Asked in: JPMORGAN
def funWithAnagrams(text):
visited_words = set()
result = []
for word in text:
// ... rest of solution available after purchase
```
To approach the "FunWithAnagrams" problem, begin by understanding the core concept of anagrams. Two words are anagrams of each other if they contain the same characters with the same frequencies, regardless of the order. This means that any two strings that are permutations of the same set of characters are anagrams. Your task is to remove any string from the array that is an anagram of a string that has already appeared before it.
The first step in solving this problem is to identify a consistent and efficient way to detect whether two strings are anagrams. A common strategy is to sort the characters in the string. When two strings are anagrams, sorting their characters will result in identical strings. For example, both "code" and "doce" become "cdeo" after sorting, which is a clear indicator that they are anagrams.
Now, consider how to apply this insight to the entire array of strings. As you iterate through the array, you want to keep track of which anagram patterns (the sorted character versions) you’ve already encountered. To do this efficiently, think about using a data structure like a set that allows for fast lookups and insertions. Each time you process a new string, sort it to create its normalized form and check whether this form already exists in your set. If it doesn’t, it means this is the first time you’re seeing this pattern, so you add the sorted version to your set and keep the original string in the result. If the normalized form is already present in the set, you skip the string because it’s an anagram of something you’ve already kept.
After processing all the strings this way, you’ll have a filtered list that contains only the first occurrence of each unique anagram group. The final requirement is to sort this filtered list in ascending order. This sorting should be done on the original string values, not the sorted anagram keys.
Let’s now break this approach down into logical steps you can follow when thinking about your implementation:
1. Initialize an empty set to store the sorted version of strings (i.e., anagram signatures).
2. Initialize an empty list to store the final result.
3. Iterate through each string in the original input array.
4. For each string, sort its characters to create a normalized version.
5. Check if this normalized version exists in the set:
- If it does not, add the normalized version to the set and the original string to the result list.
- If it does exist, it means this string is an anagram of a previously encountered string, so skip it.
6. Once all strings have been processed, sort the result list in lexicographical order.
7. Return this sorted list.
Now, consider a few key details and edge cases:
- Strings with the same characters but in different cases (e.g., "Code" and "code") may or may not be considered anagrams depending on whether the comparison is case-sensitive. The problem likely assumes case sensitivity based on the example, so you should not convert characters to a common case unless explicitly required.
- The problem wants you to retain the first occurrence of each anagram group, so the order in which you process the array matters. This means you cannot sort the array before filtering for anagrams.
- Sorting each string individually is acceptable here because strings are typically short, and sorting them takes time proportional to their length. If there are N strings of average length M, the total time spent sorting individual strings is roughly O(N * M log M), which is efficient for most practical cases.
In summary, this problem is a combination of recognizing anagram equivalence via character sorting, tracking encountered patterns using a set for fast access, and then presenting the final answer in sorted order. The overall approach is efficient and well-suited to handle a moderately large list of strings. Focus your thought process on transforming each string into a form that allows quick anagram checks, and then manage which strings to include based on that transformation.
```