"Problem: Given an input string txt consisting of alphanumeric characters and the parentheses characters '(' & ')', write a function which removes the minimum number of characters to return a version of the string with properly balanced parenthesis.
Answer: You can do this with a counter.
Psuedo-Python
Start with counter = 0
output = []
Iterate through the string, every time you encounter a '(', increment the counter. Add the character to the output.
If you encounter a ')', decrement the coun"
Michael B. - "Problem: Given an input string txt consisting of alphanumeric characters and the parentheses characters '(' & ')', write a function which removes the minimum number of characters to return a version of the string with properly balanced parenthesis.
Answer: You can do this with a counter.
Psuedo-Python
Start with counter = 0
output = []
Iterate through the string, every time you encounter a '(', increment the counter. Add the character to the output.
If you encounter a ')', decrement the coun"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
Software Engineer
Data Structures & Algorithms
+4 more
🧠 Want an expert answer to a question? Saving questions lets us know what content to make next.
"def find_first(array: List[int], num: int) -> int:
lo = 0
hi = len(array)-1
while lo = num:
hi = mid - 1
if lo == mid and array[mid] == num:
return mid
else:
array[mid] < num
lo = mid + 1
return -1
`"
Gabriele G. - "def find_first(array: List[int], num: int) -> int:
lo = 0
hi = len(array)-1
while lo = num:
hi = mid - 1
if lo == mid and array[mid] == num:
return mid
else:
array[mid] < num
lo = mid + 1
return -1
`"See full answer
"Count items between indices within compartments
compartments are delineated by by: '|'
items are identified by: '*'
input_inventory = "*||||"
inputstartidxs = [1, 4, 6]
inputendidxs = [9, 5, 8]
expected_output = [3, 0, 1]
Explanation:
"*||||"
0123456789... indices
++ + # within compartments
^ start_idx = 1
^ end_idx = 9
-- - # within idxs but not within compartments
"*||||"
0123456789... indices
"
Anonymous Unicorn - "Count items between indices within compartments
compartments are delineated by by: '|'
items are identified by: '*'
input_inventory = "*||||"
inputstartidxs = [1, 4, 6]
inputendidxs = [9, 5, 8]
expected_output = [3, 0, 1]
Explanation:
"*||||"
0123456789... indices
++ + # within compartments
^ start_idx = 1
^ end_idx = 9
-- - # within idxs but not within compartments
"*||||"
0123456789... indices
"See full answer
"Questions to ask :
Are there negative values in the input array? Interview : YES
Will the product of two number fit into 32-bit Integer. If not, will it fit 64-bit Integer. If not, then is it safe to use Big Integer? Interview : let's worry only about 32 bit Integer
What should we return if the input array is null or size (size of input array) is less than 2? Return 0
From above Information:
General approach is as follows :
a) Track smallest 2 elements in the array -> p"
Rajendra D. - "Questions to ask :
Are there negative values in the input array? Interview : YES
Will the product of two number fit into 32-bit Integer. If not, will it fit 64-bit Integer. If not, then is it safe to use Big Integer? Interview : let's worry only about 32 bit Integer
What should we return if the input array is null or size (size of input array) is less than 2? Return 0
From above Information:
General approach is as follows :
a) Track smallest 2 elements in the array -> p"See full answer
"from typing import List
def three_sum(nums: List[int]) -> List[List[int]]:
nums.sort()
triplets = set()
for i in range(len(nums) - 2):
firstNum = nums[i]
l = i + 1
r = len(nums) - 1
while l 0:
r -= 1
elif potentialSum < 0:
l += 1
"
Anonymous Roadrunner - "from typing import List
def three_sum(nums: List[int]) -> List[List[int]]:
nums.sort()
triplets = set()
for i in range(len(nums) - 2):
firstNum = nums[i]
l = i + 1
r = len(nums) - 1
while l 0:
r -= 1
elif potentialSum < 0:
l += 1
"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:
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
"
from typing import Dict, List, Optional
def max_profit(prices: Dict[str, int]) -> Optional[List[str]]:
pass # your code goes here
max = [None, 0]
min = [None, float("inf")]
for city, price in prices.items():
if price > max[1]:
max[0], max[1] = city, price
if price 0:
return [min[0], max[0]]
return None
debug your code below
prices = {'"
Rick E. - "
from typing import Dict, List, Optional
def max_profit(prices: Dict[str, int]) -> Optional[List[str]]:
pass # your code goes here
max = [None, 0]
min = [None, float("inf")]
for city, price in prices.items():
if price > max[1]:
max[0], max[1] = city, price
if price 0:
return [min[0], max[0]]
return None
debug your code below
prices = {'"See full answer
"You are given a string S and a number K as input, and your task is to print S to console output considering that, at most, you can print K characters per line.
Example:
S = "abracadabra sample"
K = 11
Output:
abracadabra
sample
Note that this problem requires the interviewee gather extra requirements from the interviewer (e.g. do we care about multiple white spaces? what if the length of a word is greater than K, ...)"
B. T. - "You are given a string S and a number K as input, and your task is to print S to console output considering that, at most, you can print K characters per line.
Example:
S = "abracadabra sample"
K = 11
Output:
abracadabra
sample
Note that this problem requires the interviewee gather extra requirements from the interviewer (e.g. do we care about multiple white spaces? what if the length of a word is greater than K, ...)"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