🚨 [CodeSignal OA Questions Leaked] 4 High-Frequency Problems Explained: Strings, Cipher, HashMap, Teleport Labyrinth!

Attention candidates preparing for TikTok, Google, Amazon and other major tech company OAs! This CodeSignal question set covers 4 classic problem types, from string processing to graph simulation, with increasing difficulty.

We consistently provide professional online assessment services for major tech companies like TikTok, Google, and Amazon, guaranteeing perfect scores. Feel free to contact us if you're interested.

✅ Question 1: Count Substrings With Vowels

Problem Description

You are developing an automatic moderation system for a chat application that detects certain linguistic patterns.

Given a string chatMessage, count the number of substrings of exactly length 3 where at least one character is a vowel ("a", "e", "i", "o", "u", or their uppercase counterparts).

Constraints: 1 ≤ chatMessage.length ≤ 1000

Examples

Input: chatMessage = "CodeSignal"
Output: 8

Explanation: Substrings of length 3: "Cod", "ode", "deS", "eSi", "Sig", "ign", "gna", "nal"
All contain at least one vowel → 8 substrings

Input: chatMessage = "StrEngThen"
Output: 5

Explanation: Substrings with vowels: "trE", "rEn", "Eng", "The", "hen" → 5 substrings

oavoservice Perfect Solution

def solution(chatMessage):
    vowels = set('aeiouAEIOU')
    count = 0
    for i in range(len(chatMessage) - 2):
        substring = chatMessage[i:i+3]
        if any(c in vowels for c in substring):
            count += 1
    return count

Time Complexity: O(n), single pass solution

✅ Question 2: Consonant Cipher Shift

Problem Description

You're implementing a simple substitution cipher that only affects consonant characters in a memo.

This cipher works by shifting every k-th consonant to the following consonant in the alphabet, while leaving vowels and other characters unchanged.

Rules:

Only consonant characters are affected: 'b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'y', 'z'
When shifting from 'z', it wraps around to 'b'
The case of each character must be preserved

Example

Input: memo = "CodeSignal", k = 3
Output: "CodeTignam"

Explanation:
Consonants: 'C', 'd', 'S', 'g', 'n', 'l'
3rd consonant 'S' → 'T'
6th consonant 'l' → 'm'

oavoservice Perfect Solution

def solution(memo, k):
    consonants = 'bcdfghjklmnpqrstvwxyz'
    consonant_set = set(consonants + consonants.upper())
    
    def next_consonant(c):
        lower = c.lower()
        idx = consonants.index(lower)
        next_c = consonants[(idx + 1) % len(consonants)]
        return next_c.upper() if c.isupper() else next_c
    
    result = list(memo)
    consonant_count = 0
    
    for i, c in enumerate(memo):
        if c in consonant_set:
            consonant_count += 1
            if consonant_count % k == 0:
                result[i] = next_consonant(c)
    
    return ''.join(result)

Key Point: Consonant counter + modulo operation to identify every k-th consonant

✅ Question 3: Molecular Bond Pairing

Problem Description

You are a chemist working in a laboratory that studies molecular compounds. You have two arrays representing the atomic weights of elements in two different compounds:

x represents the primary elements
y represents the secondary elements

Two compounds with the same balance factor (difference between primary and secondary atomic weights) can form stable bonds.

Task: Count the number of unique molecular pairings (i, j) where i < j and x[i] - y[i] == x[j] - y[j].

Example

Input: x = [2, -2, 5, 3], y = [1, 5, -1, 1]
Output: 6

Explanation:
Compute all x[i] - y[i]: → 1, -7, 6, 2
Count pairs with same difference value

oavoservice Perfect Solution

from collections import defaultdict

def solution(x, y):
    diff_count = defaultdict(int)
    count = 0
    
    for i in range(len(x)):
        diff = x[i] - y[i]
        count += diff_count[diff]
        diff_count[diff] += 1
    
    return count

Time Complexity: O(n), single pass with HashMap

Core Idea: Classic Two Sum variant using HashMap to store difference frequencies

✅ Question 4: Teleport Labyrinth

Problem Description

Imagine you're exploring a mysterious labyrinth in the shape of a rectangular matrix, which contains obstacles and teleports. Your goal is to reach the lower-right corner by only moving to the right starting from the upper-left corner.

Input:

Integers n and m: dimensions of the labyrinth
obstacles: cells that cannot be passed
teleports: list of [start_row, start_col, end_row, end_col] for one-way teleportation

Movement Rules:

You always move to the right (from [row, col] to [row, col + 1])
If you reach a cell that is a teleport start, you are moved instantly to its end cell
Obstacles block movement completely
You stop if you move outside bounds or hit an obstacle

Output:

Return the number of cells traversed, including start and teleport endpoints
Return -1 if blocked by obstacle or out of bounds
Return -2 if stuck in infinite teleport loop
Return -3 if unable to reach goal but not due to obstacle or teleport loop

Example

Input: n = 3, m = 3, obstacles = [[2, 1]], teleports = [[0, 1, 2, 0]]

Grid:
[0,0]  [0,1]*T  [0,2]
[1,0]  [1,1]    [1,2]
[2,0]  [2,1]X   [2,2]

You start at [0,0], move to [0,1] → teleport to [2,0]
Cannot move right to [2,1] because it's an obstacle
→ Return -1

oavoservice Perfect Solution

def solution(n, m, obstacles, teleports):
    obstacle_set = set(tuple(o) for o in obstacles)
    teleport_map = {(t[0], t[1]): (t[2], t[3]) for t in teleports}
    
    row, col = 0, 0
    visited = set()
    cells_traversed = 0
    
    while True:
        # Check for teleport loop
        state = (row, col)
        if state in visited:
            return -2
        visited.add(state)
        
        # Check for obstacle
        if (row, col) in obstacle_set:
            return -1
        
        cells_traversed += 1
        
        # Check if reached destination
        if row == n - 1 and col == m - 1:
            return cells_traversed
        
        # Check for teleport
        if (row, col) in teleport_map:
            row, col = teleport_map[(row, col)]
            continue
        
        # Move right
        col += 1
        
        # Check bounds
        if col >= m:
            return -3
        
        # Check obstacle
        if (row, col) in obstacle_set:
            return -1
    
    return -3

Core Focus: State machine simulation + cycle detection + multiple termination conditions

💡 Why Do These Problems Filter Out So Many Candidates?

These 4 problems may seem independent, but they test the core competencies in OA assessments:

String Processing Basics - Q1 tests sliding window thinking
Simulation & State Tracking - Q2 tests counters and character mapping
HashMap Optimization - Q3 is a classic Two Sum variant
Graph Theory & State Machines - Q4 tests complex condition simulation

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