class Solution { public String decodeString(String s) { Stack<Integer> stNum = new Stack<>(); Stack<StringBuilder> stStr = new Stack<>(); StringBuilder currStr = new StringBuilder(); int currNum = 0; for(char ch : s.toCharArray()) { if(ch == '[') { stNum.push(currNum); stStr.push(currStr); currNum = 0; currStr = new StringBuilder(); } else if(ch == ']') { int num = stNum.pop(); StringBuilder prevStr = stStr.pop(); String s1 = currStr.toString(); currStr = prevStr.append(s1.repeat(num)); } else if(ch >= '0' && ch <= '9') currNum = currNum * 10 + ((int)ch - 48); else currStr.append(ch); } return currStr.toString(); } }

Great explanation

thnks

thanku mam

2d k problem jaldi samjh nhi aata...😩

Mam I solved that by using naive approach by my own but i understand efficient one by seeing a solution i am asking that when u solve that problem first time you can able to solve it efficient logic by own

Hey👋🏻🙂

Di nice explanation ❤

Di Your videos are always simple and helpful

From your linked in , wonderful series keep updating...

Meine bahut lecture dekh liye iss question per meroko smgj nhi aaya yeh question shi se ur iss video mein bhi smj nhi paa raha hoon

overall cost with video ?

its same like my office might be vary 😆😆 btw, watching after one two year

wonderful explanation...

Completed whole Playlist in 1 go. Waiting for more medium to hard array problems.🙂

Isme swapping ka code brute force mein chaiye tha... toh aur deep mein samjh aata

Mam daily ek vedio daal do practice ho jayega ek question per day

Ma'am paid mentoring chalu karo aap .. you teach very well I'll will be your first student.

Done mam you are amazing always ❤

TQ MAM 😊

❤

Hello Didi! please do solve today's lc. Not able to solve last 2 days daily questions🥲. Need ur help for sure🙏

Great ❤

If N is rows and M is columns, then it can be solved in M*N * LogN using heap. With space complexity of N

Lovely ❤

Great explanation even it is better than striver's explanation.

Thanks for the nice explanation. Can you please solve sum of subarray minimums in simple way.

This code gives wrong answer for aabccbb

Thank You Di,amazing explanation

Amazing didi

🙂👍🏻

Madam abh strings chalu karo..

Didi it astonishes me ki aap msft mai aakar bhi dsa grind on kr rkha hai. How to get this motivation. Mera toh office se nikalte hi mann hi nhi krta aur ye sab krneka. Want some tips

this entire series will be in java ?

nice solution 🫡, a bit of a funny part is "Hey guys, welcome all or welcome back to my channel". JK ! 😂😇

hey! can u solve today's lc problem. Couldn't understand what to do

Solute to your dedication Ayush 🫡

Your logic explanations are really good. Thank you. Please upload for more questions of leetcode.

Short and Simple expalanation❤❤❤

TC O(n) SC O(1) Day-Stout-Warren (DSW) algorithm Soln in Rust -> // Definition for a binary tree node. // #[derive(Debug, PartialEq, Eq)] // pub struct TreeNode { // pub val: i32, // pub left: Option<Rc<RefCell<TreeNode>>>, // pub right: Option<Rc<RefCell<TreeNode>>>, // } // // impl TreeNode { // #[inline] // pub fn new(val: i32) -> Self { // TreeNode { // val, // left: None, // right: None // } // } // } use std::{rc::Rc, cell::RefCell}; type T = Option<Rc<RefCell<TreeNode>>>; macro_rules! borrow_mut { ($option:expr) => { $option.as_ref().unwrap().borrow_mut() }; } macro_rules! borrow { ($option:expr) => { $option.as_ref().unwrap().borrow() }; } impl Solution { pub fn balance_bst(root: T) -> T { fn create_right_skewed_vine_tree(root: T) -> T { let vine_head = Some(Rc::new(RefCell::new(TreeNode::new(0)))); borrow_mut!(vine_head).right = root; let mut current = vine_head.clone(); while borrow!(current).right.is_some() { if borrow!(borrow!(current).right).left.is_some() { let right = borrow!(current).right.clone(); rotate_right(current.clone(), right); } else { current = current.unwrap().borrow().right.clone(); } } vine_head } fn get_right_skewed_vine_tree_node_count(mut root: T) -> i32 { let mut count = 0; while root.is_some() { count += 1; root = root.unwrap().borrow().right.clone(); } count } fn rotate_right(parent: T, node: T) { let temp = borrow!(node).left.clone(); borrow_mut!(node).left = borrow_mut!(temp).right.take(); borrow_mut!(temp).right = node; parent.unwrap().borrow_mut().right = temp; } fn rotate_left(parent: T, node: T) { let temp = borrow!(node).right.clone(); borrow_mut!(node).right = borrow_mut!(temp).left.take(); borrow_mut!(temp).left = node; parent.unwrap().borrow_mut().right = temp; } fn rotation_left_by_amount(mut root: T, count: i32) { for _ in 0..count { let right = borrow!(root).right.clone(); rotate_left(root.clone(), right); root = root.unwrap().borrow().right.clone(); } } //create vine let vine_head = create_right_skewed_vine_tree(root); // get total node count let nodecount = get_right_skewed_vine_tree_node_count(borrow_mut!(vine_head).right.clone()); // get node count of perfect tree let mut perfect_tree_node_count = 2_i32.pow(((nodecount + 1) as f32).log2().floor() as u32) - 1; //shift the extra nodes to the left rotation_left_by_amount(vine_head.clone(), nodecount - perfect_tree_node_count); // make rest of the tree while perfect_tree_node_count > 1 { perfect_tree_node_count /= 2; rotation_left_by_amount(vine_head.clone(), perfect_tree_node_count); } vine_head.unwrap().borrow_mut().right.take() } }

very nice explanation

It is a great explanation by the far the best, really thanks for making this hard question an easy one

Diff number ka xor explain nhi Kiya..like 3^4

Thanks for putting in such great efforts and time for creating such in depth explanation videos it helps a lot 😀 but the java code was not really clear to me i missed somethings in it if feasible for you can you please create a short video on that part

Hey 👋 Can you start Leetcode Contest discussion It will be more helpful

apne code provide nhi kiya...🥲

didi , but dp vali approach mei aapne to code dry run krwa diya. , BUT INTUITION KAHA SE MILI?? RECURSIVE CODE KAISE LIKHA?? wo sab bhi thoda bata dijiye, help hojaegi meri , please 🙏🙏

Didi, can we message you for a referral for posting at Walmart?

👍🏻👍🏻

Dedication + Hardwork + Discipline = Ayushi