Merry Xmas!

Author: jameszu

Daily-challenge/Dec/24/README.md

@@ -30,26 +30,6 @@ Just do while loop and convert to decimal
 
 ## Code
 ```python
-class Solution:
-    def calculate(self, s: str) -> int:
-        op = '+'
-        pre = 0; num = 0; res = 0
-        for c in s+'+':
-            if c.isdigit():
-                num = 10*num + int(c)
-            elif c in '+-*/':
-                if op == '+': 
-                    res += pre
-                    pre = num
-                elif op == '-': 
-                    res += pre
-                    pre = -num
-                elif op == '*': 
-                    pre = pre*num
-                elif op == '/':
-                    pre = int(pre/num)
-                num = 0 
-                op = c
-        return res+pre
+Didn't do it
 
 ```

Daily-challenge/Dec/24/sol.py

@@ -1,21 +1 @@
-class Solution:
-    def calculate(self, s: str) -> int:
-        op = '+'
-        pre = 0; num = 0; res = 0
-        for c in s+'+':
-            if c.isdigit():
-                num = 10*num + int(c)
-            elif c in '+-*/':
-                if op == '+': 
-                    res += pre
-                    pre = num
-                elif op == '-': 
-                    res += pre
-                    pre = -num
-                elif op == '*': 
-                    pre = pre*num
-                elif op == '/':
-                    pre = int(pre/num)
-                num = 0 
-                op = c
-        return res+pre
+DIdn't do it : (

Daily-challenge/Dec/25/README.md

@@ -1,45 +1,64 @@
-# Smallest Integer Divisible by K
-Given a positive integer K, you need to find the length of the smallest positive integer N such that N is divisible by K, and N only contains the digit 1.
+# Diagonal Traverse
 
-Return the length of N. If there is no such N, return -1.
-
-Note: N may not fit in a 64-bit signed integer.
+Solution
+Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.
 
 
 
-Example 1:
+Example:
+
+Input:
+[
+ [ 1, 2, 3 ],
+ [ 4, 5, 6 ],
+ [ 7, 8, 9 ]
+]
 
-Input: K = 1
-Output: 1
-Explanation: The smallest answer is N = 1, which has length 1.
-Example 2:
+Output:  [1,2,4,7,5,3,6,8,9]
 
-Input: K = 2
-Output: -1
-Explanation: There is no such positive integer N divisible by 2.
-Example 3:
+Explanation:
 
-Input: K = 3
-Output: 3
-Explanation: The smallest answer is N = 111, which has length 3.
 
 
-Constraints:
+Note:
 
-1 <= K <= 105<br>
+The total number of elements of the given matrix will not exceed 10,000.<br>
 
 ## Idea
 Just do while loop and convert to decimal
 
 ## Code
 ```python
 class Solution:
-    def smallestRepunitDivByK(self, K: int) -> int:
-        remainder = 0
-        for length_N in range(1,K+1):
-            remainder = (remainder*10+1) % K
-            if remainder == 0:
-                return length_N
-        return -1
+    def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
+        resp = []
+        if not len(matrix): return resp
+        m = len(matrix)
+        n = len(matrix[0])
+        i, j = 0, 0
+        while len(resp) < m * n:
+            # up-right
+            while i >= 0 and j < n:
+                resp.append(matrix[i][j])
+                i -= 1
+                j += 1
+            if j < n:
+                i += 1
+            else:
+                i += 2
+                j -= 1
+
+            # down-left
+            while j >= 0 and i < m:
+                resp.append(matrix[i][j])
+                i += 1
+                j -= 1
+            if i < m:
+                j += 1
+            else:
+                j += 2
+                i -= 1
+        
+        return resp
 
 ```

Daily-challenge/Dec/25/sol.py

@@ -1,8 +1,31 @@
 class Solution:
-    def smallestRepunitDivByK(self, K: int) -> int:
-        remainder = 0
-        for length_N in range(1,K+1):
-            remainder = (remainder*10+1) % K
-            if remainder == 0:
-                return length_N
-        return -1
+    def findDiagonalOrder(self, matrix: List[List[int]]) -> List[int]:
+        resp = []
+        if not len(matrix): return resp
+        m = len(matrix)
+        n = len(matrix[0])
+        i, j = 0, 0
+        while len(resp) < m * n:
+            # up-right
+            while i >= 0 and j < n:
+                resp.append(matrix[i][j])
+                i -= 1
+                j += 1
+            if j < n:
+                i += 1
+            else:
+                i += 2
+                j -= 1
+
+            # down-left
+            while j >= 0 and i < m:
+                resp.append(matrix[i][j])
+                i += 1
+                j -= 1
+            if i < m:
+                j += 1
+            else:
+                j += 2
+                i -= 1
+        
+        return resp

Daily-challenge/Dec/26/README.md

@@ -1,25 +1,41 @@
-# Longest Substring with At Least K Repeating Characters
-Given a string s and an integer k, return the length of the longest substring of s such that the frequency of each character in this substring is less than or equal to k.
+# Decode Ways
+A message containing letters from A-Z is being encoded to numbers using the following mapping:
+
+'A' -> 1
+'B' -> 2
+...
+'Z' -> 26
+Given a non-empty string containing only digits, determine the total number of ways to decode it.
+
+The answer is guaranteed to fit in a 32-bit integer.
 
 
 
 Example 1:
 
-Input: s = "aaabb", k = 3
-Output: 3
-Explanation: The longest substring is "aaa", as 'a' is repeated 3 times.
+Input: s = "12"
+Output: 2
+Explanation: It could be decoded as "AB" (1 2) or "L" (12).
 Example 2:
 
-Input: s = "ababbc", k = 2
-Output: 5
-Explanation: The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.
+Input: s = "226"
+Output: 3
+Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
+Example 3:
+
+Input: s = "0"
+Output: 0
+Explanation: There is no character that is mapped to a number starting with '0'. We cannot ignore a zero when we face it while decoding. So, each '0' should be part of "10" --> 'J' or "20" --> 'T'.
+Example 4:
+
+Input: s = "1"
+Output: 1
 
 
 Constraints:
 
-1 <= s.length <= 104
-s consists of only lowercase English letters.
-1 <= k <= 105<br>
+1 <= s.length <= 100
+s contains only digits and may contain leading zero(s).<br>
 
 ## Idea
 Just do while loop and convert to decimal