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At companies like Spotify or Stripe, data processing tasks often require choosing the right algorithm for filtering, grouping, ranking, deduplication, and dependency analysis. Interviewers want to know whether you can identify the algorithmic building blocks behind these operations.
What algorithms do you consider essential for data processing? In your answer:
You do not need to list every algorithm in computer science. Focus on the practical core set that appears repeatedly in real data workflows, explain why each matters, and show that you can choose the right tool based on input size, ordering requirements, memory limits, and whether the task is batch or incremental.
Hash tables are essential for deduplication, frequency counting, grouping, and fast membership checks. They usually provide average O(1) insert and lookup, which makes them a default choice for one-pass processing over large datasets.
counts = {}
for x in items:
counts[x] = counts.get(x, 0) + 1
Sorting is fundamental when results must be ordered, when you need top records after global ranking, or when adjacent comparisons simplify the problem. Many interval, merge, and sweep-line problems become much easier after sorting.
items.sort(key=lambda x: x[1])
Heaps are useful when you only need the smallest or largest K elements instead of a full sort. They are especially effective in streaming or memory-constrained settings because they maintain only a bounded working set.
import heapq
heap = []
for x in nums:
heapq.heappush(heap, x)
Many data processing tasks are really dependency problems: job ordering, lineage traversal, connected components, and reachability. BFS, DFS, and topological sort are essential for reasoning about relationships between entities.
from collections import deque
q = deque([start])
When data is already sorted, two-pointer and merge-style scans can solve intersection, join-like matching, interval merging, and duplicate removal in linear time. These methods often reduce both complexity and memory usage.
i = j = 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
result.append(a[i]); i += 1; j += 1