Preparing CodeSignal Interview Questions

Given a problem, immediately produce:

1. Pattern identification (one line)
2. Brute force approach + complexity
3. Optimized approach + complexity
4. Clean, interview-ready code
5. Edge cases + dry run on the example

Example invocation: "Give me a CodeSignal-style HashMap problem" or "Here's a problem, help me solve it optimally" or paste a problem statement directly.

Progress:

• Step 1: Classify the problem's core pattern
• Step 2: State brute force solution and its complexity
• Step 3: Derive the optimized solution (the "aha" insight)
• Step 4: Write clean code (language-agnostic unless specified; default to Python or JS)
• Step 5: Dry-run against the given example
• Step 6: List edge cases and confirm code handles them
• Step 7: Summarize time/space complexity and the general pattern name (for pattern-matching future problems)

Step 1: Classify the Pattern

Map the problem to one of these recurring CodeSignal/CodeVue buckets:

If unsure, ask: what grows/shrinks as I scan left to right? That usually reveals window vs. prefix-sum vs. stack.

Step 2–3: Brute Force → Optimized

Always state brute force first (even if O(n²) or O(n³)) — CodeSignal partial credit often rewards a working brute force under time pressure. Then identify what repeated computation the brute force wastes, and eliminate it with a HashMap, pointer, or prefix structure.

Step 4: Code Style for Assessments

• Use descriptive but short variable names (count_map, left, right, prefix_sum)
• No premature abstraction — CodeSignal rewards working code over "clean architecture"
• Handle input reading explicitly if it's a data-parsing problem (don't assume pre-parsed arrays)
• Return early on trivial/empty input

Step 5–6: Dry Run + Edge Cases

Always check:

• Empty input / single element
• All identical elements
• Negative numbers (if sums involved)
• k = 0 (if subarray-sum style)
• Duplicate keys / case sensitivity (if string/hashmap)
• Malformed lines (if file parsing — missing fields, extra whitespace, inconsistent delimiters)

Example 1: Subarray Sum Equals K

Input: nums = [1, 1, 1], k = 2 Output: 2

Pattern: Prefix Sum + HashMap.

Brute force: for every (i, j) pair sum the subarray → O(n²).

Insight: sum(i..j) = prefix[j] - prefix[i-1]. We want prefix[j] - prefix[i-1] = k, i.e. prefix[i-1] = prefix[j] - k. Track how many times each prefix sum has occurred.

Python
def subarray_sum(nums, k):
    count_map = {0: 1}  # empty prefix
    prefix_sum = 0
    result = 0
    for num in nums:
        prefix_sum += num
        result += count_map.get(prefix_sum - k, 0)
        count_map[prefix_sum] = count_map.get(prefix_sum, 0) + 1
    return result

Dry run: nums=[1,1,1], k=2

• prefix=1, need -1 → 0 matches, map={0:1,1:1}
• prefix=2, need 0 → 1 match (result=1), map={0:1,1:1,2:1}
• prefix=3, need 1 → 1 match (result=2), map={...,3:1} → Output 2 ✓

Edge cases: k=0 with zeros in array (e.g. [0,0,0]), all negatives, single-element equal to k. Complexity: O(n) time, O(n) space.

Example 2: Data Parsing — Max Average Temperature by City

Input (file, each line CITY,TEMP):

NYC,70
LA,80
NYC,74
LA,78
NYC,90

Output: LA (average 79.0 beats NYC's 78.0)

Pattern: File/Data Parsing + HashMap aggregation.

Python
def max_avg_city(lines):
    totals = {}
    counts = {}
    for line in lines:
        line = line.strip()
        if not line:
            continue
        city, temp = line.split(',')
        temp = float(temp)
        totals[city] = totals.get(city, 0) + temp
        counts[city] = counts.get(city, 0) + 1

    best_city, best_avg = None, float('-inf')
    for city in totals:
        avg = totals[city] / counts[city]
        if avg > best_avg:
            best_city, best_avg = city, avg
    return best_city, best_avg

Edge cases: blank lines, trailing newline, whitespace around fields, non-numeric temp (skip or raise), single city only, ties (define tie-break rule explicitly — usually first-seen wins). Complexity: O(n) time, O(u) space where u = unique cities.

Example 3: Sliding Window — Longest Substring Without Repeating Characters

Input: "abcabcbb" Output: 3 ("abc")

Pattern: Sliding Window + HashMap (char → last index).

Python
def length_of_longest_substring(s):
    last_seen = {}
    left = 0
    best = 0
    for right, ch in enumerate(s):
        if ch in last_seen and last_seen[ch] >= left:
            left = last_seen[ch] + 1
        last_seen[ch] = right
        best = max(best, right - left + 1)
    return best

Edge cases: empty string, all unique chars, all same char. Complexity: O(n) time, O(min(n, charset)) space.