InfyTQ 2019 : Find the position from where the parenthesis is not balanced
Given a string str consisting of parenthesis from [ “(” , “)” , “{” , “}” , “[” , “]” ].
If the String is perfectly balanced return 0 else return the index(starting from 1)at which the nesting is found to be wrong.
Examples:
Input : str = “{[()]}[]”
Output : 0Input : str = “[{]()}”
Output : 3Input : str = “}[{}]”
Output : 1
Input : str = “{([]){”
Output : 7
Approach:
- First we are creating the lst1 and lst2 that will have the opening and closing brackets respectively
- Now we will create another list lst that will work as a Stack means if any closing brackets will appear in the string then the respective opening brackets present in the lst will be deleted
- If any closing brackets will appear in the string and no repective opening brackets is present in the lst then this will be the bad index we will return it
- If we reach at the end of the string while traversing and the size of the lst is not equals to 0 we will return the len(String)+1
Note : We are using python dictionary to get the respective opening brackets for a closing one.
Below is the impementation of the above approach
# Write Python3 code here # Defining the string string = "{[()]}[]" # Storing opening braces in list lst1 lst1 =['{', '(', '['] # Storing closing braces in list lst2 lst2 =['}', ')', ']'] # Creating an empty list lst lst =[] # Creating dictionary to map # closing braces to opening ones Dict ={ ')':'(', '}':'{', ']':'['} a = b = c = 0 # If first position of string contain # any closing braces return 1 if string[0] in lst2: print(1) else: # If characters of string are opening # braces then append them in a list for i in range(0, len(string)): if string[i] in lst1: lst.append(string[i]) k = i + 2 else: # When size of list is 0 and new closing # braces is encountered then print its # index starting from 1 if len(lst)== 0 and (string[i] in lst2): print(i + 1) c = 1 break else: # As we encounter closing braces we map # them with theircorresponding opening # braces using dictionary and check # if it is same as last opened braces #(last element in list) if yes then we # delete that elememt from list if Dict[string[i]]== lst[len(lst)-1]: lst.pop() else: # Otherwise we return the index # (starting from 1) at which # nesting is found wrong print(i + 1) a = 1 break # At end if the list is empty it # means the string is perfectly nested if len(lst)== 0 and c == 0: print(0) b = 1 if a == 0 and b == 0 and c == 0: print(k) |
Output:
0
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