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

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

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

Zira C. - "Writing quality code is a major part of my coding philosophy, and I take multiple steps throughout the implementation process to ensure I'm producing quality work. Firstly, I focus on readability while coding, and test small parts as I go. This can include test-driven development, reading code aloud, or pair programming with a coworker. The next step would be automated systems to ensure quality. I will run a linter, and an automated test suite. Lastly, once my PR is created a fellow engineer wil"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

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

Roger R. - "Get data from database CDC to Kafka; Write from Kafka to a Bucket; Send the batch file to Visa from time to time;"See full answer

Jack F. - "conduct direct user research through methods such as interviews, surveys, and focus groups to gather qualitative insights into user preferences, pain points, and behaviors. Additionally, I analyze quantitative data from sources like user analytics, feedback metrics, and market research to identify trends and patterns. Collaborating closely with stakeholders, including customers, internal teams, and subject matter experts, further enriches the requirements gathering process by incorporating diver"See full answer

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"See full answer

Tiago R. - "function visitChildren(node) { let leftSubtreeHeight = 0; let rightSubtreeHeight = 0; let isChildrenBalanced = true; if (node.left) { const { isBalanced, height } = visitChildren(node.left); isChildrenBalanced = isChildrenBalanced && isBalanced; leftSubtreeHeight += height + 1; } if (isChildrenBalanced && node.right) { const { isBalanced, height } = visitChildren(node.right); isChildrenBalanced = isChildrenBalanced && isBalan"See full answer