Adobe Interview Questions

Reno S. - " ✅ Passes all test cases: Tricky: stock_prices = [1, 10, 2, 3] output: 9 (Buy on day 1 at 1 and sell on day 2 at 10) func maxProfit(_ stockPrices: [Int]) -> Int { var options: [Int] = [] // min-heap var currentProfit = 0 var maxProfit = 0 for price in stockPrices { if let cheapestOption = options.last, cheapestOption < price { if currentProfit < price { currentProfit += price // greedy profit } else { "See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Victor H. - "There is a faster approach that solves the problem in O(n) time: def find_duplicates(arr1, arr2): arr1 = set(arr1) res = [] for num in arr2: if num in arr1: res.append(num) return res `"See full answer

Matthew K. - " function climbStairs(n) { // 4 iterations of Dynamic Programming solutions: // Step 1: Recursive: // if (n <= 2) return n // return climbStairs(n-1) + climbStairs(n-2) // Step 2: Top-down Memoization // const memo = {0:0, 1:1, 2:2} // function f(x) { // if (x in memo) return memo[x] // memo[x] = f(x-1) + f(x-2) // return memo[x] // } // return f(n) // Step 3: Bottom-up Tabulation // const tab = [0,1,2] // f"See full answer

Ritaban M. - " Brute Force Two Pointer Solution: from typing import List def two_sum(nums, target): for i in range(len(nums)): for j in range(i+1, len(nums)): if nums[i]+nums[j]==target: return [i,j] return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"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

Victor O. - " import java.util.*; class Solution { // Time Complexity: O(n^2) // Space Complexity: O(n) public static List> threeSum(int[] nums) { // Ensure that the array is sorted first Arrays.sort(nums); // Create the results list to return List> results = new ArrayList(); // Iterate over the length of nums for (int i = 0; i < nums.length-2; i++) { // We will have the first number in"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

Rick E. - " O(n) time from typing import List def generate_parentheses(n: int): res = [] def generate(buf, opened, closed): if len(buf) == 2 * n: if n != 0: res.append(buf) return if opened < n: generate( buf + "(", opened + 1, closed) if closed < opened: generate(buf + ")", opened, closed + 1) generate("", 0, 0) return res debug your code below print(generate_parentheses(1"See full answer

Sarvesh G. - "def calc(expr): ans = eval(expr) return ans your code goes debug your code below print(calc("1 + 1")) `"See full answer

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

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

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

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