Given a string `s`, consider all _duplicated substrings_: (contiguous) substrings of s that occur 2 or more times. The occurrences may overlap.

Return **any** duplicated substring that has the longest possible length. If `s` does not have a duplicated substring, the answer is `" "`.

**Example 1:**

**Input:** s = "banana"
**Output:** "ana"

**Example 2:**

**Input:** s = "abcd"
**Output:** ""

**Constraints:**

*   `2 <= s.length <= 3 * 104`
*   `s` consists of lowercase English letters.


from collections import Counter


def commonChars(words):
    merged = Counter()
    for word in words:
        word_count = Counter(word)
        if not merged:
            merged = word_count
        else:
            for char in merged:
                merged[char] = min(merged[char], word_count[char])

    result = []
    for char, count in merged.items():
        result.extend([char] * count)
    return result
