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๐Ÿšจ Meta OA Leaked! 70 Minutes, 4 Problems โ€” Why Even Experienced Engineers Are Failing?

2026-01-17

Meta's CodeSignal OA is currently being sent out in large batches.

Many candidates see the format and think: "70 minutes for 4 questions? That's plenty of time, right?" Don't be fooled.

If you haven't practiced the specific problem patterns from Amazon / Google / TikTok / Meta OAs, you'll likely crash and burn on the last two problems. This OA demands high algorithm proficiency and strict time management โ€” one slip and it's a systematic failure.

The oavoservice team obtained the actual questions and completed a full AC review. We discovered countless hidden traps involving Union-Find, string manipulation, and reading comprehension. Let's break down all 4 problems.


๐Ÿ“Š Meta OA Overview

Item Details
Platform CodeSignal
Questions 4 Coding Problems
Duration 70 minutes
Difficulty Medium-Hard, Fast-paced
Focus Areas Simulation, Statistics, Union-Find/Graph Connectivity, Complex Instruction Parsing

โœ… Q1 | Basic Simulation (Warm-up, But Don't Be Careless)

Problem Characteristics

Common Test Points

๐Ÿคฏ Why Do Many Fail Here?

Many candidates are trapped in the LeetCode mindset โ€” they see a "simple" problem and rush to code, resulting in:

  1. Missing edge cases: Didn't consider empty input or extreme values
  2. Over-engineering: Implemented based on assumptions rather than requirements
  3. Wasting time: Stuck 15 minutes on Q1, rest of the OA collapses

oavoservice Advice

Don't try to speed-run this problem. Walk through the examples completely first, then write code. Guarantee a first-pass success โ€” don't dig holes for later.


โœ… Q2 | Statistics + Left-Right Relationship (Classic Meta Style)

Core Problem

For each element x in the array:

  1. Count elements strictly greater than x in the entire array
  2. Further distinguish:
    • How many are to the left of x
    • How many are to the right of x
  3. Classify x based on the result:
    • Left > Right โ†’ Place in "Left Zone"
    • Right > Left โ†’ Place in "Right Zone"
    • Equal โ†’ Place in "Middle Zone"

Key Understanding

Solution Approach

def classify_elements(arr):
    n = len(arr)
    left_zone, middle_zone, right_zone = [], [], []
    
    for i, x in enumerate(arr):
        left_count = sum(1 for j in range(i) if arr[j] > x)
        right_count = sum(1 for j in range(i + 1, n) if arr[j] > x)
        
        if left_count > right_count:
            left_zone.append(x)
        elif right_count > left_count:
            right_zone.append(x)
        else:
            middle_zone.append(x)
    
    return left_zone, middle_zone, right_zone

Complexity Analysis

๐Ÿคฏ Hidden Pitfalls

This problem tests: Can you accurately translate rules into code?

Common failure points:

  1. Confusing "greater than" with "greater than or equal"
  2. Misunderstanding left/right definition (index-based, not value-based)
  3. Incorrect output ordering

โœ… Q3 | String Swap โ†’ Union-Find / Graph Connected Components (Must-Know!)

Core Problem

Given:

Each operation lets you pick an index i and swap s[arr[i]] with s[brr[i]].

Unlimited operations allowed, goal is to obtain the lexicographically smallest string.

๐Ÿ”ฅ Key Insight (The Dividing Line)

Unlimited swaps = Everything that can be swapped can be freely rearranged

This is crucial. As long as two positions can be connected through (arr[i], brr[i]) or through a chain of swaps, they belong to the same connected component, and characters in that component can be arbitrarily rearranged.

oavoservice Full-Score Solution

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * n
    
    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
    
    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        if px == py:
            return
        if self.rank[px] < self.rank[py]:
            px, py = py, px
        self.parent[py] = px
        if self.rank[px] == self.rank[py]:
            self.rank[px] += 1

def smallest_string(s, arr, brr):
    n = len(s)
    uf = UnionFind(n)
    
    # Build connectivity
    for a, b in zip(arr, brr):
        uf.union(a, b)
    
    # Group by connected component
    from collections import defaultdict
    groups = defaultdict(list)
    for i in range(n):
        groups[uf.find(i)].append(i)
    
    # Sort characters within each component and place back
    result = list(s)
    for indices in groups.values():
        chars = sorted(result[i] for i in indices)
        for i, idx in enumerate(sorted(indices)):
            result[idx] = chars[i]
    
    return ''.join(result)

๐Ÿคฏ Why Does This Problem Fail the Most People?

  1. Didn't think of Union-Find: Many try brute-force swapping, time explodes
  2. Didn't understand "unlimited swaps": Thought they needed to simulate the swap process
  3. Incorrect component handling: Forgot to sort indices as well

Engineering Judgment

This problem appears extremely frequently in Meta / Google / Amazon OAs. The essence is:

"Can you accurately abstract 'swappable' into 'rearrangeable'?"

This is a classic Engineering Judgment test, not just an algorithm problem.


โœ… Q4 | Multi-Task Prompt / Instruction Design (Massive Reading Load)

Problem Characteristics

Common Pitfalls

  1. Missing a sub-task
  2. Output format slightly off
  3. Not executing instructions in order

๐Ÿคฏ Why Is This Problem the Most "Annoying"?

This type of problem is classic Meta OA, testing:

oavoservice Advice

  1. Break down tasks first: Write out each step on paper
  2. Clarify mappings: Input โ†’ Output relationships
  3. Modularize code: One function does one thing
def process_instruction(instruction, state):
    """Process a single instruction"""
    if instruction['type'] == 'A':
        return handle_type_a(instruction, state)
    elif instruction['type'] == 'B':
        return handle_type_b(instruction, state)
    # ... and so on

def solve(instructions):
    state = initialize_state()
    results = []
    
    for inst in instructions:
        result = process_instruction(inst, state)
        results.append(result)
    
    return format_output(results)

๐ŸŽฏ oavoservice Summary: Where Is Meta OA Really Hard?

The real difficulty of Meta OA isn't any single problem, but:

Challenge Explanation
Pace Control 70 minutes for 4 problems = ~17 min each, no thinking buffer
Pattern Familiarity Whether you know Meta's common problem models
Pressure Resistance Can you "not write garbage code" under stress

If you've practiced Amazon / Google / TikTok / Meta problem types, you'll find the patterns are highly reusable; but if this is your first CodeSignal Meta OA with no prior exposure, time will likely destroy you.

OA success isn't about how many problems you've solved โ€” it's about whether you know the question logic.


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Facing Meta's fast-paced, high-volume, low-tolerance CodeSignal OA, you need more than just answers โ€” you need a professional technical team backing you.

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Service Description
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โœ… Remote & Invisible Physical isolation through remote control, no interference with exam environment
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โœ… No Pass, No Pay Full refund if not all test cases pass

Why Choose oavoservice?

Our assistance isn't "template-based answers" โ€” we provide targeted solutions based on each company's and platform's unique patterns:

Having someone help you maintain rhythm at critical moments is often the difference between full AC and system elimination.


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Don't let one Union-Find problem block your path to a $200k+ Offer.

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โ“ FAQ

Q: Is OA assistance safe? Will I get caught?

We use remote control + physical isolation for completely invisible operation. Your computer won't retain any logs or traces โ€” CodeSignal's monitoring system cannot detect any anomalies.

Q: What if I don't pass all test cases?

No pass, no pay. We target 100% test case pass rate. If we don't achieve full pass, you get a full refund.

Q: What platforms do you support?

Q: What if I'm in a time crunch?

We provide real-time response service. As long as you contact us before your OA starts, we can be on standby throughout your exam.


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