## Remove K Digits

Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible. Note: The length of num is less than 10002 and will be ≥ k. The given num does not contain any leading zero. Example 1: Input: num = "1432219", k = 3 Output: "1219" Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.

by lek tin in "algorithm" access_time 2-min read

## Backspace String Compare

Solution (build string ❌) Time: O(N) Space: O(N) class Solution: def backspaceCompare(self, S: str, T: str) -> bool: res_1 = self.helper(S) res_2 = self.helper(T) return len(res_1) == len(res_2) and res_1 == res_2 def helper(self, s): stack = [] for c in s: if c != "#": stack.append(c) else: if len(stack) > 0: stack.pop() return "".join(stack) s Solution (two pointers 👍🏼) Time: O(N) Space: O(1) from itertools import zip_longest class Solution: def backspaceCompare(self, S: str, T: str) -> bool: return all( a == b for a, b in zip_longest(self.

by lek tin in "algorithm" access_time 1-min read

## Decode String

Given an encoded string, return its decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k.

by lek tin in "algorithm" access_time 1-min read

## Validate Stack Sequence

Given two sequences pushed and popped with distinct values, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack. Example 1 Input: pushed = [1,2,3,4,5], popped = [4,5,3,2,1] Output: true Explanation: We might do the following sequence: push(1), push(2), push(3), push(4), pop() -> 4, push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1 Example 2 Input: pushed = [1,2,3,4,5], popped = [4,3,5,1,2] Output: false Explanation: 1 cannot be popped before 2.

by lek tin in "algorithm" access_time 1-min read

## Remove Duplicate Letters

Given a string which contains only lowercase letters, remove duplicate letters so that every letter appears once and only once. You must make sure your result is the smallest in lexicographical order among all possible results. Example 1 Input: "bcabc" Output: "abc" Example 2 Input: "cbacdcbc" Output: "acdb" Example 3 Input: "cba" Output: "cba" Solution class Solution: def removeDuplicateLetters(self, s: str) -> str: stack = [] seen = set() last_indices = {c: i for i, c in enumerate(s)} for i, c in enumerate(s): print(c, seen, stack) if c not in seen: while stack and c < stack[-1] and i < last_indices[stack[-1]]: seen.

by lek tin in "algorithm" access_time 1-min read

## Evaluate Reverse Polish Notation

Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Note Division between two integers should truncate toward zero. The given RPN expression is always valid. That means the expression would always evaluate to a result and there won’t be any divide by zero operation. Example 1 Input: ["2", "1", "+", "3", "*"] Output: 9 Explanation: ((2 + 1) * 3) = 9 Example 2 Input: ["4", "13", "5", "/", "+"] Output: 6 Explanation: (4 + (13 / 5)) = 6 Example 3 Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"] Output: 22 Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5 = ((10 * (6 / (12 * -11))) + 17) + 5 = ((10 * (6 / -132)) + 17) + 5 = ((10 * 0) + 17) + 5 = (0 + 17) + 5 = 17 + 5 = 22 Solution class Solution: def evalRPN(self, tokens: List[str]) -> int: if not tokens: return None stack = [] operators = set(["+", "-", "*", "/"]) N = len(tokens) curr = 0 while curr < N: while curr < N and tokens[curr] not in operators: stack.

by lek tin in "algorithm" access_time 2-min read

## Daily Temperatures

Given a list of daily temperatures T, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead. For example, given the list of temperatures T = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].

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## Minimum Remove to Make Valid Parentheses

Given a string s of '(' , ')' and lowercase English characters. Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string. Formally, a parentheses string is valid if and only if: It is the empty string, contains only lowercase characters, or It can be written as AB (A concatenated with B), where A and B are valid strings, or It can be written as (A), where A is a valid string.

by lek tin in "algorithm" access_time 2-min read

## Longest Valid Parentheses

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring. Example 1 Input: "(()" Output: 2 Explanation: The longest valid parentheses substring is "()" Example 2 Input: ")()())" Output: 4 Explanation: The longest valid parentheses substring is "()()" Solution Dynamic programming // Time: `O(n)` // Space: `O(n)` public class Solution { public int longestValidParentheses(String s) { int maxans = 0; // dp array: ith element of dp represents the length of the longest valid substring ending at ith index.

by lek tin in "algorithm" access_time 2-min read

## Prefix Postfix Conversion

Prefix : An expression is called the prefix expression if the operator appears in the expression before the operands. Simply of the form (operator operand1 operand2). Example : *+AB-CD (Infix : (A+B) * (C-D) ) Postfix: An expression is called the postfix expression if the operator appears in the expression after the operands. Simply of the form (operand1 operand2 operator). Example : AB+CD-* (Infix : (A+B * (C-D) ) Given a Prefix expression, convert it into a Postfix expression.

by lek tin in "algorithm" access_time 2-min read