Software Engineer Interview Questions

Anjali M. - "Colleague moved to a different role, so I decided to fill in and did outshine in showing results within the group."See full answer

Poha - "Design the cache, set policy like LRU or other custom policies."See full answer

Mridul J. - "Interviewer wanted to learn more about architecture and or scalability. But I was not sure what the expected answer was"See full answer

An D. - "Understanding the Basics Choosing Learning Resources Practicing and Applying Knowledge Seeking Help and Staying Updated Leveraging Modern Tools"See full answer

Gaurav K. - "I think sliding window will work here and it is the most optimized approach to solve this question."See full answer

Jagan M. - "This system design question is very small compared to other questions like design instagram, twitter, google drive etc... Since the design involves less components the level of detail we have to go in them were deep. I had to explain how to deal with all the NFR for the distributed cache system. Whether it is a push model or a pull model. Hade to do BOE calculations for the database too."See full answer

Aman G. - "Was given 90 minutes with an exhaustive set of requirements to be implemented as a full-stack coding exercise. It was supposed to have a UX, a server and a database to store and retrieve data. The IDE was supposed to be self-setup before the interview. The panel asked questions on top of the implementation around decision making from a technical perspective"See full answer

Tiago R. - "function knapsack(weights, values, cap) { const indicesByValue = Object.keys(weights).map(weight => parseInt(weight)); indicesByValue.sort((a, b) => values[b]-values[a]); const steps = new Map(); function knapsackStep(cap, sack) { if (steps.has(sack)) { return steps.get(sack); } let maxOutput = 0; for (let index of indicesByValue) { if (!sack.has(index) && weights[index] <= cap) { maxOutput ="See full answer

Shashwat K. - "Binary Search on the array and after than compare the numbers at low and the high pointers whichever is closest is the answer. Because after the binary search low will be pointing to a number which is immediate greater than x and high will be pointing to a number which is immediate lesser than x. int low = 0; int high = n-1; while(low <= high){ int mid = (low + high) / 2; if(x == arr[mid]) return mid; //if x is already present then it will be the closest else if(x < arr[mid]) high"See full answer

Anonymous Jaguar - "Code reviews, automated linting, unit tests, integration tests, validation tests"See full answer

Tiago R. - "function findRotations(nums) { if (nums.length 0 && nums[mid] > nums[mid-1]) { left = mid; } else { right = mid; } } return rig"See full answer

Tes d H. - "Here is my first shot at it. Please excuse formatting. To find the maximum depth of the dependencies given a list of nodes, each having a unique string id and a list of subnodes it depends on, you can perform a depth-first search (DFS) to traverse the dependency graph. Here's how you can implement this: Represent the nodes and their dependencies using a dictionary. Perform a DFS on each node to find the maximum depth of the dependencies. Keep track of the maximum depth encountered dur"See full answer