Find the Earliest Anagram Window

Question

The candidate is asked to find the earliest substring in a given string that contains all the characters of another string, which can be an anagram of the second string.

Codemia · Accepted Answer

To solve the problem of finding the earliest anagram window, we can use the sliding window technique combined with a hash map to count character occurrences. The idea is to maintain a window that expands and contracts over the input string while checking if it contains all characters of the target string.

1. Create a hash map to store the character counts of the target string.
2. Initialize two pointers for the sliding window and a counter to track matched characters.
3. Expand the right pointer to include characters in the window and update the hash map.
4. When the window contains all characters, contract the left pointer to find the smallest valid window.
5. Return the earliest valid window found or indicate that none exists.

The time complexity of this approach is O(n), where n is the length of the input string, since each character is processed at most twice (once by each pointer). The space complexity is O(m), where m is the size of the character set used in the target string, due to the hash map.

```python
def find_anagram_window(s: str, t: str) -> str:
    from collections import Counter
    target_count = Counter(t)
    window_count = Counter()
    left = 0
    matched = 0
    min_length = float('inf')
    start_index = 0

for right in range(len(s)):
        char = s[right]
        window_count[char] += 1
        if char in target_count and window_count[char] == target_count[char]:
            matched += 1

while matched == len(target_count):
            if right - left + 1 < min_length:
                min_length = right - left + 1
                start_index = left
            window_count[s[left]] -= 1
            if s[left] in target_count and window_count[s[left]] < target_count[s[left]]:
                matched -= 1
            left += 1

return s[start_index:start_index + min_length] if min_length != float('inf') else ""
```
This code efficiently finds the earliest anagram window by maintaining a count of characters and adjusting the window size dynamically.

Consider edge cases such as when the target string is longer than the input string, or when either string is empty. These should be handled gracefully by returning an appropriate result, such as an empty string.

Find the Earliest Anagram Window

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