Adobe Interview Questions

Jerry O. - "\# Definition for a binary tree node. class TreeNode: def init(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def maxPathSum(self, root: TreeNode) -> int: self.max_sum = float('-inf')"See full answer

Marc S. - "This is just a two line solve, I'm not sure why it's being made more complicated? function productExceptSelf(nums) { // your code goes here const product = nums.reduce((acc, val) => acc * val, 1); return nums.map(num => num === 0 ? 0 : Math.floor(product/num)); } `"See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

Tiago R. - "function findPrimes(n) { if (n < 2) return []; const primes = []; for (let i=2; i <= n; i++) { const half = Math.floor(i/2); let isPrime = true; for (let prime of primes) { if (i % prime === 0) { isPrime = false; break; } } if (isPrime) { primes.push(i); } } return primes; } `"See full answer

Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

Tiago R. - "const ops = { '+': (a, b) => a+b, '-': (a, b) => a-b, '/': (a, b) => a/b, '': (a, b) => ab, }; function calc(expr) { // Search for + or - for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if (['+', '-'].includes(char)) { return opschar), calc(expr.slice(i+1))); } } // Search for / or * for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if"See full answer

Prajwal M. - "from typing import List def maxprofitgreedy(stock_prices: List[int]) -> int: l=0 # buying r=1 # selling max_profit=0 while rstock_prices[l]: profit=stockprices[r]-stockprices[l] maxprofit=max(maxprofit,profit) else: l=r r+=1 return max_profit debug your code below print(maxprofitgreedy([7, 1, 5, 3, 6, 4])) `"See full answer

Josh L. - "I would define success by first figuring out what our goal is by building the reels feature. Are we trying to increase DAUs? Increase enagement? Etc... For the sake of this, I think to define success it makes most sense to see if there is an increase in the amount of time users are spending on instagram. If time spent per user increases, it is likely that ad spend can increase and in turn increases instagram's reveue. We need to be sure that there are guard rails in place and make sure that by"See full answer

Tiago R. - "function generateParentheses(n) { if (n < 1) { return []; } if (n === 1) { return ["()"]; } const combinations = new Set(); let previousCombinations = generateParentheses(n-1); for (let prev of previousCombinations) { for (let i=0; i < prev.length; i++) { combinations.add(prev.slice(0, i+1) + "()" + prev.slice(i+1)); } } return [...combinations]; } `"See full answer

Sridhar R. - "Arrays.sort(inputarray) sliding window with a size of 2. Check for the sum in the sliding window. subtract the start when window moves"See full answer

Rishi G. - "Problem Statement: The Fibonacci sequence is defined as F(n) = F(n-1) + F(n-2) with F(0) = 1 and F(1) = 1. The solution is given in the problem statement itself. If the value of n = 0, return 1. If the value of n = 1, return 1. Otherwise, return the sum of data at (n - 1) and (n - 2). Explanation: The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, typically starting with 0 and 1. Java Solution: public static int fib(int n"See full answer

Tiago R. - "function addChildren(root, val, inorder) { const rootInOrderIndex = inorder.indexOf(root.value); const childrenInOrderIndex = inorder.indexOf(val); if (childrenInOrderIndex < rootInOrderIndex) { if (!root.left) { root.left = new TreeNode(val); } else { addChildren(root.left, val, inorder); } } else { if (!root.right) { root.right = new TreeNode(val); } else { addChildren(root.right,"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

Exponent - "This may seem tricky at first, but this is actually an Expansion problem, since we're expanding to a new market. The only difference is this is specifically for small businesses within that market. This is the formula you should use when tackling these types of interview questions: Ask clarifying questions Perform user analysis Market risk analysis State goals Perform channel analysis Prioritize growth channels Strategy Summarize Witho"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

Aryan D. - "int main() { int a1[7]={1,2,3,4,5,6,7}; int a2[7]={1,9,10,11,12,13,14}; vectorv; v.insert(v.begin(),begin(a1),end(a1)); v.insert(v.begin(),begin(a2),end(a2)); int a3[v.size()]; sort(v.begin(),v.end()); for(int i=0;i<v.size();i++) { a3[i]=v[i]; } } `"See full answer

Seyed rasoul J. - "I would start with the company vision then assuming we have more than one product team, craft the vision for the product through a collaboration of PMs and Tech leads, then based on that I will define the scope of each product team's milestones to achieve the product vision and go from both ends to fill the gap from what we already have and what needed to achieve the milestones."See full answer