"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
"
import java.util.*;
class Solution {
// Time Complexity: O(n^2)
// Space Complexity: O(n)
public static List> threeSum(int[] nums) {
// Ensure that the array is sorted first
Arrays.sort(nums);
// Create the results list to return
List> results = new ArrayList();
// Iterate over the length of nums
for (int i = 0; i < nums.length-2; i++) {
// We will have the first number in"
Victor O. - "
import java.util.*;
class Solution {
// Time Complexity: O(n^2)
// Space Complexity: O(n)
public static List> threeSum(int[] nums) {
// Ensure that the array is sorted first
Arrays.sort(nums);
// Create the results list to return
List> results = new ArrayList();
// Iterate over the length of nums
for (int i = 0; i < nums.length-2; i++) {
// We will have the first number in"See full answer
"
from typing import List
def getnumberof_islands(binaryMatrix: List[List[int]]) -> int:
if not binaryMatrix: return 0
rows = len(binaryMatrix)
cols = len(binaryMatrix[0])
islands = 0
for r in range(rows):
for c in range(cols):
if binaryMatrixr == 1:
islands += 1
dfs(binaryMatrix, r, c)
return islands
def dfs(grid, r, c):
if (
r = len(grid)
"
Rick E. - "
from typing import List
def getnumberof_islands(binaryMatrix: List[List[int]]) -> int:
if not binaryMatrix: return 0
rows = len(binaryMatrix)
cols = len(binaryMatrix[0])
islands = 0
for r in range(rows):
for c in range(cols):
if binaryMatrixr == 1:
islands += 1
dfs(binaryMatrix, r, c)
return islands
def dfs(grid, r, c):
if (
r = len(grid)
"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:
pass # your code goes here
if head is None:
return False
previousNodes = set()
iter = head
while iter:
if iter.val in previousNodes:
return True
previousNodes.add(iter.val)
iter = iter.next;
return False
debug your code below
node1 = ListNode(1)
node2 = ListNode(2)
n"
Cagdas A. - "
class ListNode:
def init(self, val=0, next=None):
self.val = val
self.next = next
def has_cycle(head: ListNode) -> bool:
pass # your code goes here
if head is None:
return False
previousNodes = set()
iter = head
while iter:
if iter.val in previousNodes:
return True
previousNodes.add(iter.val)
iter = iter.next;
return False
debug your code below
node1 = ListNode(1)
node2 = ListNode(2)
n"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