Data Structures & Algorithms Interview Questions

Stanley Y. - "First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently. Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"See full answer

Divya R. - "public static Integer[] findLargest(int[] input, int m) { if(input==null || input.length==0) return null; PriorityQueue minHeap=new PriorityQueue(); for(int i:input) { if(minHeap.size()(int)top){ minHeap.poll(); minHeap.add(i); } } } Integer[] res=minHeap.toArray(new Integer[0]); Arrays.sort(res); return res; }"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

Akash C. - " from typing import List def find_first(array: List[int], num: int) -> int: left = 0 right = len(array) - 1 result = -1 # keep track of leftmost occurence found so far while left <= right: mid = (left + right) // 2 if array[mid] == num: result = mid #Found a potential result right = mid - 1 elif array[mid] < num: left = mid + 1 else: right = mid - 1 return result debug your code"See full answer

Vidhyadhar V. - "public class CircularBuffer { private T[] buffer; private int head; private int tail; private int size; private final int capacity; public CircularBuffer(int capacity) { this.capacity = capacity; this.buffer = (T[]) new Object[capacity]; this.head = 0; this.tail = 0; this.size = 0; } public void enqueue(T item) { if (isFull()) { throw new IllegalStateException("Buffer is full"); } buf"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

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

Rahul J. - "find total sum. assign that to rightsum traverse from left to right: keep updating left sum and right sum, when they match return the index. else if you reach end return -1 or not found"See full answer

Anonymous Wombat - "Was the statement very similar to the leetcode or was it changed and only the main idea remained?"See full answer

Rajendra D. - "Questions to ask : Are there negative values in the input array? Interview : YES Will the product of two number fit into 32-bit Integer. If not, will it fit 64-bit Integer. If not, then is it safe to use Big Integer? Interview : let's worry only about 32 bit Integer What should we return if the input array is null or size (size of input array) is less than 2? Return 0 From above Information: General approach is as follows : a) Track smallest 2 elements in the array -> p"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 - "class ListNode: def init(self, val=0, next=None): self.val = val self.next = next def has_cycle(head: ListNode) -> bool: slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: return True return False debug your code below node1 = ListNode(1) node2 = ListNode(2) node3 = ListNode(3) node4 = ListNode(4) creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"See full answer

Kargi C. - "Average case - lookup/insert/delete - o(1) -> assuming a low load factor and uniform hash distribution. Worst case - o(n) -> where are keys collide in same bucket"See full answer

Rick E. - " from typing import Dict, List, Optional def max_profit(prices: Dict[str, int]) -> Optional[List[str]]: pass # your code goes here max = [None, 0] min = [None, float("inf")] for city, price in prices.items(): if price > max[1]: max[0], max[1] = city, price if price 0: return [min[0], max[0]] return None debug your code below prices = {'"See full answer

Michael B. - "Write a function which Caesar ciphers all the strings so that the first character is "a". Use ascii code points and the modulo operator to do this. Use this function to create a hashmap between each string and the CC-a string. Then go through each key:value pair in the hashmap, and use the CC-a ciphered value as the key in a new defaultdict(list), adding the original string to the value field in the output."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