"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;
}
`"
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
"class ListNode:
def init(self, val=0, next=None):
self.val = val
self.next = next
def has_cycle(head: ListNode) -> bool:
slow, fast = head, head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False
debug your code below
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)
creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"
Anonymous Roadrunner - "class ListNode:
def init(self, val=0, next=None):
self.val = val
self.next = next
def has_cycle(head: ListNode) -> bool:
slow, fast = head, head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False
debug your code below
node1 = ListNode(1)
node2 = ListNode(2)
node3 = ListNode(3)
node4 = ListNode(4)
creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"See full answer
"const ops = {
'+': (a, b) => a+b,
'-': (a, b) => a-b,
'/': (a, b) => a/b,
'': (a, b) => ab,
};
function calc(expr) {
// Search for + or -
for (let i=expr.length-1; i >= 0; i--) {
const char = expr.charAt(i);
if (['+', '-'].includes(char)) {
return opschar), calc(expr.slice(i+1)));
}
}
// Search for / or *
for (let i=expr.length-1; i >= 0; i--) {
const char = expr.charAt(i);
if"
Tiago R. - "const ops = {
'+': (a, b) => a+b,
'-': (a, b) => a-b,
'/': (a, b) => a/b,
'': (a, b) => ab,
};
function calc(expr) {
// Search for + or -
for (let i=expr.length-1; i >= 0; i--) {
const char = expr.charAt(i);
if (['+', '-'].includes(char)) {
return opschar), calc(expr.slice(i+1)));
}
}
// Search for / or *
for (let i=expr.length-1; i >= 0; i--) {
const char = expr.charAt(i);
if"See full answer
"from typing import List
def two_sum(nums: List[int], target: int) -> List[int]:
prevMap = {}
for i, n in enumerate(nums):
diff = target - n
if diff in prevMap:
return [prevMap[diff], i]
else:
prevMap[n] = i
return []
debug your code below
print(two_sum([2, 7, 11, 15], 9))
`"
Anonymous Roadrunner - "from typing import List
def two_sum(nums: List[int], target: int) -> List[int]:
prevMap = {}
for i, n in enumerate(nums):
diff = target - n
if diff in prevMap:
return [prevMap[diff], i]
else:
prevMap[n] = i
return []
debug your code below
print(two_sum([2, 7, 11, 15], 9))
`"See full answer
Machine Learning Engineer
Data Structures & Algorithms
+5 more
🧠 Want an expert answer to a question? Saving questions lets us know what content to make next.
"Hey Grandma, you've had a lot of experience with infants, haven't you? When they were babies, you taught them how to chew in their first six months. This initial phase is like giving them data. Once they learned how to chew, they could handle any food you gave them. Next, you refined their learning by teaching them that they should only chew on food. This is like refining the data so they understand what is relevant. Then, a few months later, they started crawling and walking, learning by observ"
Hari priya K. - "Hey Grandma, you've had a lot of experience with infants, haven't you? When they were babies, you taught them how to chew in their first six months. This initial phase is like giving them data. Once they learned how to chew, they could handle any food you gave them. Next, you refined their learning by teaching them that they should only chew on food. This is like refining the data so they understand what is relevant. Then, a few months later, they started crawling and walking, learning by observ"See full answer
"A much better solution than the one in the article, below:
It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods.
shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns.
My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N))
class ListNode {
constructor(val = 0, next = null) {
th"
Guilherme F. - "A much better solution than the one in the article, below:
It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods.
shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns.
My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N))
class ListNode {
constructor(val = 0, next = null) {
th"See full answer
"
O(n) time, O(1) space
from typing import List
def maxsubarraysum(nums: List[int]) -> int:
if len(nums) == 0:
return 0
maxsum = currsum = nums[0]
for i in range(1, len(nums)):
currsum = max(currsum + nums[i], nums[i])
maxsum = max(currsum, max_sum)
return max_sum
debug your code below
print(maxsubarraysum([-1, 2, -3, 4]))
`"
Rick E. - "
O(n) time, O(1) space
from typing import List
def maxsubarraysum(nums: List[int]) -> int:
if len(nums) == 0:
return 0
maxsum = currsum = nums[0]
for i in range(1, len(nums)):
currsum = max(currsum + nums[i], nums[i])
maxsum = max(currsum, max_sum)
return max_sum
debug your code below
print(maxsubarraysum([-1, 2, -3, 4]))
`"See full answer
"from typing import List
def traprainwater(height: List[int]) -> int:
if not height:
return 0
l, r = 0, len(height) - 1
leftMax, rightMax = height[l], height[r]
res = 0
while l < r:
if leftMax < rightMax:
l += 1
leftMax = max(leftMax, height[l])
res += leftMax - height[l]
else:
r -= 1
rightMax = max(rightMax, height[r])
"
Anonymous Roadrunner - "from typing import List
def traprainwater(height: List[int]) -> int:
if not height:
return 0
l, r = 0, len(height) - 1
leftMax, rightMax = height[l], height[r]
res = 0
while l < r:
if leftMax < rightMax:
l += 1
leftMax = max(leftMax, height[l])
res += leftMax - height[l]
else:
r -= 1
rightMax = max(rightMax, height[r])
"See full answer