Added Java solutions for 17, 22 / Python solutions for 2413, 2409, 1909

java/17_Letter_Combinations_of_a_Phone_Number.java

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+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]);
+            }
+        }
+    }
+}

java/22_Generate_Parentheses.java

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+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);
+        }
+    }
+}

python/1909_Remove_One_Element_to_Make_the_Array_Strictly_Increasing.py

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+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
+

python/2409_Count_Days_Spent_Together.py

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+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)

python/2413_Smallest_Even_Multiple.py

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+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
+