Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Bytedance SWE interview threw a variant of the classic string addition problem at me, basically the same as LeetCode 415 but with raw character arrays instead of strings. Pretty standard algorithmic round, nothing too wild.

Questions Asked (1)

Given two non-negative integers represented as character arrays of digits (most significant digit first), return their sum as a character array. You cannot convert to a built-in big integer type or use any arbitrary-precision library.

Algorithms & Data Structures

Author's notes

I recognized it as an add-strings variant pretty fast, which was both good and slightly dangerous because I jumped into code before fully thinking through the carry-on-final-digit edge case.

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AI HintsAI Generated

Suggested Approach

Use a two-pointer technique starting from the least significant digit (end of arrays) and simulate manual addition with a carry. Build the result in a StringBuilder by appending digits, then reverse it to get the correct order. Handle different lengths by continuing until both pointers are exhausted and carry is zero.

Pro tip: Clarify input assumptions upfront (e.g., no leading zeros except for zero itself) and mention edge cases like empty arrays or all zeros. Also, discuss how you would handle negative numbers if the problem were extended, showing foresight.

1. Clarify and Validate Inputs

Confirm that inputs are non-negative, contain only digits, and have no leading zeros (except for '0'). Ask about empty arrays and expected output format.

2. Initialize Pointers and Carry

Set pointers i and j to the last indices of the two arrays, and initialize carry to 0. Prepare a StringBuilder to store the result digits.

3. Iterate and Compute Sum

While i >= 0 or j >= 0 or carry > 0, extract digits (0 if pointer out of bounds), compute sum = digit1 + digit2 + carry, append sum % 10 to result, and update carry = sum / 10.

4. Reverse and Return Result

After the loop, reverse the StringBuilder to get the most significant digit first, convert to a character array, and return it.

5. Analyze Complexity and Edge Cases

State time complexity O(max(n, m)) and space complexity O(max(n, m)). Discuss edge cases like one array empty, carry propagation, and all zeros.

Key Points to Mention

Two-pointer approach from least significant digit
Carry propagation and handling of different lengths
Use of StringBuilder for efficient string building and reversal
Time and space complexity analysis
Edge cases: empty arrays, leading zeros, all zeros, maximum carry
Avoiding built-in big integer types as per constraints

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.