"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
Data Engineer
Data Structures & Algorithms
+4 more
🧠 Want an expert answer to a question? Saving questions lets us know what content to make next.
"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