Coding Interview Questions

Tiago R. - "function knapsack(weights, values, cap) { const indicesByValue = Object.keys(weights).map(weight => parseInt(weight)); indicesByValue.sort((a, b) => values[b]-values[a]); const steps = new Map(); function knapsackStep(cap, sack) { if (steps.has(sack)) { return steps.get(sack); } let maxOutput = 0; for (let index of indicesByValue) { if (!sack.has(index) && weights[index] <= cap) { maxOutput ="See full answer

Frederick A. - "This problem can be solved with two approaches Iterative approach Recursive approach Quite easy to think about the iterative approach, you can make use of a while loop in that case. But what if you want to make use of previously computed values? That case going for the recursive solution is quite useful. class Collatz: def init(self) -> None: self.cache = {} self.steps = 0 def steps_from(self, n) -> int: # base case if n == 1: "See full answer

Kedar W. - "with login_data as ( select * from useractivitylog where activity_type = 'LOGIN' ) ,cte as ( select userid, timestamp as currentlogin ,lag(timestamp,1,timestamp) over (partition by user_id order by timestamp asc) as previous_login, round((julianday(timestamp) - julianday(lag(timestamp,1,timestamp) over (partition by user_id order by timestamp asc))) * 24 * 60) as minutes_elapsed from login_data ) select * from cte where currentlogin  previouslogin;"See full answer

Tiago R. - "function spiralCopy(inputMatrix) { if (inputMatrix.length === 1) return inputMatrix[0]; let lowerY = 0; let upperY = inputMatrix.length-1; let lowerX = 0; let upperX = inputMatrix[0].length-1; const output = []; while (true) { if (lowerX > upperX) break; for (let x = lowerX; x upperY) break; for (let y = lowerY; y <= upperY; y++) { output.push(inputMatrix[y][u"See full answer

Soma R. - "SELECT a.marketing_channel, AVG(a.purchasevalue) AS avgpurchase_value, SUM(CASE WHEN a.purchasevalue > 0 THEN 1 ELSE 0 END) * 100 / COUNT(a.sessionid) AS conversion_rate FROM attribution a LEFT JOIN user_sessions u ON a.sessionid = u.sessionid GROUP BY a.marketing_channel ORDER BY conversion_rate DESC; "See full answer

Akash C. - " CODE INSTRUCTIONS: 1) The method findInOrderSuccessor you're asked to implement is located at line 30. 2) Use the helper code below to implement it. 3) In a nutshell, the helper code allows you to to build a Binary Search Tree. 4) Jump to line 88 to see an example for how the helper code is used"See full answer

Aneesha K. - "Test case is wrong. It expects to sort in asc order of month_year. -- Write your query here SELECT strftime('%Y-%m', createdat) AS monthyear, COUNT(DISTINCT userid) AS numcustomers, COUNT(t.id) AS num_orders, SUM(price * quantity) AS order_amt FROM transactions t INNER JOIN products p ON t.product_id = p.id GROUP BY month_year ORDER BY month_year ; "See full answer

Anonymous Tortoise - "SELECT i.item_category, o.order_date, SUM(o.orderquantity) AS totalunits_ordered FROM orders o JOIN items i ON o.itemid = i.itemid WHERE o.order_date >= DATE('now', '-6 days') GROUP BY i.item_category, o.order_date ORDER BY i.item_category ASC, o.order_date ASC;"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

Anonymous Owl - "def getcheapestcost(rootNode): \# need to do DFS for each branch \# but this can be done recursively n = len(rootNode.children) if n == 0: return 0 else: min_cost = float('inf') for i in range(len(n)): tempcost = getcheapest_cost(rootNode.children[i]) if (tempcost < mincost): mincost = tempcost return min_cost + rootNode.cost \# A node class Node: \# Constructor to create a new node def init\(self, cost): self.cost = cost self.children = [] self.parent = None"See full answer

Tiago R. - "function findRotations(nums) { if (nums.length 0 && nums[mid] > nums[mid-1]) { left = mid; } else { right = mid; } } return rig"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Anonymous Armadillo - "I couldn't follow the solution offered here, but my solution seemed to pass 6/6 tests. Any feedback is welcome, thank you! def encrypt(word): en_word = "" for i in range(len(word)): if i == 0: en_word += chr(ord(word[0])+1) else: num = ord(word[i]) + ord(en_word[i-1]) while num > 122: num -= 26 en_word += chr(num) return en_word def decrypt(word): de_word = "" for i in range(len(word)): if i == 0: de_word += chr(ord(word[i]"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

Steven S. - "too many questions for clarification on this to start"See full answer

Alvin P. - "WITH previous AS(SELECT viewer_id, watch_hours, LAG(watchhours) OVER(PARTITION BY viewerid ORDER BY year, month) AS previous_hours, year, month FROM watch_time GROUP BY viewer_id, year, month ), streaks AS(SELECT viewer_id, SUM(CASE WHEN previoushours IS NOT NULL AND previoushours = 3 `"See full answer