"I most want to communicate a few principals of conflict resolution that I believe were integral in this situation, which are mutual respect, a results orientation, an unwavering focus on the user.
To that end, here’s how I’d like to structure this answer: First, I’ll tell you about the project we were working on, to provide some background for you. Second, I’ll describe the disagreement. Third, I’ll describe how we arrived at a solution, and finally, I’ll discuss how those 3 conflict resolut"
Ross B. - "I most want to communicate a few principals of conflict resolution that I believe were integral in this situation, which are mutual respect, a results orientation, an unwavering focus on the user.
To that end, here’s how I’d like to structure this answer: First, I’ll tell you about the project we were working on, to provide some background for you. Second, I’ll describe the disagreement. Third, I’ll describe how we arrived at a solution, and finally, I’ll discuss how those 3 conflict resolut"See full answer
"from collections import deque
def updateword(words, startword, end_word):
if end_word not in words:
return None # Early exit if end_word is not in the dictionary
queue = deque([(start_word, 0)]) # (word, steps)
visited = set([start_word]) # Keep track of visited words
while queue:
word, steps = queue.popleft()
if word == end_word:
return steps # Found the target word, return steps
for i in range(len(word)):
"
叶 路. - "from collections import deque
def updateword(words, startword, end_word):
if end_word not in words:
return None # Early exit if end_word is not in the dictionary
queue = deque([(start_word, 0)]) # (word, steps)
visited = set([start_word]) # Keep track of visited words
while queue:
word, steps = queue.popleft()
if word == end_word:
return steps # Found the target word, return steps
for i in range(len(word)):
"See full answer
"Reversing a linked list is a very popular question. We have two approaches to reverse the linked list: Iterative approach and recursion approach.
Iterative approach (JavaScript)
function reverseLL(head){
if(head === null) return head;
let prv = null;
let next = null;
let cur = head;
while(cur){
next = cur.next; //backup
cur.next = prv;
prv = cur;
cur = next;
}
head = prv;
return head;
}
Recursion Approach (JS)
function reverseLLByRecursion("
Satyam S. - "Reversing a linked list is a very popular question. We have two approaches to reverse the linked list: Iterative approach and recursion approach.
Iterative approach (JavaScript)
function reverseLL(head){
if(head === null) return head;
let prv = null;
let next = null;
let cur = head;
while(cur){
next = cur.next; //backup
cur.next = prv;
prv = cur;
cur = next;
}
head = prv;
return head;
}
Recursion Approach (JS)
function reverseLLByRecursion("See full answer
"As you know, this is the most important question for any interview. Here is a structure I like to follow,
Start with 'I'm currently a SDE/PM/TPM etc with XYZ company.... '
Mention how you got into PM/TPM/SDE field (explaining your journey)
Mention 1 or 2 accomplishments
Mention what you do outside work (blogging, volunteer etc)
Share why are you looking for a new role
Ask the interviewer if they have any questions or will like to dive deep into any of your experience"
Bipin R. - "As you know, this is the most important question for any interview. Here is a structure I like to follow,
Start with 'I'm currently a SDE/PM/TPM etc with XYZ company.... '
Mention how you got into PM/TPM/SDE field (explaining your journey)
Mention 1 or 2 accomplishments
Mention what you do outside work (blogging, volunteer etc)
Share why are you looking for a new role
Ask the interviewer if they have any questions or will like to dive deep into any of your experience"See full answer
Data Engineer
Behavioral
+9 more
🧠 Want an expert answer to a question? Saving questions lets us know what content to make next.
"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
"public static boolean isPalindrome(String str){
boolean flag = true;
int len = str.length()-1;
int j = len;
for(int i=0;i<=len/2;i++){
if(str.charAt(i)!=str.charAt(j--)){
flag = false;
break;
}
}
return flag;
}"
Sravanthi M. - "public static boolean isPalindrome(String str){
boolean flag = true;
int len = str.length()-1;
int j = len;
for(int i=0;i<=len/2;i++){
if(str.charAt(i)!=str.charAt(j--)){
flag = false;
break;
}
}
return flag;
}"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
"we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"
Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer
"Idea for solution:
Reverse the complete char array
Reverse the words separated by space. i.e. Find the space characters and the reverse the subarray between two space characters.
vector reverseSubarray(vector& arr, int s, int e)
{
while (s reverseWords(vector& arr )
{
int n = arr.size();
reverse(arr, 0, n - 1"
Rahul M. - "Idea for solution:
Reverse the complete char array
Reverse the words separated by space. i.e. Find the space characters and the reverse the subarray between two space characters.
vector reverseSubarray(vector& arr, int s, int e)
{
while (s reverseWords(vector& arr )
{
int n = arr.size();
reverse(arr, 0, n - 1"See full answer
"this solution here is much faster than the exponent reference soln. It is also far more concise and easy to understand
def moveZerosToEnd(arr: List[int]) -> List[int]:
left = 0
for right in range(len(arr)):
if arr[right] == 0:
pass
else:
if left != right:
temp = arr[left]
arr[left] = arr[right]
arr[right] = temp
left += 1
return arr
`"
Devesh K. - "this solution here is much faster than the exponent reference soln. It is also far more concise and easy to understand
def moveZerosToEnd(arr: List[int]) -> List[int]:
left = 0
for right in range(len(arr)):
if arr[right] == 0:
pass
else:
if left != right:
temp = arr[left]
arr[left] = arr[right]
arr[right] = temp
left += 1
return arr
`"See full answer
"In python
def find_duplicates(arr1: List[int], arr2: List[int]) -> List[int]:
result = list(set(arr1) & set(arr2))
return result
"
Sammy R. - "In python
def find_duplicates(arr1: List[int], arr2: List[int]) -> List[int]:
result = list(set(arr1) & set(arr2))
return result
"See full answer
"If 0's aren't a concern, couldn't we just
multiply all numbers.
and then divide product by each number in the list ?
if there's more than one zero, then we just return an array of 0s
if there's one zero, then we just replace 0 with product and rest 0s.
what am i missing?"
Sachin R. - "If 0's aren't a concern, couldn't we just
multiply all numbers.
and then divide product by each number in the list ?
if there's more than one zero, then we just return an array of 0s
if there's one zero, then we just replace 0 with product and rest 0s.
what am i missing?"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