🚨 ZipRecruiter 2026 OA: Longest Common Suffix Path — New Grad Must-Solve, Path Simplification + Suffix Matching!

ZipRecruiter OA Overview

Today let's talk about the ZipRecruiter OA New Grad online assessment experience.

For fresh graduates, ZipRecruiter OA has moderate difficulty, but it heavily tests logical thinking and detail handling ability. Having assisted many batches of ZipRecruiter OA, I've found it has a different style from other companies: it doesn't pursue complex algorithm implementations, but rather values whether you can quickly clarify your approach, correctly handle edge cases and classification problems.

ZipRecruiter OA Process

ZipRecruiter's Online Assessment (OA) usually has 1-2 rounds. Most New Grad candidates encounter a single-round OA:

Item	Details
Duration	50-70 minutes (recently mostly 70min)
Questions	2-4 problems
Platform	CodeSignal (General Coding Assessment)
Languages	Python, Java, C++ and other mainstream languages

OA Core Characteristics

Moderate question count: Unlike some big company OAs with many questions, each problem tests your approach and attention to detail
Logic is king: Rather than writing super complex algorithms, focuses on clear thinking, edge case handling, and classification ability
Data processing heavy: Arrays, matching problems, greedy strategies, sorting, and two pointers are common patterns

ZipRecruiter OA FAQ (2026 Latest Experience Summary)

Question	Answer
Q1: Platform?	CodeSignal (General Coding Assessment)
Q2: Questions & time?	4 problems, 60-70 minutes (recently mostly 70min)
Q3: Proctored?	Some batches require camera + screen sharing (proctored), invitation email will specify
Q4: Difficulty?	LeetCode Easy ~ Medium (first 2-3 easy, 4th Medium)
Q5: Common types?	Array/String, Hash Table, Simulation, Two Pointers, Simple DP (occasionally multi-part implementation)
Q6: Passing standard?	No official threshold, 700+ score is safer (speed + accuracy); high scores may still be rejected due to resume/needs
Q7: How to prepare?	Practice on CodeSignal + LeetCode Easy/Medium; practice AC 4 problems in 70min
Q8: After passing?	Recruiter Call → Technical Screen → Onsite

Problem Breakdown: Longest Common Suffix Path

🔍 Original Problem

You are given an array of strings paths containing paths to some files or directories. All file paths start with "/".

Note: File paths may contain special characters ".." to represent parent directory specifiers. For example, "/a/b/../" points to the same directory as "/a". It is guaranteed that all given file paths are valid.

Your task is to find the longest common suffix path without file masks or special characters (e.g., "..", etc.). If there isn't any common suffix path, return an empty string as "".

🎯 Example Analysis

Input:

paths = [
    "/a/folder1/../folder1/a/leaf.txt",
    "/b/folder2/../folder1/a/leaf.txt",
    "/a/folder3/folder1/folder1/../a/leaf.txt"
]

Output:

"/folder1/a/leaf.txt"

Explanation:

First simplify each path (resolve ..):

Original Path	Simplified
`/a/folder1/../folder1/a/leaf.txt`	`/a/folder1/a/leaf.txt`
`/b/folder2/../folder1/a/leaf.txt`	`/b/folder1/a/leaf.txt`
`/a/folder3/folder1/folder1/../a/leaf.txt`	`/a/folder3/folder1/a/leaf.txt`

Compare from the end:

leaf.txt ✓ same
a ✓ same
folder1 ✓ same
3rd level: a vs b vs folder3 ✗ different

Therefore, the longest common suffix path is /folder1/a/leaf.txt.

Deep Dive: Problem Essence

This problem tests two core abilities:

Test Point	Specific Content
Path Simplification	Handle `..` to go up one directory level
Suffix Matching	Find longest common part from the end
Edge Case Handling	Empty paths, all same, no common suffix
String Splitting	Split into path components by `/`

Key Insights

Path Simplification: Use stack to simulate directory traversal, pop on ..
Suffix Matching: Reverse path components, find Longest Common Prefix (LCP), reverse back
Note: Output must start with /

Solution: Core Algorithm

Solution: Path Simplification + Suffix LCP

def solution(paths: list) -> str:
    """
    ZipRecruiter OA: Longest Common Suffix Path
    
    Args:
        paths: Array of file paths
    
    Returns:
        str: Longest common suffix path
    """
    
    def simplify_path(path: str) -> list:
        """
        Simplify path: resolve .. and return component list
        
        Args:
            path: Original path string
        
        Returns:
            list: Simplified path component list
        """
        components = path.split('/')
        stack = []
        
        for comp in components:
            if comp == '' or comp == '.':
                # Empty string or current directory, skip
                continue
            elif comp == '..':
                # Parent directory, pop stack (if not empty)
                if stack:
                    stack.pop()
            else:
                # Normal directory/filename, push to stack
                stack.append(comp)
        
        return stack
    
    def longest_common_suffix(lists: list) -> list:
        """
        Find longest common suffix of multiple lists
        
        Args:
            lists: List of path component lists
        
        Returns:
            list: Longest common suffix components
        """
        if not lists:
            return []
        
        # Reverse all lists, convert to finding LCP
        reversed_lists = [lst[::-1] for lst in lists]
        
        # Find minimum list length
        min_len = min(len(lst) for lst in reversed_lists)
        
        # Compare position by position
        common_len = 0
        for i in range(min_len):
            # Use first list's i-th element as base
            base = reversed_lists[0][i]
            # Check if all lists have same i-th element
            if all(lst[i] == base for lst in reversed_lists):
                common_len += 1
            else:
                break
        
        # Return common suffix (reversed back)
        return reversed_lists[0][:common_len][::-1]
    
    # Step 1: Simplify all paths
    simplified = [simplify_path(p) for p in paths]
    
    # Step 2: Find longest common suffix
    common_suffix = longest_common_suffix(simplified)
    
    # Step 3: Build result path
    if not common_suffix:
        return ""
    
    return '/' + '/'.join(common_suffix)

🧪 Test Cases

# Test 1: Basic example
paths1 = [
    "/a/folder1/../folder1/a/leaf.txt",
    "/b/folder2/../folder1/a/leaf.txt",
    "/a/folder3/folder1/folder1/../a/leaf.txt"
]
print(solution(paths1))  # Expected: "/folder1/a/leaf.txt"

# Test 2: Identical paths
paths2 = [
    "/a/b/c/file.txt",
    "/a/b/c/file.txt",
    "/a/b/c/file.txt"
]
print(solution(paths2))  # Expected: "/a/b/c/file.txt"

# Test 3: No common suffix
paths3 = [
    "/a/file1.txt",
    "/b/file2.txt",
    "/c/file3.txt"
]
print(solution(paths3))  # Expected: ""

# Test 4: Only filename matches
paths4 = [
    "/a/b/file.txt",
    "/x/y/file.txt",
    "/m/n/file.txt"
]
print(solution(paths4))  # Expected: "/file.txt"

# Test 5: Multiple ..
paths5 = [
    "/a/b/c/../d/../e/file.txt",
    "/x/y/../z/e/file.txt"
]
print(solution(paths5))  # Expected: "/e/file.txt"

# Test 6: Single path
paths6 = ["/a/b/c/file.txt"]
print(solution(paths6))  # Expected: "/a/b/c/file.txt"

# Test 7: .. at root
paths7 = [
    "/../a/b/file.txt",
    "/c/d/file.txt"
]
print(solution(paths7))  # Expected: "/file.txt"

# Test 8: Empty array
paths8 = []
print(solution(paths8))  # Expected: ""

Complexity Analysis

Item	Complexity
Time Complexity	O(n × m), n = number of paths, m = average path length
Space Complexity	O(n × m), storing simplified paths

🤯 Interviewer's Followup Traps

Followup 1: Return the common suffix from original paths?

Question: Not the simplified suffix, but the corresponding part in original strings?

Answer: Need to maintain mapping between original and simplified paths.

def solution_with_original_mapping(paths):
    """Return common suffix from original paths"""
    # Track original position for each simplified component
    # Implementation complexity is higher, requires character-by-character tracking
    pass

Followup 2: How to handle symbolic links (symlink)?

Question: If paths contain symbolic links, .. behavior might differ?

Answer: Need to distinguish "logical path" vs "physical path". Problems usually assume logical path handling.

Followup 3: What if paths are very long (1 million characters)?

Answer: Use streaming processing, don't load entire path at once.

def simplify_path_streaming(path_iterator):
    """Streaming path simplification"""
    stack = []
    for char in path_iterator:
        # Process character by character
        pass
    return stack

Followup 4: How to optimize space complexity?

Answer: Don't store complete simplified paths, compare directly from end.

def solution_optimized(paths):
    """Space-optimized version: simplify and compare simultaneously"""
    # Start from end, compare component by component
    # Need to reverse-traverse paths
    pass

🔥 Why Most People Fail

Common Mistake	Correct Approach
Forget to handle `..`	Use stack to simulate directory traversal
Directly compare string suffixes	Split by `/` into components then compare
Forget result needs leading `/`	Add leading `/` when joining
No empty path handling	Check empty array, empty component list
`..` at root directory	Don't pop when stack is empty

Code Template (Ready to Use)

def solution(paths: list) -> str:
    def simplify(path):
        stack = []
        for comp in path.split('/'):
            if comp == '..':
                if stack: stack.pop()
            elif comp and comp != '.':
                stack.append(comp)
        return stack
    
    if not paths:
        return ""
    
    simplified = [simplify(p) for p in paths]
    
    # Reverse to find LCP
    rev = [s[::-1] for s in simplified]
    min_len = min(len(r) for r in rev)
    
    common = 0
    for i in range(min_len):
        if all(r[i] == rev[0][i] for r in rev):
            common += 1
        else:
            break
    
    suffix = rev[0][:common][::-1]
    return '/' + '/'.join(suffix) if suffix else ""

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