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Author: leetlab11

04-Stack/16-00020-valid-parentheses.py

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+# Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
+# An input string is valid if:
+# Open brackets must be closed by the same type of brackets.
+# Open brackets must be closed in the correct order.
+# Every close bracket has a corresponding open bracket of the same type.
+ 
+# Example 1:
+# Input: s = "()"
+# Output: true
+
+# Example 2:
+# Input: s = "()[]{}"
+# Output: true
+
+# Example 3:
+# Input: s = "(]"
+# Output: false
+
+# Example 4:
+# Input: s = "([])"
+# Output: true
+
+# Constraints:
+# 1 <= s.length <= 104
+# s consists of parentheses only '()[]{}'.
+
+# -------------------------------------------------------------------------------------------------------------------------------------------------
+
+# STACK:
+
+# We can use a stack to store characters. Iterate through the string. For an opening bracket, push it onto the stack. 
+# If the bracket is a closing type, check for the corresponding opening bracket at the top of the stack. 
+# If we don't find the corresponding opening bracket, immediately return false. 
+# This works because in a valid parenthesis expression, every opening bracket must have a corresponding closing bracket. 
+# The stack is used to process the valid string, and it should be empty after the entire process. 
+# This ensures that there is a valid substring between each opening and closing bracket.
+
+# create an empty list(stack)
+# create a hashmap manually; where close brackets are keys and open are values
+# iterate through the string
+# 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)
+# if so, pop the bracket from the stack
+# if we encounter an open bracket, add it to the stack(append)
+# if stack is empty at the end, it means all brackets matched, so parenthesis are valied- return True
+# if stack is not empty, return False
+
+
+class Solution:
+    def isValid(self, s: str) -> bool:
+
+        closeToOpen = {')':'(', ']':'[', '}':'{'}
+        stack = []
+
+        for c in s:
+            if c in closeToOpen:
+                if stack and stack[-1] == closeToOpen[c]:
+                    stack.pop()
+                else:
+                    return False
+            else:
+                stack.append(c)
+        
+        return True if not stack else False
+
+Time Complexity = O(n)
+Space Complexity = O(n)
+
+# Companies:
+# Amazon- 18
+# Google- 17
+# Bloomberg- 14
+# Meta- 13
+# Microsoft- 7
+# LLinkedIn- 7
+# Apple- 6
+# Intuit- 4
+# Turing- 4
+# IBM- 3
+# Tiktok- 5
+# TCS- 4
+# Zoho- 4
+# Tripadvisor- 4
+# Walmart Labs- 3
+# eBay- 3
+# Blackrock- 3
+# Infosys- 2
+# Oracle- 2
+# Epic Systems- 2
+# Yandex- 36
+# Adobe- 34
+# Uber- 27
+# Yahoo- 20
+# JP Morgan- 18
+# ServiceNow- 12
+# Ozon- 10
+# Accenture- 9
+# EPAM Systems- 8
+# Goldman Sachs- 8