If you are preparing for Snowflake 2026 OA, or have already made it halfway on HackerRank and started doubting yourself, then you have most likely realized one problem:
Solving many LeetCode problems doesn't guarantee you'll pass the OA.
Many candidates have a common misconception: "I've already done 300+ LeetCode problems, I can solve Medium questions in seconds, so Snowflake OA shouldn't be a problem."
However, the reality is different: As a quintessential Bar Raiser-level company, Snowflake's OA isn't meant to test whether you "know how to solve problems"—it's designed to evaluate whether you can write production-ready code while operating under significant time constraints.
What Snowflake OA Really Tests
Snowflake’s OA has several very stable and highly selective characteristics, and the difficulty of the Snowflake OA has surged recently:
- The problem description is concise.
- The algorithms involved are all commonly used and well-known algorithms.
- The problems are extremely detail-dense, with strict requirements on boundary handling and state management.
It doesn’t eliminate candidates by using obscure algorithms. Instead, it relies on hidden test cases. If you can pass all the visible test cases correctly, your chances of getting an interview are very high. These issues are rarely tested systematically during everyday practice, but in an OA environment, they are intentionally amplified and rigorously checked.
Real-world case: even with a solid foundation, it’s still possible to stumble.
A recent real-world case we supported illustrates this well. One of our candidates with a CMU background had a very solid algorithmic foundation, yet showed clear signs of nervousness at the very beginning of the Snowflake OA, starting from the first question.
In the initial solution, their instinctive approach was to use recursion for state transitions. Without timely intervention, this implementation would have been very likely to cause stack overflow or TLE once the test scale increased.
Q1: Word count restricted by consecutive vowels (DP)
Problem recap
You are given a word of length n, composed of letters from the alphabet. The constraint is that no position may contain more than m consecutive vowels. How many different valid words can be formed under this constraint?
The difficulty of this problem does not lie in the implementation, but in whether the abstraction is done correctly.
Many candidates initially approach this problem using permutations, case analysis, or the multiplication rule, which leads to overly complex logic and uncontrollable states, making edge-case errors very likely.
Problem-solving ideas
What Snowflake is trying to assess is not mathematical tricks, but the ability to control states. There is no need to care about which specific letters are used; you only need to focus on one thing: how many consecutive vowels appear at the end of the current string. This is a classic state-machine dynamic programming problem.
Status definition
dp[j]: the number of ways to form a string of the current length that ends with exactlyjconsecutive vowels
State transition
- Place a consonant
- All states are reset to
j = 0 - Multiply by 21 (the number of consonants in the English alphabet)
- All states are reset to
- Place a vowel
- Transitions are only allowed from
j - 1toj - Multiply by 5 (the number of vowels)
- Transitions are only allowed from
Common error: repeatedly computing sum(dp) in every iteration without optimization, forgetting to apply the modulo, and other issues that require extra care during implementation.
Reference implementation (Python)
def count_valid_words(n: int, m: int) -> int:
MOD = 10**9 + 7
dp = [0] * (m + 1)
dp[0] = 1
for _ in range(n):
new_dp = [0] * (m + 1)
total = sum(dp) % MOD
new_dp[0] = total * 21 % MOD
for j in range(1, m + 1):
new_dp[j] = dp[j - 1] * 5 % MOD
dp = new_dp
return sum(dp) % MOD
Additional note: In some variants of the problem, the value of n can be extremely large. If one fails to recognize that matrix fast exponentiation can be used for optimization, it is very easy to run into a time limit exceeded (TLE).
Q2: Count the number of pairs in a strictly increasing array that satisfy the product constraint (two pointers)
Problem recap
Based on the given rules, generate a strictly increasing array s. How many pairs (i, j) satisfy i < j and s[i] * s[j] <= a? The only truly valuable condition here is that s is strictly increasing. Once the array is monotonic, your first instinct should be: the two-pointer technique.
Problem-solving ideas
Use two pointers: let i move from left to right, and j shrink from right to left. When i is fixed:
- Try to keep j as large as possible.
- When
s[i] * s[j] > a, we decreasejby 1. - Once the condition is satisfied, all pairs from
(i, i + 1)to(i, j)are valid, so they can be accumulated in one step.
Main Examination Points
Whether you’re sensitive to the “strictly increasing” condition in the two-pointer one-pass interval counting mindset, and whether you can turn a brute-force O(n²) solution into O(n).
Q3: Maximum profit from weighted non-overlapping intervals (Hard)
Problem recap
Given n intervals, each with an associated value, choose a subset of non-overlapping intervals such that the total value is maximized. This is a high-frequency, comprehensive problem type commonly seen in Snowflake, Google, and Airbnb interviews.
Its value lies in simultaneously evaluating:
- arrange in order
- binary search
- Dynamic programming
- Understanding interval boundaries
Problem solving steps
This is a classic problem: Weighted Interval Scheduling. If you’ve practiced this type of problem before, this approach should come to mind almost immediately.
1）Sort:First, sort the intervals by their right endpoints in ascending order.
2）Define the DP:dp[k] represents: the maximum profit achievable when considering only the first k intervals.
3）State transition:For the k-th interval, there are two possible choices:
- Do not choose it:
dp[k-1] - choose
value[k] + dp[p]Here, p is the index of the last interval whose right endpoint is < the current interval’s left endpoint. You can find p by binary searching in the intervals after sorting.
4）What if the interval coordinates are very large?The interval endpoints going up to 1e9 make no difference, because the DP is done on interval indices, not on the actual coordinates, so there’s no need to allocate an array based on the coordinate values.
Snowflake OA is not well-suited for a “solo run.”
The overall TC for Snowflake NG / Intern typically falls within:$180k – $220k.
And the reality of the OA is:
- There is no partial credit.
- One bug means total failure.
- Failure often means waiting another year.
If you don’t want to leave an opportunity like Snowflake to chance, or risk seeing all your efforts wasted because of a single hidden test case, real-time OA support plus end-to-end guidance can significantly reduce unnecessary risks. Contact us to turn your OA into a controlled, predictable performance, rather than a gamble.
