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"""
Question
Is Anagram
Easy
Given two strings s and t, return true if the two strings are anagrams of each other, otherwise return false.
An anagram is a string that contains the exact same characters as another string, but the order of the characters can be different.
Example 1:
Input: s = "racecar", t = "carrace"
Output: true
Example 2:
Input: s = "jar", t = "jam"
Output: false
Constraints:
- s and t consist of lowercase English letters.
"""
class Solution:
def hashmap(self, s: str, t: str) -> bool:
if len(s) != len(t):
return False
countS: dict = {}
countT: dict = {}
for i, si in enumerate(s):
countS[si] = 1 + countS.get(si, 0)
countT[t[i]] = 1 + countT.get(t[i], 0)
return countS == countT
caseS1 = "racecar"
caseT1 = "carrace"
caseS2 = "jar"
caseT2 = "jam"
caseS3 = "asneaxl"
caseT3 = "jinyedp"
solution = Solution()
print(f"hashmap case1: {solution.hashmap(caseS1, caseT1)}")
print(f"hashmap case2: {solution.hashmap(caseS2, caseT2)}")
print(f"hashmap case3: {solution.hashmap(caseS3, caseT3)}")
"""
Solution
url: https://neetcode.io/problems/is-anagram
video: https://youtu.be/9UtInBqnCgA
1. brute force
time: O(n)
space: O(1)
2. hashmap
time: O(s+t)
space: O(s+t)
code:
```python
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
if len(s) != len(t):
return False
countS, countT = {}, {}
for i in range(len(s)):
countS[s[i]] = 1 + countS.get(s[i], 0)
countT[t[i]] = 1 + countT.get(t[i], 0)
return countS == countT
```
"""
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