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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, 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)
for c in countS:
if countS[c] != countT.get(c, 0):
return False
return True
case1S = "apple"
case1T = "plpea"
case2S = "asneaxl"
case2T = "yejindp"
case3S = "iloveyou"
case3T = "youlovei"
solution = Solution()
print(f"hashmap case1: {solution.hashmap(case1S, case1T)}")
print(f"hashmap case2: {solution.hashmap(case2S, case2T)}")
print(f"hashmap case3: {solution.hashmap(case3S, case3T)}")
"""
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:
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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