🧑‍💻 Interviewer Guide: Longest Common Prefix (UMPIRE Method)

https://leetcode.com/problems/longest-common-prefix/

📌 Problem Statement

Given an array of strings strs.
Return the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string "".

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🧭 U — Understand the Problem

Interviewer Prompts & Possible Answers:

• "Can you restate the problem in your own words?"
  → "We need to find the longest string that appears at the beginning of every string in the array. All strings must start with this prefix."

• "What does 'common prefix' mean exactly?"
  → "A prefix is a substring that starts at index 0. A common prefix is a prefix shared by all strings. For example, 'fl' is a common prefix for ['flower', 'flow', 'flight']."

• "What should we return if there's no common prefix?"
  → "Return an empty string ''."

• "What if the array has only one string?"
  → "That string itself is the longest common prefix since it's the only string."

• "What if the array is empty?"
  → "We should handle this edge case by returning an empty string, though the problem typically guarantees at least one string."

• "Can the strings contain special characters or just letters?"
  → "The strings can contain any characters. We just need to match them character by character."

• "If one string is a prefix of another, what happens?"
  → "The shorter string would be the common prefix. For example, ['flow', 'flower'] → 'flow'."

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🧪 M — Match with Patterns

Interviewer Prompts & Possible Answers:

• "What type of problem is this?"
  → "This is a string comparison problem that requires finding commonality across multiple strings."

• "What approaches come to mind?"
  → "Horizontal scanning (compare pairs), vertical scanning (compare column by column), divide and conquer, or binary search on the prefix length."

• "Which approach seems most intuitive?"
  → "Vertical scanning is intuitive: check each character position across all strings until we find a mismatch."

• "How is this different from finding the longest common substring?"
  → "This only checks prefixes (beginning of strings), not substrings anywhere in the string. This makes it simpler."

• "What determines when we can stop searching?"
  → "We stop when we find a character that doesn't match across all strings, or when we reach the end of the shortest string."

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🛠️ P — Plan the Solution

Interviewer Prompts & Possible Answers:

• "Walk me through your approach step by step."
  → "Use the first string as a reference. Compare each character position across all strings. When we find a position where characters differ, or reach the end of any string, return the prefix up to that point."

• "What's your strategy for comparing characters?"
  → "Iterate through character positions (0, 1, 2, ...). At each position, check if all strings have the same character. If not, we've found where the common prefix ends."

• "How do you handle strings of different lengths?"
  → "The common prefix can't be longer than the shortest string, so we either check the shortest string's length upfront, or stop when any string runs out of characters."

• "What's the early termination condition?"
  → "Stop as soon as we find a character that doesn't match across all strings, or when we've processed the entire shortest string."

• "Could you use sorting to help?"
  → "Yes! After sorting, we only need to compare the first and last strings. If they share a prefix, all middle strings must share it too."

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⌨️ I — Implement

Main Solution (Vertical Scanning)

def longestCommonPrefix(strs: list[str]) -> str:
    if not strs:
        return ""
    
    # Use first string as reference
    for i in range(len(strs[0])):
        char = strs[0][i]
        
        # Check if this character matches in all other strings
        for string in strs[1:]:
            # If we've reached the end of this string or characters don't match
            if i >= len(string) or string[i] != char:
                return strs[0][:i]
    
    # If we've checked all characters in first string, it's the common prefix
    return strs[0]

Alternative Implementation (Horizontal Scanning)

def longestCommonPrefix(strs: list[str]) -> str:
    if not strs:
        return ""
    
    # Start with first string as prefix
    prefix = strs[0]
    
    # Compare with each string and reduce prefix
    for string in strs[1:]:
        # Reduce prefix until it matches the beginning of current string
        while not string.startswith(prefix):
            prefix = prefix[:-1]
            if not prefix:
                return ""
    
    return prefix

Optimized Implementation (Sorting)

def longestCommonPrefix(strs: list[str]) -> str:
    if not strs:
        return ""
    
    # Sort the array
    strs.sort()
    
    # Only need to compare first and last strings
    first = strs[0]
    last = strs[-1]
    
    # Find common prefix between first and last
    i = 0
    while i < len(first) and i < len(last) and first[i] == last[i]:
        i += 1
    
    return first[:i]

Implementation Using Min/Max

def longestCommonPrefix(strs: list[str]) -> str:
    if not strs:
        return ""
    
    # Get lexicographically smallest and largest strings
    # They will differ most, so if they share a prefix, all do
    min_str = min(strs)
    max_str = max(strs)
    
    # Find common prefix
    for i in range(len(min_str)):
        if min_str[i] != max_str[i]:
            return min_str[:i]
    
    return min_str

Implementation Using Zip (Pythonic)

def longestCommonPrefix(strs: list[str]) -> str:
    if not strs:
        return ""
    
    # Zip groups characters at same position across all strings
    prefix = []
    for chars in zip(*strs):
        # If all characters at this position are the same
        if len(set(chars)) == 1:
            prefix.append(chars[0])
        else:
            break
    
    return ''.join(prefix)

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🔁 R — Review

Interviewer Prompts & Possible Answers:

• "What edge cases should we test?"
  → "Empty array, array with one string, no common prefix, all strings identical, one string is prefix of others, strings with different lengths, empty string in array."

• "Can you trace through an example?"
  → "Let's trace ['flower', 'flow', 'flight']: Position 0: f=f=f ✓, Position 1: l=l=l ✓, Position 2: o=o=i ✗ → return 'fl'."

• "Trace through when there's no common prefix?"
  → "['dog', 'racecar', 'car']: Position 0: d≠r → return ''."

• "How does the sorting approach work?"
  → "After sorting ['flower', 'flow', 'flight'] → ['flight', 'flow', 'flower'], compare first 'flight' and last 'flower'. They differ at position 2, so common prefix is 'fl'."

• "Could you write some test cases?"

def test_longestCommonPrefix():
    # Test case 1: Normal case with common prefix
    assert longestCommonPrefix(["flower", "flow", "flight"]) == "fl"
    
    # Test case 2: No common prefix
    assert longestCommonPrefix(["dog", "racecar", "car"]) == ""
    
    # Test case 3: Single string
    assert longestCommonPrefix(["alone"]) == "alone"
    
    # Test case 4: All strings identical
    assert longestCommonPrefix(["test", "test", "test"]) == "test"
    
    # Test case 5: Empty array
    assert longestCommonPrefix([]) == ""
    
    # Test case 6: One string is prefix of all others
    assert longestCommonPrefix(["flower", "flow", "flo"]) == "flo"
    
    # Test case 7: Empty string in array
    assert longestCommonPrefix(["abc", "", "ab"]) == ""
    
    # Test case 8: Single characters, no match
    assert longestCommonPrefix(["a", "b", "c"]) == ""
    
    # Test case 9: Single characters, match
    assert longestCommonPrefix(["a", "a", "a"]) == "a"
    
    # Test case 10: Different lengths, common prefix
    assert longestCommonPrefix(["interspecies", "interstellar", "interstate"]) == "inters"

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🧹 E — Evaluate & Optimize

Interviewer Prompts & Possible Answers:

• "What's the time and space complexity of the vertical scanning approach?"
  → "Time: O(S) where S is the sum of all characters in all strings. In worst case (all strings identical), we check every character. Space: O(1) for the algorithm itself."

• "What about the horizontal scanning approach?"
  → "Time: O(S) as well, but it might be slower in practice because of repeated string operations. Space: O(1) if we don't count the prefix variable."

• "What's the complexity of the sorting approach?"
  → "Time: O(S * log n) where n is the number of strings, due to sorting. After sorting, comparison is O(m) where m is the length of the shortest string. Space: O(1) if sorting in-place, or O(S) for the sort."

• "Which approach is best?"
  → "Vertical scanning is usually best for this problem: it's intuitive, has optimal O(S) time complexity, and terminates early. Sorting is elegant but has overhead. The zip approach is Pythonic but may be less readable for some."

• "When would sorting be advantageous?"
  → "When the strings are already sorted or nearly sorted, or when you need to perform multiple queries on the same dataset."

• "How do these compare in the best case?"
  → "Best case: when the first character differs or first string is empty. Vertical scanning: O(n) just checks first char of each string. Horizontal: O(1) finds mismatch immediately. Sorting: O(S * log n) can't avoid sorting."

• "What about the worst case?"
  → "Worst case: all strings are identical. All approaches need to check all characters, so O(S)."

• "Any optimization for very long strings?"
  → "Binary search on the prefix length: check if a prefix of length mid is common, then search left or right half. Time complexity: O(S * log m) where m is minimum string length."

Advanced: Binary Search Implementation (Optional)

def longestCommonPrefix(strs: list[str]) -> str:
    if not strs:
        return ""
    
    def is_common_prefix(length):
        """Check if prefix of given length is common to all strings"""
        prefix = strs[0][:length]
        return all(s.startswith(prefix) for s in strs)
    
    # Binary search on prefix length
    min_len = min(len(s) for s in strs)
    left, right = 0, min_len
    
    while left <= right:
        mid = (left + right) // 2
        if is_common_prefix(mid):
            left = mid + 1
        else:
            right = mid - 1
    
    return strs[0][:right]

Time Complexity: O(S * log m) where S is sum of all characters, m is minimum length
Space Complexity: O(1)

Comparison Table

Approach	Time	Space	Pros	Cons
Vertical Scanning	O(S)	O(1)	Optimal, early termination	-
Horizontal Scanning	O(S)	O(1)	Simple logic	More string operations
Sorting	O(S * log n)	O(1)/O(S)	Elegant, only compare 2 strings	Sorting overhead
Zip	O(S)	O(m)	Pythonic	Less intuitive, creates tuples
Binary Search	O(S * log m)	O(1)	Good for very long strings	More complex

Recommendation: Use vertical scanning for most cases due to its optimal complexity and clarity.