algocode-hashmap-practice-438 solution

algocode/hash_tables/anagram.py

@@ -0,0 +1,68 @@
+"""
+Given two strings s and t, return true if t is an anagram of s, and false otherwise.
+
+Example 1:
+Input: s = "anagram", t = "nagaram"
+Output: true
+
+Example 2:
+Input: s = "rat", t = "car"
+Output: false
+
+Constraints:
+1 <= s.length, t.length <= 5 * 104
+s and t consist of lowercase English letters.
+
+Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?
+"""
+
+"""
+Анаграмма - слово составленное из одинакового кол-ва букв
+Поулчается что по сути - тут надо делать счетчик букв в каждом слове через хэшмапу
+"""
+from collections import defaultdict
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+        word_1 = defaultdict(int)
+        word_2 = defaultdict(int)
+        for char in s:
+            word_1[char] += 1
+        for char in t:
+            word_2[char] += 1
+        for char, count in word_2.items():
+            """
+            {
+                "n": 1,
+                "a": 3,
+                "g": 1,
+                "r": 1
+            }
+            """
+            if not word_1.get(char) == count:
+                return False
+        return True
+
+
+sc = Solution()
+
+s = "anagram"
+t = "nagaram"
+result = True
+solution = sc.isAnagram(s, t)
+assert solution == result, f"Solution: {solution}, result: {result}"
+
+s = "rat"
+t = "car"
+result = False
+solution = sc.isAnagram(s, t)
+assert solution == result, f"Solution: {solution}, result: {result}"
+
+s = "ab"
+t = "b"
+result = False
+solution = sc.isAnagram(s, t)
+assert solution == result, f"Solution: {solution}, result: {result}"
+

algocode/hash_tables/find_all_anagrams_in_a_string.py

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+"""
+Working solution from 7 try
+Inerview passed: False
+"""
+from typing import List, Optional
+from collections import defaultdict
+
+class Solution:
+    def findAnagrams(self, s: str, p: str):
+        if len(p) > len(s):
+            return []
+        p_len = len(p)
+        p_count = defaultdict(int)
+        for char in p:
+            p_count[char] += 1
+        s_len = len(s)
+        res = [] 
+        cur_pos = 0
+        cur_count = defaultdict(int)
+        # Заполняем предварительный словарь
+        for i in range(0, p_len):
+            cur_count[s[i]] += 1
+            cur_pos += 1
+        if cur_count == p_count:
+            res.append(0)
+        cur_pos = 1
+        while cur_pos <= s_len - p_len:
+            # Delete char from prev iteration
+            d_char = s[cur_pos-1]
+            cur_count[d_char] -= 1
+            if cur_count.get(d_char) == 0:
+                del cur_count[d_char]
+            # Adding current iteration last char
+            a_char = s[cur_pos + p_len - 1]
+            cur_count[a_char] += 1
+            # Compare
+            print(f"{cur_count}\n{p_count}\n----")
+            if cur_count == p_count:
+                res.append(cur_pos)
+            # Next
+            cur_pos += 1
+        return res
+
+c = Solution()
+s = "cbaebabacd"
+p = "abc"
+print(c.findAnagrams(s, p))
+
+s = "af"
+p = "be"
+print(c.findAnagrams(s, p))