"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
"This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array.
We are not going to mo"
Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array.
We are not going to mo"See full answer
"public class sample {
public int [] merge(int [] a, int [] b)
{
if(a == null || a.length == 0 || b == null || b.length == 0) return null;
int i = 0, j = 0, index = -1;
int [] merged = new int[a.length + b.length];
while (i < a.length && j < b.length)
{
if(a[i] < b[i]) merged[++index] = a[i++];
else merged[++index] = b[j++];
}
while (i < a.length)
{
merged[++index] = a[i++];
}
"
Nikhil R. - "public class sample {
public int [] merge(int [] a, int [] b)
{
if(a == null || a.length == 0 || b == null || b.length == 0) return null;
int i = 0, j = 0, index = -1;
int [] merged = new int[a.length + b.length];
while (i < a.length && j < b.length)
{
if(a[i] < b[i]) merged[++index] = a[i++];
else merged[++index] = b[j++];
}
while (i < a.length)
{
merged[++index] = a[i++];
}
"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
Data Engineer
Data Structures & Algorithms
+4 more
🧠Want an expert answer to a question? Saving questions lets us know what content to make next.
"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
"function isPalindrome(s, start, end) {
while (s[start] === s[end] && end >= start) {
start++;
end--;
}
return end <= start;
}
function longestPalindromicSubstring(s) {
let longestPalindrome = '';
for (let i=0; i < s.length; i++) {
let j = s.length-1;
while (s[i] !== s[j] && i <= j) {
j--;
}
if (s[i] === s[j]) {
if (isPalindrome(s, i, j)) {
const validPalindrome = s.substring(i, j+1"
Tiago R. - "function isPalindrome(s, start, end) {
while (s[start] === s[end] && end >= start) {
start++;
end--;
}
return end <= start;
}
function longestPalindromicSubstring(s) {
let longestPalindrome = '';
for (let i=0; i < s.length; i++) {
let j = s.length-1;
while (s[i] !== s[j] && i <= j) {
j--;
}
if (s[i] === s[j]) {
if (isPalindrome(s, i, j)) {
const validPalindrome = s.substring(i, j+1"See full answer
"Make current as root.
2 while current is not null,
if p and q are less than current,
go left.
If p and q are greater than current,
go right.
else return current.
return null"
Vaibhav D. - "Make current as root.
2 while current is not null,
if p and q are less than current,
go left.
If p and q are greater than current,
go right.
else return current.
return null"See full answer
"Leetcode 347: Heap + Hashtable
Follow up question: create heap with the length of K instead of N (more time complexity but less space )"
Chen J. - "Leetcode 347: Heap + Hashtable
Follow up question: create heap with the length of K instead of N (more time complexity but less space )"See full answer
"import java.util.Arrays;
import java.util.stream.Collectors;
class Main
{
// Recursive function to print all combinations of numbers from \i\ to \n\
// having sum \n. The index\ denotes the next free slot in the output array \out\
public static void printCombinations(int i, int n, int[] out, int index)
{
// if the sum becomes n, print the combination
if (n == 0)
{
System.out.println(Arrays.stream(out).limit(index)
.boxed().collect(Collectors.toList()));
}
// start from the previous e"
Relynn may silver B. - "import java.util.Arrays;
import java.util.stream.Collectors;
class Main
{
// Recursive function to print all combinations of numbers from \i\ to \n\
// having sum \n. The index\ denotes the next free slot in the output array \out\
public static void printCombinations(int i, int n, int[] out, int index)
{
// if the sum becomes n, print the combination
if (n == 0)
{
System.out.println(Arrays.stream(out).limit(index)
.boxed().collect(Collectors.toList()));
}
// start from the previous e"See full answer
"The height of a binary tree is the maximum number of edges from the root node to any leaf node. To calculate the height of a binary tree, we can use a recursive approach. The basic idea is to compare the heights of the left and right subtrees of the root node, and return the maximum of them plus one."
Prashant Y. - "The height of a binary tree is the maximum number of edges from the root node to any leaf node. To calculate the height of a binary tree, we can use a recursive approach. The basic idea is to compare the heights of the left and right subtrees of the root node, and return the maximum of them plus one."See full answer