Merge pull request #3239 from giggling-ginger/ginger

1397 solution

Author: Ykhan799

Diff for: python/0338-counting-bits.py

@@ -8,3 +8,17 @@ def countBits(self, n: int) -> List[int]:
                 offset = i
             dp[i] = 1 + dp[i - offset]
         return dp
+
+# Another dp solution
+class Solution2:
+    def countBits(self, n: int) -> List[int]:
+        res = [0] * (n + 1)
+        for i in range(1, n + 1):
+            if i % 2 == 1:
+                res[i] = res[i - 1] + 1
+            else:
+                res[i] = res[i // 2]
+        return res
+# This solution is based on the division of odd and even numbers. 
+# I think it's easier to understand.
+# This is my full solution, covering the details: https://leetcode.com/problems/counting-bits/solutions/4411054/odd-and-even-numbers-a-easier-to-understanding-way-of-dp/

Diff for: python/1397-find-all-good-strings-i.py

@@ -0,0 +1,48 @@
+class Solution:
+    def findGoodStrings(self, n: int, s1: str, s2: str, evil: str) -> int:
+        a = ord('a')
+        z = ord('z')
+        
+        arr_e = list(map(ord, evil))
+        len_e = len(evil)
+        next = [0] * len_e
+
+        for i in range(1, len_e):
+            j = next[i - 1]
+            while j > 0 and evil[i] != evil[j]:
+                j = next[j - 1]
+            if evil[i] == evil[j]:
+                next[i] = j + 1
+
+        def good(s):
+            arr = list(map(ord, s))
+            len_a = len(arr)
+
+            @cache
+            def f(i, skip, reach, e):
+                if e == len_e:
+                    return 0
+                if i == len_a:
+                    return 0 if skip else 1
+
+                limit = arr[i] if reach else z
+                ans = 0
+
+                if skip:
+                    ans += f(i + 1, True, False, 0)
+
+                for c in range(a, limit + 1):
+                    ee = e
+                    while ee > 0 and arr_e[ee] != c:
+                        ee = next[ee - 1] 
+
+                    if arr_e[ee] == c:
+                        ee += 1
+
+                    ans += f(i + 1, False, reach and c == limit, ee)
+
+                return ans % int(1e9 + 7)
+
+            return f(0, True, True, 0)
+
+        return (good(s2) - good(s1) + int(evil not in s1)) % int(1e9 + 7)