Problem Description

The problem presents two input strings, s and t, and asks to determine whether t is an anagram of s. An anagram is defined as a word or phrase that is created by rearranging the letters of another word or phrase, using all the original letters exactly once. The goal is to return true if t is an anagram of s, and false otherwise. This implies that if t is an anagram of s, both strings should have the same letters with the exact amount of occurrences for each letter.

Intuition

To solve this problem efficiently, we can use a counting approach. The intuition comes from the definition of an anagram – the strings must have the same characters in the same quantities. First, we can quickly check if s and t are of the same length. If they're not, t cannot be an anagram of s and we can immediately return false.

Once we've established that s and t are of equal length, the next step is to count the occurrences of each character in s. We can do this efficiently by using a hash table (in Python, this could be a Counter from the collections module). The hash table will map each character to the number of times it appears in s.

After setting up the counts for s, we traverse the string t. For each character in t, we decrease its corresponding count in the hash table by one. If at any point we find that a character's count goes below zero, this means that t contains more of that character than s does, which violates the anagram property. In this case, we can return false.

If we can successfully traverse all characters in t without any counts going below zero, it means that t has the same characters in the same quantities as s. Hence, t is an anagram of s, and we return true.

Learn more about Sorting patterns.

Solution Approach

The solution approach utilizes a hash table to keep track of the characters in string s. The use of a hash table allows us to efficiently map characters to the number of times they occur in s. This is a common pattern when we need to count instances of items, such as characters in a string.

The solution proceeds with the following steps:

1. Check Length: We compare the lengths of s and t. If they are different, we immediately return false because an anagram must have the same number of characters.

2. Initialize Counter: We initialize a Counter from the collections module in Python for the first string s. This Counter object effectively creates the hash table that maps each character to its count in the string.

3. Character Traversal & Counting: We then iterate over each character c in the second string t and decrement the count of c in our Counter by one for each occurrence. The Counter object allows us to do this in constant time for each character.

4. Check Negative Counts: We check if the decremented count of any character goes below zero. If this happens, it means that t has an extra occurrence of a character that does not match the count in s, and we return false.

5. Return True: If we go through the entire string t without encountering a negative count, this means that t has the exact same characters in the same amounts as s, confirming that t is an anagram of s. Thus, we return true.

By using a hash table, we achieve a linear time complexity of O(n), where n is the length of the strings, since we go through each string once. This makes the algorithm efficient for large strings. The solution combines the steps of setting up the Counter, iterating over t, and verifying the counts into a concise and effective approach.

Example Walkthrough

Let's consider an example where s = "listen" and t = "silent". We want to determine if t is an anagram of s using the solution approach described above.

1. Check Length: We check the lengths of both s and t. In this case, both strings are of length 6. As they are equal, we proceed to the next step.

2. Initialize Counter: We utilize a Counter from the collections module in Python to tally the characters in string s. So, we get the counts as follows:

   {'l': 1, 'i': 1, 's': 1, 't': 1, 'e': 1, 'n': 1}

3. Character Traversal & Counting: We iterate over each character in string t.

   • Decrement count for 's': {'l': 1, 'i': 1, 's': 0, 't': 1, 'e': 1, 'n': 1}
   • Decrement count for 'i': {'l': 1, 'i': 0, 's': 0, 't': 1, 'e': 1, 'n': 1}
   • Decrement count for 'l': {'l': 0, 'i': 0, 's': 0, 't': 1, 'e': 1, 'n': 1}
   • Decrement count for 'e': {'l': 0, 'i': 0, 's': 0, 't': 1, 'e': 0, 'n': 1}
   • Decrement count for 'n': {'l': 0, 'i': 0, 's': 0, 't': 1, 'e': 0, 'n': 0}
   • Decrement count for 't': {'l': 0, 'i': 0, 's': 0, 't': 0, 'e': 0, 'n': 0}

4. Check Negative Counts: Throughout the traversal, we check the counts after each decrement. No count goes below zero, meaning there are no extra occurrences of any character.

5. Return True: After traversing t, all the counts are exactly zero, which means that t has precisely the same characters in the same amounts as s. Therefore, according to the solution approach, we return true, confirming that t is an anagram of s.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n) where n represents the length of the string s (and t, since they are of the same length for the comparison to be valid). This complexity arises because the code iterates over each character in both s and t exactly once.

The space complexity of the code is O(C) where C is the size of the character set used in the strings. In this context, since we are typically dealing with lowercase English letters, C is equal to 26. The space complexity comes from the use of a counter to store the frequency of each character in string s.

Learn more about how to find time and space complexity quickly using problem constraints.