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 : 0
Input : str = “[{]()}” 
Output : 3
Input : str = “}[{}]” 
Output : 1
Input : str = “{([]){” 
Output : 7 

Approach:

1. First we are creating the lst1 and lst2 that will have the opening and closing brackets respectively
2. 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
3. If any closing brackets will appear in the string and no respective opening brackets is present in the lst then this will be the bad index we will return it
4. 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

Below is the implementation of the above approach 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

int main()
{

    // Defining the string
    string String = "{[()]}[]";

    // Storing opening braces in list lst1
    vector<char> lst1 ={'{', '(', '['};

    // Storing closing braces in list lst2
    vector<char> lst2 ={'}', ')', ']'};

    // Creating an empty list lst
    vector<char> lst;

    int k;

    // Creating dictionary to map
    // closing braces to opening ones
    map<char,char> Dict;
    Dict.insert(pair<int, int>(')', '('));
    Dict.insert(pair<int, int>('}', '{'));
    Dict.insert(pair<int, int>(']', '['));

    int a = 0, b = 0, c = 0;

    // If first position of string contain
    // any closing braces return 1
    if(count(lst2.begin(), lst2.end(), String[0]))
    {
        cout << 1 << endl;
    }
    else
    {
        // If characters of string are opening
        // braces then append them in a list
        for(int i = 0; i < String.size(); i++)
        {
            if(count(lst1.begin(), lst1.end(), String[i]))
            {
                lst.push_back(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(lst.size()== 0 && (count(lst2.begin(), lst2.end(), String[i])))
                {
                    cout << (i + 1) << endl;
                    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 element from list
                    if(Dict[String[i]]== lst[lst.size()-1])
                    {
                        lst.pop_back();
                    }
                    else
                    {
                        // Otherwise we return the index
                        // (starting from 1) at which
                        // nesting is found wrong
                        break;
                        cout << (i + 1) << endl;
                        a = 1;
                    }
                }
             }
        }

        // At end if the list is empty it
        // means the string is perfectly nested
        if(lst.size() == 0 && c == 0)
        {
            cout << 0 << endl;
            b = 1;
        }

        if(a == 0 && b == 0 && c == 0)
        {
            cout << k << endl;
        }
    }

    return 0;
}

// This code is contributed by decode2207.

Java

// Java implementation of the approach
import java.util.*;
public class Main
{
    public static void main(String[] args)
    {

        // Defining the string
        String string = "{[()]}[]";

        // Storing opening braces in list lst1
        char[] lst1 = {'{', '(', '['};

        // Storing closing braces in list lst2
        char[] lst2 = {'}', ')', ']'};

        // Creating an empty list lst
        Vector<Character> lst = new Vector<Character>();

        // Creating dictionary to map
        // closing braces to opening ones
        HashMap<Character, Character> Dict = new HashMap<>();
        Dict.put(')', '(');
        Dict.put('}', '{');
        Dict.put(']', '[');

        int a = 0, b = 0, c = 0;

        // If first position of string contain
        // any closing braces return 1
        if(Arrays.asList(lst2).contains(string.charAt(0)))
        {
            System.out.println(1);
        }
        else
        {
            int k = 0;

            // If characters of string are opening
            // braces then append them in a list
            for(int i = 0; i < string.length(); i++)
            {
                if(Arrays.asList(lst1).contains(string.charAt(i)))
                {
                    lst.add(string.charAt(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(lst.size()== 0 && Arrays.asList(lst2).contains(string.charAt(i)))
                    {
                        System.out.println((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 element from list
                        if(lst.size() > 0 && Dict.get(string.charAt(i))== lst.get(lst.size() - 1))
                        {
                            lst.remove(lst.size() - 1);
                        }
                        else
                        {
                            // Otherwise we return the index
                            // (starting from 1) at which
                            // nesting is found wrong
                            a = 1;
                            break;
                        }
                    }
                 }
            }

            // At end if the list is empty it
            // means the string is perfectly nested
            if(lst.size() == 0 && c == 0)
            {
                System.out.println(0);
                b = 1;
            }

            if(a == 0 && b == 0 && c == 0)
            {
                System.out.println(k);
            }
        }
    }
}

// This code is contributed by divyesh072019.

Python3

# 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 element 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)

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
  static void Main()
  {

    // Defining the string
    string String = "{[()]}[]";

    // Storing opening braces in list lst1
    char[] lst1 = {'{', '(', '['};

    // Storing closing braces in list lst2
    char[] lst2 = {'}', ')', ']'};

    // Creating an empty list lst
    List<char> lst = new List<char>();

    // Creating dictionary to map
    // closing braces to opening ones
    Dictionary<char, char> Dict = new Dictionary<char, char>();
    Dict[')'] = '(';
    Dict['}'] = '{';
    Dict[']'] = '[';

    int a = 0, b = 0, c = 0;

    // If first position of string contain
    // any closing braces return 1
    if(Array.Exists(lst2, element => element == String[0]))
    {
        Console.WriteLine(1);
    }
    else
    {
        int k = 0;

        // If characters of string are opening
        // braces then append them in a list
        for(int i = 0; i < String.Length; i++)
        {
            if(Array.Exists(lst1, element => element == String[i]))
            {
                lst.Add(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(lst.Count== 0 && Array.Exists(lst2, element => element == String[i]))
                {
                    Console.WriteLine((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 element from list
                    if(lst.Count > 0 && Dict[String[i]]== lst[lst.Count - 1])
                    {
                        lst.RemoveAt(lst.Count - 1);
                    }
                    else
                    {
                        // Otherwise we return the index
                        // (starting from 1) at which
                        // nesting is found wrong
                        a = 1;
                        break;
                    }
                }
             }
        }

        // At end if the list is empty it
        // means the string is perfectly nested
        if(lst.Count == 0 && c == 0)
        {
            Console.WriteLine(0);
            b = 1;
        }

        if(a == 0 && b == 0 && c == 0)
        {
            Console.WriteLine(k);
        }
    }
  }
}

// This code is contributed by divyeshrabadiya07.

Javascript

<script>
    // Javascript implementation of the approach

    // Defining the string
    let string = "{[()]}[]";

    // Storing opening braces in list lst1
    let lst1 =['{', '(', '['];

    // Storing closing braces in list lst2
    let lst2 =['}', ')', ']'];

    // Creating an empty list lst
    let lst =[];

    // Creating dictionary to map
    // closing braces to opening ones
    let Dict ={ ')':'(', '}':'{', ']':'['}

    let a = 0, b = 0, c = 0;

    // If first position of string contain
    // any closing braces return 1
    if(string[0]  in lst2)
    {
        document.write(1 + "</br>");
    }
    else
    {
        // If characters of string are opening
        // braces then append them in a list
        for(let i = 0; i < string.length; i++)
        {
            if(string[i] in lst1)
            {
                lst.push(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(lst.length== 0 && (string[i] in lst2))
                {
                    document.write((i + 1) + "</br>");
                    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 element from list
                    if(Dict[string[i]]== lst[lst.length-1])
                    {
                        lst.pop();
                    }
                    else
                    {
                        // Otherwise we return the index
                        // (starting from 1) at which
                        // nesting is found wrong
                        break;
                        document.write((i + 1) + "</br>");
                        a = 1;
                    }
                }
             }
        }

        // At end if the list is empty it
        // means the string is perfectly nested
        if(lst.length == 0 && c == 0)
        {
            document.write(0 + "</br>");
            b = 1;
        }

        if(a == 0 && b == 0 && c == 0)
        {
            document.write(k + "</br>");
        }
    }

    // This code is contributed by rameshtravel07.
</script>

Output:

0