(Leetcode) Valid parentheses

Question

This is a Leetcode problem -

Given a string 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.

Note that an empty string is also considered valid.

Example 1 -

Input: "()"
Output: True

Example 2 -

Input: "()[]{}"
Output: True

Example 3 -

Input: "(]"
Output: False

Example 4 -

Input: "([)]"
Output: False

Example 5 -

Input: "{[]}"
Output: True

---

Here is my solution to this challenge -

def is_valid(s):
 if len(s) == 0:
 return True
 parentheses = ['()', '[]', '{}']
 flag = False
 while len(s) > 0:
 i = 0
 while i < 3:
 if parentheses[i] in s:
 s = s.replace(parentheses[i], '')
 i = 0
 flag = True
 else:
 i += 1
 if len(s) == 0:
 return True
 else:
 flag = False
 break
 return False

So I would like to know whether I could improve performance and make my code shorter.

• moondra: Is there a reason you are using flags? I removed the flags and it still seems to work, unless I made a mistake. Thank you.

Answer 1

Not a performance suggestion, but you can make use of the fact that an empty collection is Falsey.

if len(s) == 0:

Is functionally the same as just:

if not s:

And similarly,

while len(s) > 0:

Can be just:

while s:

Relevant PEP entry (search for "For sequences" under the linked heading).

Answer 2

You can make your code shorter and much faster by using stack -

Stack works on the principle of \$"\$Last-in, first-out \$"\$. Also, the inbuilt functions in Python make the code short and simple. To add an item to the top of the list, i.e., to push an item, we use the append() function and to pop out an element we use the pop() function. These functions work quite efficiently and fast in end operations.

Here's a visual representation -

enter image description here

^{Source - https://www.geeksforgeeks.org/stack-data-structure/}

Here's a shorter and faster version of your program using stack -

def is_valid(s):
 stack = []
 mapping = {
 ")" : "(",
 "}" : "{",
 "]" : "["
 }
 if len(s) != 0: 
 for char in s: 
 if char in mapping: 
 if len(stack) == 0: 
 return False
 else: 
 top_element = stack.pop() 
 if top_element != mapping[char]:
 return False
 else:
 stack.append(char)
 return len(stack) == 0
 return True

Let's compare Leetcode timings (76 test cases) -

Your Leetcode result -

enter image description here

---

Leetcode result for stack solution (76 test cases) -

enter image description here

Also, you don't need any flag here (Thanks to @moondra for pointing it out) -

parentheses = ['()', '[]', '{}']
flag = False
while len(s) > 0:
 # rest of the code

Or here -

s = s.replace(parentheses[i], '')
i = 0
flag = True

Or here -

 else:
 flag = False
 break
return False

The program will work without the flags.

---

• 301_Moved_Permanently: There is no need to check for len(s) == 0 as, if it is the case, the for loop wouldn't execute and len(stack) would be 0. Also, if not stack: inside the loop feels more pythonic.
• Justin: @MathiasEttinger - Could you include this in an answer? This seems interesting.