In the realm of programming, particularly in the domain of string manipulation, challenges often arise that require efficient solutions. One such problem is determining whether two strings are valid anagrams of each other. In this blog post, we'll delve into this problem, understand its significance, and provide a Python solution to tackle it effectively.

## Understanding the Problem 🧠

The task at hand is straightforward: given two strings `s`

and `t`

, we need to determine if `t`

is an anagram of `s`

. An anagram, as defined, is a word or phrase formed by rearranging the letters of another word or phrase, using all the original letters exactly once.

### Example Scenarios 📝

Let's consider a few scenarios to illustrate the problem:

**Input:**`s = "anagram"`

,`t = "nagaram"`

**Output:**`True`

Explanation: Both`s`

and`t`

contain the same letters arranged differently, making`t`

an anagram of`s`

.**Input:**`s = "rat"`

,`t = "car"`

**Output:**`False`

Explanation: The letters in`s`

and`t`

differ, making`t`

not an anagram of`s`

.

### Approach to Solution 🔍

To solve this problem, we can employ a simple approach using Python's dictionary data structure. Here's how it works:

- First, we check if the lengths of the two strings
`s`

and`t`

are equal. If not, they cannot be anagrams, and we return`False`

. - Next, we create two dictionaries
`countS`

and`countT`

to store the count of each character in`s`

and`t`

, respectively. - We iterate through each character in the strings
`s`

and`t`

, updating the respective counts in the dictionaries. - Finally, we compare the dictionaries
`countS`

and`countT`

. If they are equal, it means that`t`

is indeed an anagram of`s`

, and we return`True`

; otherwise, we return`False`

.

### Python Implementation 🐍

Here's the Python implementation of the solution:

```
def isAnagram(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
```

### Complexity Analysis ⏰

**Time Complexity:**The time complexity of this solution is O(n), where n is the length of the input strings`s`

and`t`

. This is because we iterate through both strings once to build the dictionaries.**Space Complexity:**The space complexity is also O(n) as we store the counts of characters in dictionaries, which could potentially contain all unique characters from both strings.

### Handling Unicode Characters 🌐

What if the inputs contain Unicode characters? The solution remains largely the same. Python's dictionary data structure can handle Unicode characters without any modifications. Therefore, the provided solution would seamlessly extend to scenarios involving Unicode characters.

### Conclusion 🎉

In this blog post, we explored the Valid Anagram problem, its significance, and a Python solution to solve it efficiently. By leveraging dictionary data structures and a straightforward comparison approach, we can determine whether two strings are valid anagrams of each other with ease. This problem serves as a great exercise in string manipulation and algorithmic thinking, showcasing the power and versatility of Python in tackling such challenges. 🚀

For more information and practice, you can visit the problem page on LeetCode. 📚

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