How to Crack Any Coding Interview

A repeatable framework for tackling interview problems under pressure

The coding interview isn't just about knowing algorithms, it's about communicating your thinking. Here's the framework that top engineers use, every time.

The 5-step interview framework:
1. Clarify, Ask questions. What are the constraints? Can there be duplicates? Can input be empty?
2. Example, Draw a small example on paper. Input → expected output. Walk through it manually.
3. Brute force, Describe the simplest solution out loud. Don't code yet.
4. Optimise, Think about whether you can do better. "Can I avoid the nested loop?"
5. Code, Now write the code. Clean, with variable names that make sense.

Talk out loud the entire time. Interviewers want to hear your thinking. Silence is the enemy. Even saying "I'm not sure yet, let me think about this step..." is better than silence.

Classic interview pattern: Sliding Windowpython

# Problem: Find the maximum sum of a subarray of size k
def max_subarray_sum(arr, k):
    # Window: track the running sum of k elements
    window_sum = sum(arr[:k])   # sum of first window
    max_sum = window_sum

    for i in range(k, len(arr)):
        window_sum += arr[i]         # add new element
        window_sum -= arr[i - k]     # remove old element
        max_sum = max(max_sum, window_sum)

    return max_sum

print(max_subarray_sum([2, 1, 5, 1, 3, 2], 3))  # 9 (5+1+3)

Practice Exercises

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Exercise 1 of 2medium

⏱ 00:00

Valid Parentheses

Write a function is_valid(s) that checks if a string of brackets is valid.
Rules:- Open brackets must be closed in the correct order- "()" → True- "()[]{}"" → True- "(]" → False- "([)]" → False
Hint: Use a stack (list)! Push opening brackets, pop when you see a closing bracket.

solution.py

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Exercise 2 of 2hard

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Maximum Subarray (Kadane's Algorithm)

Write a function max_subarray(nums) that finds the contiguous subarray with the largest sum.
Example: max_subarray([-2,1,-3,4,-1,2,1,-5,4]) → 6 (subarray [4,-1,2,1])
This is a classic interview problem. The trick: keep a running sum, reset to 0 when it goes negative.

solution.py

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