92 questions from 11 interviews · updated from reports 2019-2024
About
PhonePe is an Indian digital payments and financial services company that offers UPI payments, money transfers, bill payments, recharges, and related consumer finance services. In India, it hires technical roles such as Software Engineer, SDE-2, and software engineering interns or full-time graduates.
The roles that come up most are Software Engineer, SDE-2 and Software Engineer (Intern / Full-Time). This covers 11 candidate interviews reported from 2019 to 2024. Most sat it at entry level (8 of 10 that recorded a level), with 1 internship interviews alongside. Among the 7 that recorded either route, arrivals split between campus drives (7, 100%). Most questions fall under DSA and CS fundamentals.
Ans. Stack memory is used for function calls and local variables, while heap memory is used for dynamically allocated objects that may live beyond a single function call. Stack allocation is fast and automatically cleaned up when a function returns. Heap allocation is more flexible, but slower and needs garbage collection or manual deallocation.
Ans. Use a stack to decode the string by scanning left to right, building the current number and current substring. When you see [, push the previous substring and repeat count. When you see ], pop them and expand. This handles nesting correctly. Time complexity is O(n + output size), space is O(n).
Ans. Merge sort guarantees O(n log n) time and is stable, while quick sort is usually faster in practice but has O(n²) worst-case time. Merge sort needs extra memory for merging, typically O(n). Quick sort sorts mostly in place, using O(log n) stack space on average with good pivot selection.
Ans. The Singleton pattern ensures a class has exactly one instance and provides a global access point to it. It is usually implemented with a private constructor and a static method or property returning the instance. The key detail is thread safety, especially if the instance is created lazily in a multi-threaded program.
Ans. A trie is a tree data structure used to store strings by sharing common prefixes. Each node represents a character, and paths from the root form words or keys. Tries support fast prefix search, insertion, and lookup in O(L) time, where L is string length. Common uses include autocomplete, spell checkers, dictionaries, and IP routing.
Ans. An operating system manages computer hardware and provides services for programs. Its fundamentals include process and thread management, memory management, file systems, device and I/O handling, scheduling, security, and networking. The key idea is abstraction: it hides hardware details and safely shares limited resources between many programs and users.
Ans. Hashing is a one-way transformation used to produce a fixed-size digest, while encryption is a reversible transformation used to protect data confidentiality. Encrypted data can be decrypted with the correct key. A hash cannot practically be reversed, so it is used for integrity checks, password storage, and fast lookups.
Ans. Hybrid sorting combines two or more sorting algorithms to get better practical performance than using one alone. It is used in standard library sorts and real systems, for example introsort combines quicksort, heapsort and insertion sort. The key idea is switching algorithm based on input size, recursion depth, or data pattern.
Ans. Use a min-heap to repeatedly take the smallest current element from the k arrays and append it to the result. Store each heap entry as the value plus its array index and position, then push the next element from that array. For N total elements, time is O(N log k) and space is O(k).
Ans. The roots are the tree centroids, found by repeatedly removing all current leaves until only one or two nodes remain. Build an adjacency list and degree array, push all degree one nodes into a queue, then trim layer by layer. The remaining nodes are the minimum height roots. This takes O(n) time and O(n) space.
Ans. HTTP methods describe the action a client wants to perform, while response status codes describe the result of that request. Common methods are GET to read, POST to create or submit, PUT or PATCH to update, and DELETE to remove. Status codes are grouped as 2xx success, 3xx redirect, 4xx client error, and 5xx server error.
Ans. Use backtracking to split the string into four segments, each of length one to three, and keep only segments between 0 and 255 with no leading zero unless the segment is exactly "0". Store the current path in a small list. There are at most 3^4 splits, so time is effectively O(1), with O(1) extra space.
Ans. Use tables for restaurants, menus, menu_sections, menu_items, item_prices, allergens, item_allergens, modifiers, modifier_options, and orders if ordering is in scope. Restaurants own menus, menus contain sections, sections contain items, and items have prices, availability, and dietary metadata. The key detail is versioning menus or prices so past orders keep correct historical values.
Ans. I use the UNIX command line to inspect files, manage processes, control permissions, search text, and automate routine tasks with shell scripts. The most important concepts are the filesystem hierarchy, pipes and redirection, process management with ps, top, kill, permissions with chmod and chown, and debugging using logs and standard streams.
Ans. Use SELECT to choose columns, FROM to name tables, WHERE to filter rows, JOIN to combine related tables, GROUP BY with aggregate functions to summarise, HAVING to filter groups, and ORDER BY to sort results. The key detail is matching join keys correctly so results are accurate and duplicates are controlled.
Ans. I would solve each problem by confirming inputs, outputs and constraints, then deriving a correct baseline before optimising. The key detail is choosing the right data structure, such as a hash map for fast lookup, a heap for priorities, or a stack for nested state, and stating the time and space complexity clearly.
Ans. Use distinct scores, sort them in descending order, then take one row after skipping the first nine scores. In SQL, this is typically done with DISTINCT, ORDER BY score DESC, LIMIT 1 and OFFSET 9. This returns the 10th highest unique score. Sorting dominates the cost, usually O(n log n).
Ans. Use a 2D prefix sum matrix to answer each submatrix sum query in constant time after preprocessing. Store prefix[i][j] as the sum from the top left to cell i, j. For query rectangle r1, c1 to r2, c2, use inclusion exclusion. Preprocessing is O(rows * cols), each query is O(1).
Ans. Indexing in databases is creating an additional data structure, commonly a B-tree or hash index, that lets the database find rows without scanning the whole table. It improves read performance by narrowing searches, joins, sorting, and filtering to relevant records. The trade-off is extra storage and slower writes because indexes must be maintained.
Ans. Use a monotonic increasing stack and count each element’s contribution as the minimum of subarrays. For each index, find the distance to the previous strictly smaller element and the next smaller or equal element. Its contribution is value times left distance times right distance. Sum all contributions. This runs in O(n) time.
Ans. Race the rats in four groups of four, using the four cakes as the four lanes or targets. Keep only the winner from each group. Then race those four winners against each other. The winner of that final race is the fastest rat overall, because every other rat has already lost to one of the finalists.
Ans. Use backtracking to build strings one character at a time, adding an opening bracket if fewer than n have been used, and a closing bracket only if it would not exceed the number of openings. Store the current string and counts in recursion. Time is proportional to the Catalan number times n.
Ans. PhonePe should use layered, real-time fraud controls: device fingerprinting, velocity limits, behavioural anomaly detection, beneficiary risk scoring, graph analysis of linked accounts, and ML models trained on confirmed fraud. The most important detail is to stop risky payments before authorisation using step-up verification, transaction limits, cooling periods, and manual review for high-risk cases.
Ans. Use a single lift controller with current floor, direction, door state and two ordered request sets: floors above and floors below. Serve requests in the current direction first, stopping at each requested floor, then reverse when none remain. A 1000-bit bitmap per direction is enough, giving fast lookup and small memory.
Ans. Use backtracking to build each string, adding an opening bracket if fewer than n have been used, and adding a closing bracket only if it would not exceed the number of openings. Store the current sequence in a list or string builder. The time complexity is proportional to the nth Catalan number.
Ans. Virtual functions are member functions resolved at runtime through a base pointer or reference, enabling polymorphism. A virtual class usually means a virtual base class, used in multiple inheritance to ensure only one shared base subobject exists. They are used in interfaces, inheritance hierarchies, callbacks, and avoiding diamond inheritance duplication.
Ans. SQL databases are best for structured data, strong consistency and complex queries, while NoSQL databases are best for flexible schemas, high scale and varied data models. I would choose SQL for transactions, reporting and relational data, such as banking. I would choose NoSQL for large, fast-changing or distributed data, such as logs, feeds or document storage.
Ans. Use binary search on the speed and return the smallest speed whose simulated journey time is at most the given hour. For a candidate speed, add ceil(distance / speed) for every leg except the last, because departures are hourly, then add the exact final leg time. If none works, return -1. Complexity is O(n log M).
Ans. I solve advanced DSA problems by first identifying the pattern, then choosing the right data structure and proving the complexity. For example, graph problems often need BFS, DFS, Dijkstra, or union find, while range queries may need segment trees or Fenwick trees. I explain invariants, edge cases, and time and space complexity.
Ans. Sort the array, then use a sliding window to find the largest group that can be made equal to the rightmost value using at most k increments. Maintain the window sum and shrink from the left while nums[right] times window size minus sum exceeds k. The answer is the maximum window size. Time complexity is O(n log n).
Ans. I manage branching by keeping main stable, creating short-lived feature branches, and regularly pulling or rebasing from the target branch. For merge conflicts, I inspect the conflicting files, understand both changes, keep the correct combined version, run tests, and commit the resolution. Clear commits and early communication reduce painful conflicts.
Ans. A deadlock is a state where two or more processes are permanently blocked because each is waiting for a resource held by another. The necessary conditions are mutual exclusion, hold and wait, no preemption, and circular wait. For example, process A holds a file lock and waits for a database lock, while process B holds the database lock and waits for the file lock.
Ans. Give each cake to a group of three rats for the same short time. The cake with the most eaten must be the group containing the faster rat, because all other rats eat at the same speed. Then give the three suspects equal pieces of remaining cake and watch which piece is eaten first. That rat is faster.
Ans. A production-ready Tiny URL service would take about 8 to 12 weeks for a small team. Use an API to create links, generate a unique base62 code from an ID or random token, store mappings in a durable database, cache hot redirects, and add expiry, rate limits, monitoring, analytics, abuse controls, backups, and multi-zone deployment.
Ans. It is possible to finish all courses if the prerequisite graph has no directed cycle. Model courses as nodes and prerequisites as edges, then use topological sorting with an adjacency list and indegree array. Repeatedly process zero-indegree courses. If you process numCourses courses, return true; otherwise false. Time and space are O(V + E).
Ans. Find the path from the root to the target node, then return the element k positions before the target in that path, or -1 if it does not exist. A DFS with a list or recursion stack is enough. The time complexity is O(n), and the extra space is O(h) to O(n).
Ans. Use a synchronisation mechanism so only one thread or process updates shared state at a time, or route all writes through a thread-safe queue. The key detail is to make the shared operation atomic, using locks, mutexes, semaphores, transactions, or message passing, and to keep the critical section small.
Ans. I would first clarify whether z should be supplied by the backend, derived by the frontend, or optional. If z is required business data, update the API contract and backend response. Until then, the frontend should handle the missing field safely with validation, defaults, or an error state.
Ans. Check every pair of points, compute the slope, and keep the pair with the largest value, then draw or return the line segment between them. If two points have the same x-coordinate, the slope is vertical and effectively infinite, so that pair is the maximum. This uses the input array and takes O(n²) time.
Ans. Use BFS from the target, treating the tree as an undirected graph. First traverse the tree to store each node’s parent in a map. Then run BFS using left child, right child and parent as neighbours, marking visited nodes. Each BFS level is one unit of time. Time complexity is O(n), space is O(n).
Ans. Scan each element as the middle index and query the best valid value on its left and right. Coordinate compress values, use segment trees or Fenwick trees storing minimum and maximum candidates for indices before and after j. For each j, test the four left/right extremes, keep the largest product and indices. Time is O(n log n).
Ans. Growth followed by decline usually means early product market fit weakened, acquisition quality dropped, competition caught up, retention was poor, or operations failed to scale. I would diagnose cohort retention, unit economics, churn reasons, funnel conversion, and market changes, then prioritise fixes such as product improvements, pricing changes, stronger onboarding, cost control, or repositioning.
Ans. Because that location enables unusually many very short rides, drivers can hit X rides cheaply and quickly, then collect benefits disproportionate to real work. Check ride counts, trip distances, fares, repeat rider-driver pairs, and timing by area. If one spot has clustered short trips around the threshold, the incentive is being gamed there.
Ans. Use aggregate functions to summarise rows, GROUP BY to summarise per category, ORDER BY to sort results, and window functions to rank or number rows without collapsing them. For example, calculate total sales per customer, sort by highest total, and use ROW_NUMBER or ROWNUM to return the top N customers.
Ans. Start with core entities, their responsibilities, and relationships, then define interfaces before implementation details. Use simple classes, clear ownership, and patterns only when they reduce coupling, such as strategy for variable behaviour or factory for creation. Discuss trade-offs around extensibility, complexity, concurrency, storage, and failure handling. State time and space costs for key operations.
Ans. The backend treats the return as another payment or refund, updating ledgers by debiting the sender and crediting the recipient. The key detail is that both transfers remain linked in audit, risk and AML systems, so the pattern can still be flagged even if the user’s visible balance looks restored.
Ans. Yes: sort the array in ascending order, then check each consecutive triple to see if the two smaller values sum to more than the largest. For sorted values, this single inequality is enough. Ignore non-positive values if side lengths must be valid. Time complexity is O(n log n).
Ans. Use the relations as a directed graph and check whether they imply a unique topological order. Convert each comparison to one direction, such as taller to shorter. The information is sufficient only if topological sorting always has exactly one available zero indegree node at each step, and no cycle exists. Time complexity is O(V + E).
Ans. Convert the numbers to strings, build a frequency map, then try every split of X’s string into a non-empty prefix and suffix. For each split, add frequency[prefix] multiplied by frequency[suffix], because pairs are ordered. If i and j must be distinct, subtract frequency[prefix] when prefix equals suffix. Time is linear in input digits.
Ans. Sort the city pairs by their north bank position, then find the longest increasing subsequence of their south bank positions. That LIS length is the maximum number of non-crossing bridges. Sorting fixes one side’s order, and the increasing subsequence keeps the other side in the same order. Time complexity is O(n log n).
Ans. Use classes ParkingLot, Level, ParkingSpot, Vehicle, Ticket, Gate, DisplayBoard and ParkingAssignmentService. ParkingLot has Levels, Level has Spots, Ticket links Vehicle to Spot. Keep available spots in priority queues by vehicle type, ordered by nearest distance to entry gate. On entry, pop the best spot, issue ticket, update displays and guide by level, row and spot signs.
Ans. Use a hash map from element to frequency and another hash map from frequency to a set of elements, while maintaining current minFreq and maxFreq. On add or remove, move the element between frequency buckets and delete empty buckets. This gives O(1) average add, remove, find minimum frequency, and find maximum frequency.
Ans. Sort the array, then try every split where smaller elements get +k and larger elements get -k, keeping the minimum possible range. For split i, the new minimum is min(a[0] + k, a[i+1] - k) and maximum is max(a[i] + k, a[n-1] - k). Time complexity is O(n log n).
Ans. Compute all possible sauce subset costs with a bitset, then for each bread cost b, look for the sauce sum closest to B minus b and update the best total. The key detail is that each sauce is used once by shifting the bitset once per sauce. Time is O(nS + mS) conceptually, where S is total sauce cost.
Ans. Use a max heap of empty intervals, keyed by the best distance achievable in that interval. For a middle interval choose its midpoint; for an edge interval choose seat 1 or N. After allocating, split the interval and push remaining parts back. Initial sorting costs O(K log K); each query costs O(log K).
Ans. Model the game with Board, Player, Dice, GameEngine and a SquareEffect interface implemented by Snake, Ladder or future obstacles. Store effects in a map from start square to effect, and players in a queue. Each turn rolls dice, moves, applies any effect, checks win. Turn time is constant, setup is linear.
Ans. Use a two-pass greedy approach to assign the minimum candies. Give every child one candy, scan left to right and increase candies when a rating is higher than the left neighbour, then scan right to left and fix higher-than-right cases using the maximum. Sum the array. Time is O(N), space is O(N).
Ans. A strong answer should pick a finance setting where technology clearly improved speed, control, access or decision making, such as payments, fraud detection, trading, reporting or customer service. Emphasise risk management, data quality, regulation, cyber security and customer trust. Interviewers listen for balanced thinking, not blind enthusiasm, and awareness that technology supports judgement.
Ans. Pick a practical problem where you took responsibility, understood constraints, involved others, and produced a clear result. It could be improving a team process, resolving a customer issue, organising an event, or handling a logistical challenge. Emphasise your judgement, communication, initiative, and measurable impact. Interviewers listen for ownership and transferable problem-solving skills.
Ans. Pick a real, low-drama work disagreement where the stakes mattered and you behaved constructively. Emphasise listening, clarifying facts, staying calm, and finding a practical compromise or agreed decision. Interviewers listen for self-awareness, respect, ownership, and the ability to disagree without damaging relationships or slowing delivery.
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Candidate interviews most often cover DSA (46%) and CS fundamentals (37%).
Candidate interviews show an average of 3.3 rounds per experience, with a typical sequence of Online test → Technical → Technical → Managerial. Individual interview paths can vary.
Among questions with a recorded difficulty, the mix is easy 21%, medium 64%, hard 15%.
