"This system design question is very small compared to other questions like design instagram, twitter, google drive etc... Since the design involves less components the level of detail we have to go in them were deep. I had to explain how to deal with all the NFR for the distributed cache system. Whether it is a push model or a pull model. Hade to do BOE calculations for the database too."
Jagan M. - "This system design question is very small compared to other questions like design instagram, twitter, google drive etc... Since the design involves less components the level of detail we have to go in them were deep. I had to explain how to deal with all the NFR for the distributed cache system. Whether it is a push model or a pull model. Hade to do BOE calculations for the database too."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
"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);
}
`"
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
"
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
Software Engineer
Data Structures & Algorithms
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