"
The Situation
A few months ago, our trading platform started experiencing significant latency issues during peak trading hours. This latency was affecting our ability to process real-time market data and execute trades efficiently, potentially leading to substantial financial losses and missed opportunities.
Identifying the Problem
The first step was to identify the root cause of the latency. I organized a team meeting with our data engineers, DevOps, and network specialists to gather"
Scott S. - "
The Situation
A few months ago, our trading platform started experiencing significant latency issues during peak trading hours. This latency was affecting our ability to process real-time market data and execute trades efficiently, potentially leading to substantial financial losses and missed opportunities.
Identifying the Problem
The first step was to identify the root cause of the latency. I organized a team meeting with our data engineers, DevOps, and network specialists to gather"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
Software Engineer
Data Structures & Algorithms
+4 more
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"First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently.
Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"
Stanley Y. - "First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently.
Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"See full answer
"In 2019, I was given a very important problem to solve. In a team of 3 we had to build a mobility assist device. The customer segment we would go for was something we could decide. The project was very close to me as I had lost someone I loved because of cancer and I saw how reduced mobility was a huge pain point in not being able to do physical activities. My team could only think of elderly people as the main target market.
As the Head of Product what I did was:
1) I helped them dive even d"
Soumya S. - "In 2019, I was given a very important problem to solve. In a team of 3 we had to build a mobility assist device. The customer segment we would go for was something we could decide. The project was very close to me as I had lost someone I loved because of cancer and I saw how reduced mobility was a huge pain point in not being able to do physical activities. My team could only think of elderly people as the main target market.
As the Head of Product what I did was:
1) I helped them dive even d"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
"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
"A few months ago I joined a micro-services platform engineering team as their manager, at that time my team was struggling to deliver towards an upcoming production deadline for a customer facing product. Production date had been moved 5 times already and there were about 40% of product features which were remaining to be tested and signed off to move to production . I was made responsible to deliver the release of this product within the deadline and turnaround the software delivery throughput."
Shuchi A. - "A few months ago I joined a micro-services platform engineering team as their manager, at that time my team was struggling to deliver towards an upcoming production deadline for a customer facing product. Production date had been moved 5 times already and there were about 40% of product features which were remaining to be tested and signed off to move to production . I was made responsible to deliver the release of this product within the deadline and turnaround the software delivery throughput."See full answer
"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
"#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
"public static void sortBinaryArray(int[] array) {
int len = array.length;
int[] res = new int[len];
int r=len-1;
for (int value : array) {
if(value==1){
res[r]= 1;
r--;
}
}
System.out.println(Arrays.toString(res));
}
`"
Nitin P. - "public static void sortBinaryArray(int[] array) {
int len = array.length;
int[] res = new int[len];
int r=len-1;
for (int value : array) {
if(value==1){
res[r]= 1;
r--;
}
}
System.out.println(Arrays.toString(res));
}
`"See full answer