Top Software Engineer Interview Questions

Praveena P. - "I have been with Cisco, I am trying to move into a technical project manager role so that I will not loose the touch of latest technologies and also that aligns my passion."See full answer

GalacticInterviewer - "Approach 1: Use sorting and return the kth largest element from the sorted list. Time complexity: O(nlogn) Approach 2: Use max heap and then select the kth largest element. time complexity: O(n+logn) Approach 3: Quickselect. Time complexity O(n) I explained my interviewer the 3 approaches. He told me to solve in a naive manner. Used Approach 1 had some time left so coded approach 3 also The average time complexity of Quickselect is O(n), making it very efficient for its purpose. However, in"See full answer

Rasika D. - "Message queue is like a buffer. Its has 3 parts: Producer -> which creates a message and sends it to the queue. Queue -> which acta like a buffer pipe and stores the message. Subscriber -> which processes and releases messages one by one in FIFO manner."See full answer

Sambangi C. - "#simple solution 1.firstly find the node in the bst (O(logn) time complexity it take) 2.now removing the node consists of 3 cases: 1.if the node is leaf (no children): (keep track of parent and do) parent.left or parent.right=NULL simply remove the node () 2.if(has one child) replace the node with its child 3.if has both childs we replace the node with either inorder predesor(max of left tree)or inorder succesor and remove the node wh"See full answer

Lakshman B. - "As we can pass info to only one child at a time, I told that from any given node, we have to pass the info to that child(of this node) which has the largest subtree rooted at it. To calculate the subtree sizes, I used DFS. And then to calculate the minimum time to pass info to all the nodes, I used BFS picking the largest subtree child first at every node. I couldn't write the complete code in the given time and also made a mistake in telling the overall time complexity of my approach. I think t"See full answer

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

Reno S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["See full answer

Scott S. - "As a Data Engineer working in the derivatives domain, my role involves managing and optimizing data pipelines, ensuring data quality, and supporting data analytics for trading strategies. The fast-paced nature of the trading environment requires both proactive and reactive data management practices. Task: My daily tasks are focused on maintaining the integrity and availability of data, developing new data pipelines to support evolving trading strategies, and collaborating with both data scienti"See full answer

Divya R. - "public Double calculateRatio(String source, String destination) { Double ratio=1.0; while(graph.containsKey(source) && !visited.contains(source)) { visited.add(source); Map valueMap=graph.get(source); if(valueMap.containsKey(destination)) { return ratio*=valueMap.get(destination); } Map.Entry firstEntry=valueMap.entrySet().iterator().next(); source=firstEntry.getKey(); ratio*=firstEntry.getValue(); System.out.println("Entered"); } return null; }"See full answer

Anthony O. - "What excites me at work is being able to work on new and challenging projects where I have to be allowed to be creative and work with little supervision. I'm demotivated by that constant supervision and breathing down on one's neck every minute of the day."See full answer

Mohith J. - "Recursion: 0 if NULL, else 1+max(height(left), height(right))"See full answer

Chen J. - "Leetcode 347: Heap + Hashtable Follow up question: create heap with the length of K instead of N (more time complexity but less space )"See full answer

Chris R. - "It was like say we have a library A which has a library B as a dependency and so on, how would we determine in the dependency chain that whether there is a circular depedency?"See full answer

Sh R. - "i responded using a multi sourced BFS and in place marking, then i checked the final grid to see if any free spots were left unmarked."See full answer

Ramachandra N. - "from collections import deque from typing import List def longestsubarraydifflessthan_n(nums: List[int], N: int) -> int: """ Find the length of the longest contiguous subarray such that the difference between any two elements in the subarray is less than N. Equivalent condition: max(subarray) - min(subarray) < N Approach (Optimal): Sliding window with two monotonic deques: max_d: decreasing deque of indices (front is index of current max"See full answer