Data Structures & Algorithms Interview Questions

Anonymous Raccoon - "def find_primes(n): lst=[] for i in range(2,n+1): is_prime=1 for j in range(2,int(i**0.5)+1): if i%j==0: is_prime=0 break if is_prime: lst.append(i) return lst "See full answer

Mridul J. - "We have a list of documents. We want to build an index that maps keywords to documents containing them. Then, given a query keyword, we can efficiently retrieve all matching documents. docs = [ "Python is great for data science", "C++ is a powerful language", "Python supports OOP and functional programming", "Weather today is sunny", "Weather forecast shows rain" ]"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. - "Problem: Given an input string txt consisting of alphanumeric characters and the parentheses characters '(' & ')', write a function which removes the minimum number of characters to return a version of the string with properly balanced parenthesis. Answer: You can do this with a counter. Psuedo-Python Start with counter = 0 output = [] Iterate through the string, every time you encounter a '(', increment the counter. Add the character to the output. If you encounter a ')', decrement the coun"See full answer

Anonymous Unicorn - "Count items between indices within compartments compartments are delineated by by: '|' items are identified by: '*' input_inventory = "*||||" inputstartidxs = [1, 4, 6] inputendidxs = [9, 5, 8] expected_output = [3, 0, 1] Explanation: "*||||" 0123456789... indices ++ + # within compartments ^ start_idx = 1 ^ end_idx = 9 -- - # within idxs but not within compartments "*||||" 0123456789... indices "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

Sean L. - " import pandas as pd def findaveragedistance(gps_data: pd.DataFrame) -> pd.DataFrame: #0. IMPORTANT: get the unordered pairs gpsdata['city1']=gpsdata[['origin','destination']].min(axis=1) gpsdata['city2']=gpsdata[['origin','destination']].max(axis=1) #1. get the mean distance by cities avgdistance=gpsdata.groupby(['city1','city2'], as_index=False)['distance'].mean().round(2) avgdistance.rename(columns={'distance':"averagedistance"}, inplace=True) "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

Jasmine Y. - "It depends on the size of the dataset. You want enough samples in both the testing, training and evaluation sets. If there is enough data, 70/20/10 is a good split"See full answer

Ankita G. - "Merge Sort"See full answer

TreeOfWisdom - "Should this question be BST, not just BT? Otherwise it would not be possible to reconstruct the tree solely based on the array regardless of its order"See full answer

Yash N. - "import time class Task: def init\(self, description, interval=None): self.description = description self.interval = interval self.next_run = time.time() class SimpleTaskScheduler: def init\(self): self.tasks = [] def add_task(self, description, interval=None): self.tasks.append(Task(description, interval)) def run(self, duration=60): end_time = time.time() + duration while time.time() < end_time: curr"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

Robert W. - "def check_byte(octet): _""" Checks if the given string \octet\ represents a valid byte (0-255). """_ Check for empty string if not octet: return False Check if the string has non-digit characters if not octet.isdigit(): return False Check for leading zeroes in multi-digit numbers if len(octet) > 1 and octet[0] == '0': return False Check if the integer value is between 0 and 255 return 0 <= int(octet) <= 255 def va"See full answer

Anonymous Condor - "Yes, I need to compare the first half of the first string with the reverse order of the second half of the second string. Repeat this process to the first half of the second string and the second half of the first string."See full answer

Steve M. - "def split_count(s): return 2**(len(s)-1) `"See full answer