Check for balanced parentheses in an expression O(1) space O(N^2) time complexity in Python

Beyond Basic Programming - Intermediate Python

Most Popular

36 Lectures 3 hours

Mohammad Nauman

More Detail

Practical Machine Learning using Python

Best Seller

91 Lectures 23.5 hours

MANAS DASGUPTA

More Detail

Practical Data Science using Python

22 Lectures 6 hours

MANAS DASGUPTA

More Detail

Suppose we have a string str containing these brackets '(', ')', '{', '}', '[' and ']', we have to check whether brackets are balanced or not. We can say brackets are balanced when Opening and closing bracket types are of same type. Brackets are closed in correct order.

So, if the input is like {([])}, then the output will be True.

To solve this, we will follow these steps −

• cnt := 0
• i := 0
• j := -1
• Define a function solve() . This will take s, temp
• cnt := cnt - 1
• s := a new list from s
• if j > -1 and s[j] is same as temp, then
  • s[i] := '#'
  • s[j] := '#'
  • while j >= 0 and s[j] is same as '#', do
    • j := j - 1
  • i := i + 1>
  • return 1
• otherwise,
  • return 0
• From the main method, do the following −
• if size of s is same as 0, then
  • return True
• otherwise,
  • ans := False
  • while i < size of s is non-zero, do
    • if s[i] is same as '}', then
      • ans := solve(s, '{')
      • if ans is same as 0, then
        • return False
      • otherwise when s[i] is same as ')', then
        • ans := solve(s, '(')
        • if ans is same as 0, then
          • return False
      • otherwise when s[i] is same as ']', then
        • ans := solve(s, '[')
        • if ans is same as 0, then
          • return False
      • otherwise,
        • j := i
        • i := i + 1
        • cnt := cnt + 1
  • if cnt is not same as 0, then
    • return False
  • return True

Example

Let us see the following implementation to get better understanding −

 Live Demo

cnt = 0
i = 0
j = -1
def solve(s, temp):
   global i, j, cnt
   cnt -= 1
   s = list(s)
   if j > -1 and s[j] == temp:
      s[i] = '#'
      s[j] = '#'
      while j >= 0 and s[j] == '#':
         j -= 1
      i += 1
      return 1
   else:
      return 0
def bracketOrderCheck(s):
   global i, j, cnt
   if len(s) == 0:
      return True
   else:
      ans = False
      while i < len(s):
         if s[i] == '}':
            ans = solve(s, '{')
            if ans == 0:
               return False
         elif s[i] == ')':
            ans = solve(s, '(')
            if ans == 0:
               return False
         elif s[i] == ']':
            ans = solve(s, '[')
            if ans == 0:
               return False
         else:
            j = i
            i += 1
            cnt += 1
      if cnt != 0:
         return False
      return True
print(bracketOrderCheck("{([])}"))

Input

"{(()[])}"

Output

True

Arnab Chakraborty

Updated on 27-Aug-2020 13:04:35

• Related Questions & Answers
• Check for balanced parentheses in an expression - O(1) space - O(N^2) time complexity in C++
• Find median of BST in O(n) time and O(1) space in Python
• Find duplicates in O(n) time and O(1) extra space - Set 1 in C++
• Find median of BST in O(n) time and O(1) space in C++
• Find the maximum repeating number in O(n) time and O(1) extra space in Python
• Check for balanced parentheses in an expression in C++
• Print left rotation of array in O(n) time and O(1) space in C Program.
• Rearrange positive and negative numbers in O(n) time and O(1) extra space in C++
• Count Fibonacci numbers in given range in O(Log n) time and O(1) space in C++
• Check if array elements are consecutive in O(n) time and O(1) space (Handles Both Positive and negative numbers) in Python
• Find maximum in a stack in O(1) time and O(1) extra space in C++
• Count frequencies of all elements in array in O(1) extra space and O(n) time in C++
• Find duplicate in an array in O(n) and by using O(1) extra space in C++
• Check for balanced parentheses in Python
• Print n x n spiral matrix using O(1) extra space in C Program.