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500+ curated questions across DSA, System Design, OOP, SQL, and more — with step-by-step explanations, optimised code, and complexity analysis.
def max_subarray(nums):
# Track current and global max
cur = res = nums[0]
for n in nums[1:]:
cur = max(n, cur + n)
res = max(res, cur)
return res
# [-2,1,-3,4,-1,2,1,-5,4] → 6Comprehensive coverage from fundamental DSA to distributed system design — aligned with what top companies actually ask in 2025.
Arrays, Trees, Graphs, DP, Sorting — the backbone of every technical interview.
180 questionsDesign URL shorteners, chat apps, payment systems, and real-world scalable architectures.
60 questionsSOLID principles, Singleton, Factory, Observer and 20+ patterns with real examples.
75 questionsJoins, Indexing, Normalization, Transactions, NoSQL vs SQL deep dives.
65 questionsProcesses, Threads, Deadlocks, TCP/IP, HTTP, DNS — essential for backend roles.
55 questionsSTAR-method answers for leadership, conflict, teamwork and situational questions.
65 questionsEvery question includes a concept explanation, optimised code, and full complexity breakdown.
Click any question to expand the full answer
n, check if target - n already exists in the map. If yes, we found our pair. This avoids the brute-force O(n²) nested loop.
def two_sum(nums, target):
seen = {}
for i, n in enumerate(nums):
diff = target - n
if diff in seen:
return [seen[diff], i]
seen[n] = il / r. Expand right; when a duplicate is found, shrink from the left until the window is valid. Track the max window size.
def length_of_longest(s):
seen, l, res = set(), 0, 0
for r in range(len(s)):
while s[r] in seen:
seen.remove(s[l]); l += 1
seen.add(s[r])
res = max(res, r - l + 1)
return resbeginWord, swapping each character with a–z and checking if the result exists in the word set. Return the level count when endWord is reached.
from collections import deque
def ladder_length(begin, end, words):
ws = set(words)
q = deque([(begin, 1)])
while q:
word, steps = q.popleft()
for i in range(len(word)):
for c in 'abcdefghijklmnopqrstuvwxyz':
nw = word[:i] + c + word[i+1:]
if nw == end: return steps + 1
if nw in ws:
ws.discard(nw)
q.append((nw, steps+1))
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