Coding Interview Questions

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

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

Salome L. - "SELECT items.item_category, SUM(orders.orderquantity) AS totalunitsorderedlast7days FROM orders JOIN items ON orders.itemid = items.itemid WHERE orders.order_date BETWEEN DATE('now', '-6 days') AND DATE('now') GROUP BY items.item_category `"See full answer

Akshay D. - "SELECT name, type, CASE WHEN type = 'Electronic' THEN price-(0.10*price) WHEN type = 'Clothing' THEN price-(0.20*price) WHEN type = 'Grocery' THEN price-(0.05*price) WHEN type = 'Book' THEN price-(0.15*price) ELSE price END AS discounted_price FROM products; `"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

Gabriele G. - "def find_first(array: List[int], num: int) -> int: lo = 0 hi = len(array)-1 while lo = num: hi = mid - 1 if lo == mid and array[mid] == num: return mid else: array[mid] < num lo = mid + 1 return -1 `"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

Dinar M. - "Even more faster and vectorized version, using np.linalg.norm - to avoid loop and np.argpartition to select lowest k. We dont need to sort whole array - we need to be sure that first k elements are lower than the rest. import numpy as np def knn(Xtrain, ytrain, X_new, k): distances = np.linalg.norm(Xtrain - Xnew, axis=1) k_indices = np.argpartition(distances, k)[:k] # O(N) selection instead of O(N log N) sort return int(np.sum(ytrain[kindices]) > k / 2.0) `"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

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

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

Alina G. - "Order the result in descending month is not applied in the solution"See full answer