Snowflake·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Snowflake coding interview with a string manipulation problem that looks deceptively simple but has a few layers to it. The palindrome-via-swaps angle threw me a bit.

Questions Asked (1)

Given a string containing only two distinct characters, determine if it can be rearranged into a palindrome using only adjacent swaps. If yes, return the minimum number of adjacent swaps needed; otherwise return -1.

Algorithms & Data Structures

Author's notes

I got the feasibility check pretty fast (odd-length strings can have one character with an odd count, even-length need both even) but the minimum swap part is where I started fumbling.

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AI HintsAI Generated

Suggested Approach

First, check if a palindrome is possible by verifying that at most one character has an odd count. Then, use a two-pointer greedy strategy: fix the leftmost character by finding its matching counterpart from the right and swapping it into place, counting swaps; if no match exists (the odd character), swap it one step toward the center and continue. This yields the minimum number of adjacent swaps.

Pro tip: Emphasize that the greedy approach is optimal for two distinct characters because each swap fixes at least one character's final position, and the odd-count character (if any) must end up in the middle. Also, mention that the problem can be solved in O(n^2) time with O(1) extra space, which is acceptable for typical constraints.

1. Check feasibility

Count the frequency of each character. If more than one character has an odd count, return -1 immediately.

2. Initialize pointers and swap counter

Set left pointer to 0, right pointer to n-1, and swaps to 0. Convert the string to a mutable list for easy swapping.

3. Greedy matching from ends

While left < right, find the rightmost occurrence of the character at left within the range [left, right]. If found, swap it to the right position, incrementing swaps by the distance. If not found (this is the odd character), swap it one step toward the center and continue without moving the pointers.

4. Return the total swaps

After the loop, return the accumulated swaps as the minimum number of adjacent swaps needed.

Key Points to Mention

Palindrome feasibility condition: at most one character with odd frequency.
Greedy two-pointer approach: fix characters from the outside in.
Handling the odd-frequency character by moving it toward the center.
Time complexity: O(n^2) due to swaps, space complexity: O(1) if using a mutable array.
Proof of optimality: each swap reduces the number of mismatched pairs by at least one.
Edge cases: empty string, single character, already palindrome, impossible cases.

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