Recent Software Engineer Interview Questions

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"See full answer

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Jorge G. - " from typing import List def two_sum(nums: List[int], target: int) -> List[int]: """ Iterate the list Create a hashmap for tracking seen complements For each element: Check if target - current was seen If so, return the index of the current and the index of the complement If not, add the current number in the hashmap as key, and the index of the current number as value. Return an empty array if the loop e"See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Anonymous Eel - "String commonStr(String str1, String str2) { int len1 = str1.length(); int len2 = str2.length(); if (len1 == 0 || len2 == 0) return ""; // let dpx reprsent the longest common str of 0...x int dp = new int len1 + 1; int maxLen = 0; int endIndex = 0; for (int i = 1; i <= len1; i++) { for (int j = 1; j <= len2; j++) { if (str1.charAt(i-1) == str2.charAt(j-1)) { dpi = dpi-1 + 1; "See full answer

Tiago R. - "function isPalindrome(s, start, end) { while (s[start] === s[end] && end >= start) { start++; end--; } return end <= start; } function longestPalindromicSubstring(s) { let longestPalindrome = ''; for (let i=0; i < s.length; i++) { let j = s.length-1; while (s[i] !== s[j] && i <= j) { j--; } if (s[i] === s[j]) { if (isPalindrome(s, i, j)) { const validPalindrome = s.substring(i, j+1"See full answer

Ayinde B. - "I was working for my friend building streams at venues across the Chicago land area for FGC (fighting game tournaments), I adjusted and engineered his equipment to be set up permanently that's until covid came around at least. I used OBS to give visual appearances to stream watchers. So we're talking about subscribe, follow, and donation notifications and things of that nature for viewers to know they contributed in one of those ways. I set up proper sign-up scheduling for participants to lock t"See full answer

Saket S. - "Project Schedule is a living document which is prepared based on inputs from the entire team including product, engineering, finance, sales, marketing, legal etc. It is owned by Project/Program Manager but it reflects progress of the whole team. Inputs required for the development of schedule includes project charter, list of stakeholders, project scope, WBS, freeze calendar, vendor SOWs etc. I will use the following approach for schedule development: Copy Milestone level plan from Project"See full answer

Umang S. - "Just a bit of context on what could be a risk in Product or Software Development. We talk about the Scope, Quality, and Time Triangle. IMHO the only variables that can matter and can pose risk are these. Try to form an answer where you used the Effort, Value, Risk framework to quantify the risk. The answer can be similar to short term sacrifices for long term gains, as you do take in mind the risk the short term sacrifices pose. Situation - I was leading the team to complete an overhaul of the"See full answer

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"See full answer

Nitin P. - "public static void sortBinaryArray(int[] array) { int len = array.length; int[] res = new int[len]; int r=len-1; for (int value : array) { if(value==1){ res[r]= 1; r--; } } System.out.println(Arrays.toString(res)); } `"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer

Ramachandra N. - "def mergeTwoListsRecursive(l1, l2): if not l1 or not l2: return l1 or l2 if l1.val < l2.val: l1.next = mergeTwoListsRecursive(l1.next, l2) return l1 else: l2.next = mergeTwoListsRecursive(l1, l2.next) return l2 "See full answer

Piyush D. - "Honestly, In addition to having the right expertise and skills required for this job, I am really looking forward to being part of this organisation’s journey. I am really confident that I will really exceed your expectations in achieving both mine and the company’s goals. I have worked with different teams in different capacities, managed a team of engineers successfully from very early part of my career and that has made me a very easy person to work with. I get things done easily. I am passio"See full answer

Dean Z. - "I will use below 4 pillars to rate and prioritize those 5 projects: Analyze the impact of those 5 projects, to our team and to the whole company maybe Find out if those 5 projects have any dependency between each other Clarify the timeline of each project Identity the status of resources to work on those 5 projects and also the status of their dependencies"See full answer

Ash I. - "Follow STAR Example: In my most recent role at x, we've been working every quarter to set our quarterly goals or even decide for the year but given in 2022 we've been through a great resignation phase there was a high attrition because of which we have overloaded engineering teams. During last quarterly planning, I started to question with my manager about giving a focus to our engineering, so we complete one thing absolutely fine which is more impactful for our customers and add immediate val"See full answer

Ushita S. - "A clarifying question: Is this question asking about when I met a tight deadline in a project or how did I manage a project that had a tight deadline? The answer uploaded to this question is good, I would also add 'creating a critical path from overall project schedule and then making sure that none of the deliverables in the critical path are sacrificed in order to meet the tight deadline' as an action taken."See full answer

Shuchi A. - "A few months ago I joined a micro-services platform engineering team as their manager, at that time my team was struggling to deliver towards an upcoming production deadline for a customer facing product. Production date had been moved 5 times already and there were about 40% of product features which were remaining to be tested and signed off to move to production . I was made responsible to deliver the release of this product within the deadline and turnaround the software delivery throughput."See full answer