"For any project based questions, it is important to structure your response clearly, showcasing your thought process, technical skills, problem-solving abilities, and how your work added value. Besides the STAR method, you can also use this kind of framework:
1. Start by selecting a relevant project (related to the role)
Give the project background and what specific problem it solved.
2. Align the project's objective and your role
Be specific about your role: were you the le"
Malay K. - "For any project based questions, it is important to structure your response clearly, showcasing your thought process, technical skills, problem-solving abilities, and how your work added value. Besides the STAR method, you can also use this kind of framework:
1. Start by selecting a relevant project (related to the role)
Give the project background and what specific problem it solved.
2. Align the project's objective and your role
Be specific about your role: were you the le"See full answer
"
Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between:
Skipping the current house and taking the maximum money stolen up to the previous house.
Robbing the current house and adding its value to the maximum money stolen up to the house two steps back.
package main
import (
"fmt"
)
// rob function calculates the maximum money a robber can steal
func maxRob(nums []int) int {
ln"
VContaineers - "
Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between:
Skipping the current house and taking the maximum money stolen up to the previous house.
Robbing the current house and adding its value to the maximum money stolen up to the house two steps back.
package main
import (
"fmt"
)
// rob function calculates the maximum money a robber can steal
func maxRob(nums []int) int {
ln"See full answer
"We can use dictionary to store cache items so that our read / write operations will be O(1).
Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1)
Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"
Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1).
Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1)
Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer
"Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."
Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer
Machine Learning Engineer
Data Structures & Algorithms
+4 more
🧠Want an expert answer to a question? Saving questions lets us know what content to make next.
"def is_valid(s: str) -> bool:
stack = []
closeToOpen = { ")" : "(", "]" : "[", "}" : "{" }
for c in s:
if c in closeToOpen:
if stack and stack[-1] == closeToOpen[c]:
stack.pop()
else:
return False
else:
stack.append(c)
return True if not stack else False
debug your code below
print(is_valid("()[]"))
`"
Anonymous Roadrunner - "def is_valid(s: str) -> bool:
stack = []
closeToOpen = { ")" : "(", "]" : "[", "}" : "{" }
for c in s:
if c in closeToOpen:
if stack and stack[-1] == closeToOpen[c]:
stack.pop()
else:
return False
else:
stack.append(c)
return True if not stack else False
debug your code below
print(is_valid("()[]"))
`"See full answer
"Is it bad to get the answer a different way? Will they mark that as not knowing Bayes Theorem or just correct as it is an easier way to get the answer?
The way I went is to look at what happens when the factory makes 100 light bulbs. Machine A makes 60 of which 3 are faulty, Machine B makes 40 of which 1.2 are faulty. Therefore the pool of faulty lightbulbs is 3/4.2 = 5/7 from machine A and 1.2/4.2 = 3/7 from Machine B."
Will I. - "Is it bad to get the answer a different way? Will they mark that as not knowing Bayes Theorem or just correct as it is an easier way to get the answer?
The way I went is to look at what happens when the factory makes 100 light bulbs. Machine A makes 60 of which 3 are faulty, Machine B makes 40 of which 1.2 are faulty. Therefore the pool of faulty lightbulbs is 3/4.2 = 5/7 from machine A and 1.2/4.2 = 3/7 from Machine B."See full answer
"\# 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')"
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
"
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"
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
"#include
// Naive method to find a pair in an array with a given sum
void findPair(int nums[], int n, int target)
{
// consider each element except the last
for (int i = 0; i < n - 1; i++)
{
// start from the i'th element until the last element
for (int j = i + 1; j < n; j++)
{
// if the desired sum is found, print it
if (nums[i] + nums[j] == target)
{
printf("Pair found (%d, %d)\n", nums[i], nums[j]);
return;
}
}
}
// we reach here if the pair is not found
printf("Pair not found");
}
"
Gundala tarun,cse2020 V. - "#include
// Naive method to find a pair in an array with a given sum
void findPair(int nums[], int n, int target)
{
// consider each element except the last
for (int i = 0; i < n - 1; i++)
{
// start from the i'th element until the last element
for (int j = i + 1; j < n; j++)
{
// if the desired sum is found, print it
if (nums[i] + nums[j] == target)
{
printf("Pair found (%d, %d)\n", nums[i], nums[j]);
return;
}
}
}
// we reach here if the pair is not found
printf("Pair not found");
}
"See full answer
"def traprainwater(height: List[int]) -> int:
n = len(height)
totalwaterlevel = 0
for i in range(n):
j = i+1
while j = n:
break
rows = j - i -1
intrwaterlevel = min(height[j], height[i]) * rows
for k in range(i+1, j):
intrwaterlevel -= height[k]
totalwaterlevel += intrwaterlevel
i = j
return totalwaterlevel"
Manoj R. - "def traprainwater(height: List[int]) -> int:
n = len(height)
totalwaterlevel = 0
for i in range(n):
j = i+1
while j = n:
break
rows = j - i -1
intrwaterlevel = min(height[j], height[i]) * rows
for k in range(i+1, j):
intrwaterlevel -= height[k]
totalwaterlevel += intrwaterlevel
i = j
return totalwaterlevel"See full answer
"Less efficient version, yet effective for the interview:
def is_palindrome(s: str) -> bool:
dim = len(s)
if dim str:
dim = len(s)
if dim < 2:
return s
left = 0
longest = ""
while left < dim:
righ"
Gabriele G. - "Less efficient version, yet effective for the interview:
def is_palindrome(s: str) -> bool:
dim = len(s)
if dim str:
dim = len(s)
if dim < 2:
return s
left = 0
longest = ""
while left < dim:
righ"See full answer