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| 1 | +# Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. |
| 2 | +# An input string is valid if: |
| 3 | +# Open brackets must be closed by the same type of brackets. |
| 4 | +# Open brackets must be closed in the correct order. |
| 5 | +# Every close bracket has a corresponding open bracket of the same type. |
| 6 | + |
| 7 | +# Example 1: |
| 8 | +# Input: s = "()" |
| 9 | +# Output: true |
| 10 | + |
| 11 | +# Example 2: |
| 12 | +# Input: s = "()[]{}" |
| 13 | +# Output: true |
| 14 | + |
| 15 | +# Example 3: |
| 16 | +# Input: s = "(]" |
| 17 | +# Output: false |
| 18 | + |
| 19 | +# Example 4: |
| 20 | +# Input: s = "([])" |
| 21 | +# Output: true |
| 22 | + |
| 23 | +# Constraints: |
| 24 | +# 1 <= s.length <= 104 |
| 25 | +# s consists of parentheses only '()[]{}'. |
| 26 | + |
| 27 | +# ------------------------------------------------------------------------------------------------------------------------------------------------- |
| 28 | + |
| 29 | +# STACK: |
| 30 | + |
| 31 | +# We can use a stack to store characters. Iterate through the string. For an opening bracket, push it onto the stack. |
| 32 | +# If the bracket is a closing type, check for the corresponding opening bracket at the top of the stack. |
| 33 | +# If we don't find the corresponding opening bracket, immediately return false. |
| 34 | +# This works because in a valid parenthesis expression, every opening bracket must have a corresponding closing bracket. |
| 35 | +# The stack is used to process the valid string, and it should be empty after the entire process. |
| 36 | +# This ensures that there is a valid substring between each opening and closing bracket. |
| 37 | + |
| 38 | +# create an empty list(stack) |
| 39 | +# create a hashmap manually; where close brackets are keys and open are values |
| 40 | +# iterate through the string |
| 41 | +# if we encounter a closed bracket, check if the stack9(stack) is not null and the top element of the stack(stack[-1]) is equal to the value of the key we just encountered(opening bracket of the closing bracket we just encountered) |
| 42 | +# if so, pop the bracket from the stack |
| 43 | +# if we encounter an open bracket, add it to the stack(append) |
| 44 | +# if stack is empty at the end, it means all brackets matched, so parenthesis are valied- return True |
| 45 | +# if stack is not empty, return False |
| 46 | + |
| 47 | + |
| 48 | +class Solution: |
| 49 | + def isValid(self, s: str) -> bool: |
| 50 | + |
| 51 | + closeToOpen = {')':'(', ']':'[', '}':'{'} |
| 52 | + stack = [] |
| 53 | + |
| 54 | + for c in s: |
| 55 | + if c in closeToOpen: |
| 56 | + if stack and stack[-1] == closeToOpen[c]: |
| 57 | + stack.pop() |
| 58 | + else: |
| 59 | + return False |
| 60 | + else: |
| 61 | + stack.append(c) |
| 62 | + |
| 63 | + return True if not stack else False |
| 64 | + |
| 65 | +Time Complexity = O(n) |
| 66 | +Space Complexity = O(n) |
| 67 | + |
| 68 | +# Companies: |
| 69 | +# Amazon- 18 |
| 70 | +# Google- 17 |
| 71 | +# Bloomberg- 14 |
| 72 | +# Meta- 13 |
| 73 | +# Microsoft- 7 |
| 74 | +# LLinkedIn- 7 |
| 75 | +# Apple- 6 |
| 76 | +# Intuit- 4 |
| 77 | +# Turing- 4 |
| 78 | +# IBM- 3 |
| 79 | +# Tiktok- 5 |
| 80 | +# TCS- 4 |
| 81 | +# Zoho- 4 |
| 82 | +# Tripadvisor- 4 |
| 83 | +# Walmart Labs- 3 |
| 84 | +# eBay- 3 |
| 85 | +# Blackrock- 3 |
| 86 | +# Infosys- 2 |
| 87 | +# Oracle- 2 |
| 88 | +# Epic Systems- 2 |
| 89 | +# Yandex- 36 |
| 90 | +# Adobe- 34 |
| 91 | +# Uber- 27 |
| 92 | +# Yahoo- 20 |
| 93 | +# JP Morgan- 18 |
| 94 | +# ServiceNow- 12 |
| 95 | +# Ozon- 10 |
| 96 | +# Accenture- 9 |
| 97 | +# EPAM Systems- 8 |
| 98 | +# Goldman Sachs- 8 |
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