"SELECT
u.user_id,
u.user_name,
u.email,
ROUND(AVG(CASE WHEN b.status = 'Unmatched' THEN 1.0 ELSE 0 END), 2) AS avgunmatchedbookings
FROM
users u
LEFT JOIN
bookings b ON u.userid = b.userid
GROUP BY
u.user_id,
u.user_name,
u.email;
`"
Akshay D. - "SELECT
u.user_id,
u.user_name,
u.email,
ROUND(AVG(CASE WHEN b.status = 'Unmatched' THEN 1.0 ELSE 0 END), 2) AS avgunmatchedbookings
FROM
users u
LEFT JOIN
bookings b ON u.userid = b.userid
GROUP BY
u.user_id,
u.user_name,
u.email;
`"See full answer
"Question: An array of n integers is given, and a positive integer k, where k << n. k indicates that the absolute difference between each element's current index (icurrent) and the index in the sorted array (isorted) is less than k (|icurr - isorted| < k).
Sort the given array.
The most common solution is with a Heap:
def solution(arr, k):
min_heap = []
result = []
for i in range(len(arr))
heapq.heappush(min_heap, arr[i])
"
Guilherme M. - "Question: An array of n integers is given, and a positive integer k, where k << n. k indicates that the absolute difference between each element's current index (icurrent) and the index in the sorted array (isorted) is less than k (|icurr - isorted| < k).
Sort the given array.
The most common solution is with a Heap:
def solution(arr, k):
min_heap = []
result = []
for i in range(len(arr))
heapq.heappush(min_heap, arr[i])
"See full answer
"#simple solution
1.firstly find the node in the bst (O(logn) time complexity it take)
2.now removing the node consists of 3 cases:
1.if the node is leaf (no children):
(keep track of parent and do)
parent.left or parent.right=NULL
simply remove the node ()
2.if(has one child) replace the node with its child
3.if has both childs
we replace the node with either inorder predesor(max of left tree)or inorder succesor and remove
the node wh"
Sambangi C. - "#simple solution
1.firstly find the node in the bst (O(logn) time complexity it take)
2.now removing the node consists of 3 cases:
1.if the node is leaf (no children):
(keep track of parent and do)
parent.left or parent.right=NULL
simply remove the node ()
2.if(has one child) replace the node with its child
3.if has both childs
we replace the node with either inorder predesor(max of left tree)or inorder succesor and remove
the node wh"See full answer
"class Node:
def init(self, value):
self.value = value
self.children = []
def inorder_traversal(root):
if not root:
return []
result = []
n = len(root.children)
for i in range(n):
result.extend(inorder_traversal(root.children[i]))
if i == n // 2:
result.append(root.value)
if n == 0:
result.append(root.value)
return result
Example usage:
root = Node(1)
child1 = Node(2)
chil"
Teddy Y. - "class Node:
def init(self, value):
self.value = value
self.children = []
def inorder_traversal(root):
if not root:
return []
result = []
n = len(root.children)
for i in range(n):
result.extend(inorder_traversal(root.children[i]))
if i == n // 2:
result.append(root.value)
if n == 0:
result.append(root.value)
return result
Example usage:
root = Node(1)
child1 = Node(2)
chil"See full answer
"Required output in the solution not the one requested from the question. only customerid, firstname, last_name and years were required. Please this needs to be very clear.
Otherwise my answer is
with totalorderyear as (
SELECT
o.customer_id,
c.first_name,
c.last_name,
EXTRACT(YEAR FROM o.orderdate) AS orderyear,
COUNT(o.orderid) AS totalorders
FROM orders o
LEFT JOIN customers c
ON c.customerid = o.customerid
GROUP BY o.customerid, c.firstname, c.last"
Gloriose H. - "Required output in the solution not the one requested from the question. only customerid, firstname, last_name and years were required. Please this needs to be very clear.
Otherwise my answer is
with totalorderyear as (
SELECT
o.customer_id,
c.first_name,
c.last_name,
EXTRACT(YEAR FROM o.orderdate) AS orderyear,
COUNT(o.orderid) AS totalorders
FROM orders o
LEFT JOIN customers c
ON c.customerid = o.customerid
GROUP BY o.customerid, c.firstname, c.last"See full answer
"Any cycle would cause the prerequisite to be greater than the course. This passes all the tests:
function canFinish(_numCourses, prerequisites) {
for (const [a, b] of prerequisites) {
if (b > a) return false
}
return true
}
`"
Jeremy D. - "Any cycle would cause the prerequisite to be greater than the course. This passes all the tests:
function canFinish(_numCourses, prerequisites) {
for (const [a, b] of prerequisites) {
if (b > a) return false
}
return true
}
`"See full answer
"Initialize left pointer: Set a left pointer left to 0.
Iterate through the array: Iterate through the array from left to right.
If the current element is not 0, swap it with the element at the left pointer and increment left.
Time complexity: O(n). The loop iterates through the entire array once, making it linear time.
Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures.
"
Avon T. - "Initialize left pointer: Set a left pointer left to 0.
Iterate through the array: Iterate through the array from left to right.
If the current element is not 0, swap it with the element at the left pointer and increment left.
Time complexity: O(n). The loop iterates through the entire array once, making it linear time.
Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures.
"See full answer
"select sum(orderquantity) as totalunitsorderedyesterday
from orders as ord join items as it
on ord.itemid=it.itemid
where order_date="2023-10-14""
Rudra pratap S. - "select sum(orderquantity) as totalunitsorderedyesterday
from orders as ord join items as it
on ord.itemid=it.itemid
where order_date="2023-10-14""See full answer
"Let me try to explain it with simple life analogy
You're cooking dinner in the kitchen. Multithreading is when you've got a bunch of friends helping out. Each friend does a different job—like one chops veggies while another stirs a sauce. Everyone focuses on their task, and together, you all make the meal faster.
In a computer, it's like different jobs happening all at once, making stuff happen quicker, just like having lots of friends helping makes dinner ready faster."
Praveen D. - "Let me try to explain it with simple life analogy
You're cooking dinner in the kitchen. Multithreading is when you've got a bunch of friends helping out. Each friend does a different job—like one chops veggies while another stirs a sauce. Everyone focuses on their task, and together, you all make the meal faster.
In a computer, it's like different jobs happening all at once, making stuff happen quicker, just like having lots of friends helping makes dinner ready faster."See full answer
"Abstract class
A class that can have Abstract methods - without implementations and Concerete Methods i.e with implementation.
Can have private, protected and public access modifiers.
Supports Single inheritance i.e a class can extend only 1 abstract class
Can have constructors
Mainly used when sharing common behaviors
Interface Class
A collection of abstract methods ( can have static and default methods also - onwards of java 8)
Public, static, final are the access"
Sue G. - "Abstract class
A class that can have Abstract methods - without implementations and Concerete Methods i.e with implementation.
Can have private, protected and public access modifiers.
Supports Single inheritance i.e a class can extend only 1 abstract class
Can have constructors
Mainly used when sharing common behaviors
Interface Class
A collection of abstract methods ( can have static and default methods also - onwards of java 8)
Public, static, final are the access"See full answer
"--country names are UPPERCASE but the table in the in the question showing lowercase. That's why it took me a while to figure it out until I ran the country column
WITH RECURSIVE Hierarchy AS (
SELECT
e.Emp_ID,
CONCAT(e.FirstName, ' ', e.MiddleName, ' ', e.LastName) AS FullName,
e.Manager_ID,
0 AS Level,
CASE
WHEN e.Country = 'IRELAND' THEN s.Salary * 1.09
WHEN e.Country = 'INDIA' THEN s.Salary * 0.012
ELSE s.Salary
"
Victor N. - "--country names are UPPERCASE but the table in the in the question showing lowercase. That's why it took me a while to figure it out until I ran the country column
WITH RECURSIVE Hierarchy AS (
SELECT
e.Emp_ID,
CONCAT(e.FirstName, ' ', e.MiddleName, ' ', e.LastName) AS FullName,
e.Manager_ID,
0 AS Level,
CASE
WHEN e.Country = 'IRELAND' THEN s.Salary * 1.09
WHEN e.Country = 'INDIA' THEN s.Salary * 0.012
ELSE s.Salary
"See full answer
"find total sum. assign that to rightsum
traverse from left to right: keep updating left sum and right sum, when they match return the index.
else if you reach end return -1 or not found"
Rahul J. - "find total sum. assign that to rightsum
traverse from left to right: keep updating left sum and right sum, when they match return the index.
else if you reach end return -1 or not found"See full answer
"import string
from collections import defaultdict
def mostcommonwords(text):
d = defaultdict(int)
s = text.translate(str.maketrans('', '', string.punctuation))
for w in s.lower().split():
d[w] = d[w] + 1
return sorted(sorted(d.items()), reverse=True, key=lambda x: x[1])
`"
Michael S. - "import string
from collections import defaultdict
def mostcommonwords(text):
d = defaultdict(int)
s = text.translate(str.maketrans('', '', string.punctuation))
for w in s.lower().split():
d[w] = d[w] + 1
return sorted(sorted(d.items()), reverse=True, key=lambda x: x[1])
`"See full answer
"You can ask some clarifying questions like
1) Ask if the list is already sorted or not
2) is zero included in the list ?
3) Natural numbers are usually positive numbers ( clarify they are non negatives)
Solution :
1) If sorted use two pointers and sort them in O(N)
2) if not sorted , -ve / only +ve numbers in the list doesn't matter - the easiest solution is
Use a priority queue and push the number and its square in each iteration
Finally return the list returned by the priority Queue. N"
Bless M. - "You can ask some clarifying questions like
1) Ask if the list is already sorted or not
2) is zero included in the list ?
3) Natural numbers are usually positive numbers ( clarify they are non negatives)
Solution :
1) If sorted use two pointers and sort them in O(N)
2) if not sorted , -ve / only +ve numbers in the list doesn't matter - the easiest solution is
Use a priority queue and push the number and its square in each iteration
Finally return the list returned by the priority Queue. N"See full answer
"python:
def justifywords(wordslist, width):
result = []
currlinechar_count = 0
curr_words = []
for word in words_list:
if curr_words:
space_needed = len(word) + 1 # Space needed for the word and a preceding space
else:
space_needed = len(word)
if currlinecharcount + spaceneeded > width:
result.append(' '.join(curr_words))
curr_words = [word]
currlinechar_count = len("
Anonymous Unicorn - "python:
def justifywords(wordslist, width):
result = []
currlinechar_count = 0
curr_words = []
for word in words_list:
if curr_words:
space_needed = len(word) + 1 # Space needed for the word and a preceding space
else:
space_needed = len(word)
if currlinecharcount + spaceneeded > width:
result.append(' '.join(curr_words))
curr_words = [word]
currlinechar_count = len("See full answer