is 0, idx 2 is the first element that has a previous element. 
+            # In other cases, idx 1 is the one.
+            for i in range(1 if bannedIdx != 0 else 2, len(nums)):
+                # if i is bannedIdx than just skip it.
+                if i == bannedIdx:
+                    continue
+
+                # if the previous element is banned.
+                # compare [i] with [i - 2]
+                if i - 1 == bannedIdx:
+                    if nums[i] <= nums[i - 2]:
+                        Flag = 0
+                        break
+                    continue
+
+                # compare [i] with [i - 1]
+                if nums[i] <= nums[i - 1]:
+                    Flag = 0
+                    break
+
+            # end of loop we will get Flag that has a 0 or 1 value.
+            canBe[bannedIdx] = Flag
+
+        if sum(canBe) > 0:
+            return True
+        return False
+                
diff --git a/python/2409_Count_Days_Spent_Together.py b/python/2409_Count_Days_Spent_Together.py
new file mode 100644
index 0000000..13747f1
--- /dev/null
+++ b/python/2409_Count_Days_Spent_Together.py
@@ -0,0 +1,38 @@
+class Solution:
+    def countDaysTogether(self, arriveAlice: str, leaveAlice: str, arriveBob: str, leaveBob: str) -> int:
+        # split the dates to month and day.
+        arriveAliceMonth, arriveAliceDay = map(int, arriveAlice.split("-"))
+        leaveAliceMonth, leaveAliceDay = map(int, leaveAlice.split("-"))
+        arriveBobMonth, arriveBobDay = map(int, arriveBob.split("-"))
+        leaveBobMonth, leaveBobDay = map(int, leaveBob.split("-"))
+
+        # prefixOfCalendar : initialize the calendar and in the past we will use this to convert month to day, index is 1 - based
+        # spentTogether, aliceSpent : work as cache list. and index is 1 - based
+        calendar = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
+        prefixOfCalendar = [0] * 13
+        totalDates = sum(calendar)
+        spentTogether, aliceSpent = [0] * (totalDates + 1), [0] * (totalDates + 1)
+
+        # calculate the prefix of calendar
+        for i in range(1, len(calendar)):
+            prefixOfCalendar[i] = prefixOfCalendar[i - 1] + calendar[i]
+
+        # if the string is "01-15", it can be treat as 15 days.
+        # if the string is "02-27", it can be treat as 58 days.
+        # So, it can be "prefixOfCalendar[month - 1] + day"
+        # and in the problem it includes the leaveDate so +1 need to be in .
+        arriveAliceTotal = prefixOfCalendar[arriveAliceMonth - 1] + arriveAliceDay
+        leaveAliceTotal = prefixOfCalendar[leaveAliceMonth - 1] + leaveAliceDay
+        for i in range(arriveAliceTotal, leaveAliceTotal + 1):
+            aliceSpent[i] += 1
+
+        # check the aliceSpent[i] is True.
+        # if it is, they spentTogether is True too.
+        arriveBobTotal = prefixOfCalendar[arriveBobMonth - 1] + arriveBobDay
+        leaveBobTotal = prefixOfCalendar[leaveBobMonth - 1] + leaveBobDay
+        for i in range(arriveBobTotal, leaveBobTotal + 1):
+            if aliceSpent[i]:
+                spentTogether[i] += 1
+
+        # I used list because of this sum function.
+        return sum(spentTogether)
diff --git a/python/2413_Smallest_Even_Multiple.py b/python/2413_Smallest_Even_Multiple.py
new file mode 100644
index 0000000..02b723f
--- /dev/null
+++ b/python/2413_Smallest_Even_Multiple.py
@@ -0,0 +1,15 @@
+class Solution:
+    def smallestEvenMultiple(self, n: int) -> int:
+        """
+            n : positive integer
+            return : smallest positive integer that is a multiple of both 2 and n
+        """
+        if n % 2 == 0:
+            # if n is alreay muliply by 2 
+            # return itself
+            return n
+        
+        # if previous condition is false 
+        # n * 2 is the smallest positive integer.
+        return n * 2
+        
diff --git a/python/2420_Find_All_Good_Indices.py b/python/2420_Find_All_Good_Indices.py
new file mode 100644
index 0000000..44498cc
--- /dev/null
+++ b/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
\ No newline at end of file
diff --git a/python/2429_Minimize_XOR.py b/python/2429_Minimize_XOR.py
new file mode 100644
index 0000000..2405feb
--- /dev/null
+++ b/python/2429_Minimize_XOR.py
@@ -0,0 +1,44 @@
+class Solution:
+    def minimizeXor(self, num1: int, num2: int) -> int:
+        # remove "0b" in front of num1, num2
+        num1, num2 = bin(num1)[2:], bin(num2)[2:]
+        lenNum1, lenNum2 = len(num1), len(num2)
+        ones = num2.count("1")
+        maxLen = max(lenNum1, lenNum2)
+
+        # ans list have elements same as the maxLen
+        ans = []
+        for _ in range(maxLen):
+            ans.append("0")
+
+        # add "0" in front of the binary numbers to make indexing easier
+        for _ in range(maxLen - lenNum1):
+            num1 = "0" + num1
+
+        for _ in range(maxLen - lenNum2):
+            num2 = "0" + num2
+
+        # now make "x XOR num1" minimal
+        # fill the ans list from index "0"
+        # because XOR give 0 when the elements are same.
+        for i in range(len(num1)):
+            if num1[i] == "1" and ones:
+                ans[i] = "1"
+                ones -= 1
+
+        # if we still got "1" to fill in the ans list.
+        # "1" need to be fill from the back of ans list.
+        # to maintain the number small.
+        for i in range(len(ans) - 1, -1, -1):
+            if ones < 1:
+                break
+
+            if ans[i] == "1":
+                continue
+
+            ans[i] = "1"
+            ones -= 1
+
+        # make the ans in string
+        ans = "".join(ans)
+        return int(ans, 2)
diff --git a/python/392_Is_Subsequence.py b/python/392_Is_Subsequence.py
new file mode 100644
index 0000000..5cf66c0
--- /dev/null
+++ b/python/392_Is_Subsequence.py
@@ -0,0 +1,11 @@
+class Solution:
+    def isSubsequence(self, s: str, t: str) -> bool:
+        for a in s:
+            if a in t:
+                for b in range(0, len(t)):
+                    if a==t[b]:
+                        t=t[b+1:]
+                        break
+            else:
+                return(False)
+        return(True)
diff --git a/python/523_Continuous_Subarray_Sum.py b/python/523_Continuous_Subarray_Sum.py
new file mode 100644
index 0000000..d37ceeb
--- /dev/null
+++ b/python/523_Continuous_Subarray_Sum.py
@@ -0,0 +1,20 @@
+class Solution:
+    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
+        # remeainders[0] = 0 is for when x == 0
+        remainders = dict()
+        remainders[0] = 0
+        pre_sum = 0
+
+        for idx, item in enumerate(nums):
+            pre_sum += item
+            remaind = pre_sum % k
+
+            # remainder doesnt exist then it has to be init
+            # if it exists, then check the prev one has the same remainder
+            if remaind not in remainders:
+                remainders[remaind] = idx + 1
+            elif remainders[remaind] < idx:
+                return True
+
+        return False
+    
diff --git a/python/732_My_Calendar_III.py b/python/732_My_Calendar_III.py
new file mode 100644
index 0000000..6347108
--- /dev/null
+++ b/python/732_My_Calendar_III.py
@@ -0,0 +1,17 @@
+from sortedcontainers import SortedDict
+
+
+class MyCalendarThree:
+  def __init__(self):
+    self.timeline = SortedDict()
+
+  def book(self, start: int, end: int) -> int:
+    self.timeline[start] = self.timeline.get(start, 0) + 1
+    self.timeline[end] = self.timeline.get(end, 0) - 1
+
+    ans = 0
+    activeEvents = 0
+
+    for count in self.timeline.values():
+      activeEvents += count
+      ans = max(ans, activeEvents)