The "Sum of Skills" problem appearing in Amazon SDE Intern OA 2026 has roughly 16% hit rate, making it a mid-to-high frequency problem. On the surface, it's an array summation; in practice, it tests prefix sum optimization + edge case handling. This article gives three solution paths, complete Python implementations, and the hidden requirements of Amazon's OA grading system.
Problem Statement
Amazon's hiring team needs to form a project team. Given an integer array skills of length n representing each candidate's skill score:
- A team is a contiguous subarray of k employees
- Team total skill = sum of subarray elements
- Return the list of skill sums for every contiguous subarray of length k, ordered by starting index
Constraints
1 ≤ n ≤ 10^51 ≤ k ≤ n-10^4 ≤ skills[i] ≤ 10^4
Example
Input: skills = [3, -1, 4, 1, 5, 9, 2, 6], k = 3
Output: [6, 4, 10, 15, 16, 17]
Walkthrough:
- [3, -1, 4] = 6
- [-1, 4, 1] = 4
- [4, 1, 5] = 10
- [1, 5, 9] = 15
- [5, 9, 2] = 16
- [9, 2, 6] = 17
Three Approaches Compared
| Approach | Time | Space | Recommendation |
|---|---|---|---|
| Brute force | O(n × k) | O(1) | ⭐ |
| Prefix sum | O(n) | O(n) | ⭐⭐⭐⭐ |
| Sliding window | O(n) | O(1) | ⭐⭐⭐⭐⭐ |
Approach 1: Brute Force (Baseline)
def team_skills_brute(skills, k):
n = len(skills)
result = []
for start in range(n - k + 1):
window_sum = sum(skills[start:start + k])
result.append(window_sum)
return result
Time: O(n × k); times out at n=10^5, k=10^4.
Approach 2: Prefix Sum (Best for Variants)
Precompute prefix[i] = skills[0] + ... + skills[i-1], then any range [i, j) sum = prefix[j] - prefix[i].
def team_skills_prefix(skills, k):
n = len(skills)
prefix = [0] * (n + 1)
for i in range(n):
prefix[i + 1] = prefix[i] + skills[i]
result = []
for start in range(n - k + 1):
result.append(prefix[start + k] - prefix[start])
return result
Time: O(n); Space: O(n).
Prefix sum is the only O(1)-per-query approach when variant problems require arbitrary range lookups.
Approach 3: Sliding Window (OA's Preferred)
Fixed window size k. Each step: subtract the leftmost element, add the new rightmost.
def team_skills(skills, k):
n = len(skills)
if k > n:
return []
window_sum = sum(skills[:k])
result = [window_sum]
for i in range(k, n):
window_sum += skills[i] - skills[i - k]
result.append(window_sum)
return result
Time: O(n); Space: O(1) excluding output.
Why the Grading System Prefers Sliding Window
Amazon's hidden test cases:
- Large input (n = 10^5) for speed → eliminates brute force
- Extreme values (10^4 max) for integer overflow—Python is fine, C++/Java need long long
- k = n (entire array) → single output value
- k = 1 (each element its own team) → minimum window
Sliding window passes all of these.
Complete Solution Template
def get_skill_sums(skills, k):
n = len(skills)
if n == 0 or k == 0 or k > n:
return []
window_sum = sum(skills[:k])
result = [window_sum]
for i in range(k, n):
window_sum += skills[i] - skills[i - k]
result.append(window_sum)
return result
def test():
assert get_skill_sums([3, -1, 4, 1, 5, 9, 2, 6], 3) == [6, 4, 10, 15, 16, 17]
assert get_skill_sums([1, 2, 3], 3) == [6]
assert get_skill_sums([5], 1) == [5]
assert get_skill_sums([], 1) == []
assert get_skill_sums([1, 2, 3], 5) == []
assert get_skill_sums([-1, -2, -3, -4], 2) == [-3, -5, -7]
print("All tests passed.")
test()
Common Variants in OA
Variant 1: Maximum team skill
def max_team_skill(skills, k):
sums = get_skill_sums(skills, k)
return max(sums) if sums else 0
Variant 2: Starting index of the maximum team
def max_team_start_index(skills, k):
n = len(skills)
if k > n:
return -1
window_sum = sum(skills[:k])
best_sum = window_sum
best_start = 0
for i in range(k, n):
window_sum += skills[i] - skills[i - k]
if window_sum > best_sum:
best_sum = window_sum
best_start = i - k + 1
return best_start
Variant 3: Count teams whose sum ≥ target
def count_teams_above_target(skills, k, target):
n = len(skills)
if k > n:
return 0
window_sum = sum(skills[:k])
count = 1 if window_sum >= target else 0
for i in range(k, n):
window_sum += skills[i] - skills[i - k]
if window_sum >= target:
count += 1
return count
OA Strategy
- Write sliding window first—save time for the second problem
- Domain-style variable names:
team_sizeis preferred overk - Add boundary asserts: Amazon's hidden tests cover many corners; write 4-5 asserts proactively
- Skip explanatory comments—the grader doesn't read them, save time
FAQ
How frequently does Sum of Skills appear in Amazon OA?
In 2026 Q1-Q2 SDE Intern OA, hit rate is roughly 16% (per 1point3acres aggregations). It joins Robot Coordination and Warehouse Restock as one of the "three usual suspects" of Amazon Intern OA.
Is the time limit tight?
Amazon Intern OA is typically 70 min for 2 problems. Sum of Skills is usually problem 1—finish in 15-20 minutes, leaving ~50 min for a tougher second problem (often DP or graphs).
Does the OA throttle Python?
Not significantly. Amazon's OA gives Python 3-5× more time than C++. Sliding window over n=10^5 finishes in 100ms in Python—safe.
What to watch for in C++?
sum accumulation can overflow int. With n=10^5 and max value 10^4, sum reaches 10^9—just barely fits but use long long for safety.
When to choose sliding window vs prefix sum?
Fixed window → sliding window (O(1) space). Variable-length / multi-query → prefix sum (O(1) per query after O(n) preprocessing). The OA accepts both, but sliding window is faster to write.
Preparing for Amazon Intern OA?
The Amazon SDE Intern OA question bank is relatively stable, but new variants appear every year. oavoservice maintains an Amazon Intern OA problem bank with 30+ frequent problems including Sum of Skills, Robot Coordination, and Warehouse Restock—each with full solutions and variant drills.
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