Check if the Depth of Parentheses is correct in the given String

Given a string S consisting of opening parentheses, closing parentheses and integers, the task is to print Yes if, in this string, the integers correctly indicates its depth. Depth refers to the number of nested sets of parentheses surrounding that integer. Print No otherwise.
Examples:

Input: S = “((2)((3)))” 
Output: Yes 
Explanation: 
No of Opening & Closing Parentheses around 2 = 2 
No of Opening & Closing Parentheses around 3 = 3 
Input: S = “((35)(2))” 
Output: No 
Explanation: 
No of Opening & Closing Parentheses around 35 = 2 
No of Opening & Closing Parentheses around 2 = 2 

Approach:

1. Iterate the string character by character
2. If it is opening or closing parenthesis then append into an array arr[]
3. If the character is digit, then iterate till the entire number is read and then append into the arr[]
4. Iterate the array arr[] element by element 
   • If the element is ‘(‘ increment depth and open by 1
   • If the element is ‘)’ decrement depth by 1 and increment close by 1
   • If the element is an Integer, check if “Integer != depth”, if true 
     set flag to 0 and break the loop
5. After iterating the entire string, check if open not equals to close, if true set flag = 0

Below is the implementation of the above approach 

C++

// Function to check if the Depth
// of Parentheses is correct
// in the given String

#include <bits/stdc++.h>
using namespace std;

bool Formatted(string s)
{
    vector<char> k;
    int i = 0;

    while (i < s.size())
    {
        // Appending if the
        // Character is not integer
        if (s[i]==')' or s[i]=='(')
        {
            k.push_back(s[i]);
            i += 1;
        }
        else
        {
            // Iterating till the entire
            // Digit is read
            char st;
            while (s[i]!=')' and s[i]!=')')
            {
                st = s[i];
                i = i + 1;
            }
            k.push_back(st);
        }
    }

    int depth = 0, flag = 1;
    int open = 0, close = 0;

    for (char i:k)
    {

        // Check if character is '('
        if (i == '(')
        {
            // Increment depth by 1
            depth += 1;

            // Increment open by 1
            open += 1;

        // Check if character is ')'
    }else if (i == ')'){

            // Decrement depth by 1
            depth -= 1;

            // Increment close by 1
            close += 1;
        }
        else{
            if (i-'0' != depth){
                flag = 0;
                break;
            }
        }
    }

    // Check if open parentheses
    // NOT equals close parentheses
    if (open != close)
        flag = 0;

    return (flag == 1)?true:false;
}

// Driver Code
int main()
{
    string s = "((2)((3)))";
    bool k = Formatted(s);
    if (k == true)
        printf("Yes");
    else
        printf("No");
    return 0;
}

// This code is contributed by mohit kumar 29

Java

// Function to check if the Depth
// of Parentheses is correct
// in the given String
import java.util.*;

class GFG{

static boolean Formatted(String s)
{
    Vector<Character> k = new Vector<Character>();
    int i = 0;

    while (i < s.length())
    {
        // Appending if the
        // Character is not integer
        if (s.charAt(i)==')' || s.charAt(i)=='(')
        {
            k.add(s.charAt(i));
            i += 1;
        }
        else
        {
            // Iterating till the entire
            // Digit is read
            char st = 0;
            while (s.charAt(i)!=')' && s.charAt(i)!=')')
            {
                st = s.charAt(i);
                i = i + 1;
            }
            k.add(st);
        }
    }

    int depth = 0, flag = 1;
    int open = 0, close = 0;

    for (char i2 : k)
    {

        // Check if character is '('
        if (i2 == '(')
        {
            // Increment depth by 1
            depth += 1;

            // Increment open by 1
            open += 1;

        // Check if character is ')'
    }else if (i2 == ')'){

            // Decrement depth by 1
            depth -= 1;

            // Increment close by 1
            close += 1;
        }
        else{
            if (i2-'0' != depth){
                flag = 0;
                break;
            }
        }
    }

    // Check if open parentheses
    // NOT equals close parentheses
    if (open != close)
        flag = 0;

    return (flag == 1)?true:false;
}

// Driver Code
public static void main(String[] args)
{
    String s = "((2)((3)))";
    boolean k = Formatted(s);
    if (k == true)
        System.out.printf("Yes");
    else
        System.out.printf("No");
}
}

// This code is contributed by Rajput-Ji

Python3

# Function to check if the Depth
# of Parentheses is correct
# in the given String

def Formatted(s):
    k = []
    i = 0

    while i < len(s):

        # Appending if the
        # Character is not integer
        if s[i].isdigit() == False:

            k.append(s[i])
            i += 1
        else:

            # Iterating till the entire
            # Digit is read
            st = ""
            while s[i].isdigit():
                st += s[i]
                i = i + 1
            k.append(int(st))

    depth, flag = 0, 1
    open, close = 0, 0

    for i in k:

        # Check if character is '('
        if i == '(':

            # Increment depth by 1
            depth += 1

            # Increment open by 1
            open += 1

        # Check if character is ')'
        elif i == ')':

            # Decrement depth by 1
            depth -= 1

            # Increment close by 1
            close += 1
        else:
            if i != depth:
                flag = 0
                break

    # Check if open parentheses
    # NOT equals close parentheses
    if open != close:
        flag = 0

    return True if flag == 1 else False

# Driver Code
if __name__ == '__main__':
    s = '((2)((3)))'
    k = Formatted(s)
    if k == True:
        print("Yes")
    else:
        print("No")

C#

// Function to check if the Depth
// of Parentheses is correct
// in the given String
using System;
using System.Collections.Generic;

class GFG
{

static bool Formatted(String s)
{
    List<char> k = new List<char>();
    int i = 0;

    while (i < s.Length)
    {
        // Appending if the
        // char is not integer
        if (s[i]==')' || s[i]=='(')
        {
            k.Add(s[i]);
            i += 1;
        }
        else
        {
            // Iterating till the entire
            // Digit is read
            char st = '\x0000';
            while (s[i] != ')' && s[i] != ')')
            {
                st = s[i];
                i = i + 1;
            }
            k.Add(st);
        }
    }

    int depth = 0, flag = 1;
    int open = 0, close = 0;

    foreach (char i2 in k)
    {

        // Check if character is '('
        if (i2 == '(')
        {
            // Increment depth by 1
            depth += 1;

            // Increment open by 1
            open += 1;

        // Check if character is ')'
    }else if (i2 == ')'){

            // Decrement depth by 1
            depth -= 1;

            // Increment close by 1
            close += 1;
        }
        else{
            if (i2-'0' != depth){
                flag = 0;
                break;
            }
        }
    }

    // Check if open parentheses
    // NOT equals close parentheses
    if (open != close)
        flag = 0;

    return (flag == 1)?true:false;
}

// Driver Code
public static void Main(String[] args)
{
    String s = "((2)((3)))";
    bool k = Formatted(s);
    if (k == true)
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Function to check if the Depth
// of Parentheses is correct
// in the given String
function Formatted(s)
{
    let k = [];
    let i = 0;

    while (i < s.length)
    {
        // Appending if the
        // Character is not integer
        if (s[i]==')' || s[i]=='(')
        {
            k.push(s[i]);
            i += 1;
        }
        else
        {
            // Iterating till the entire
            // Digit is read
            let st = 0;
            while (s[i]!=')' && s[i]!=')')
            {
                st = s[i];
                i = i + 1;
            }
            k.push(st);
        }
    }

    let depth = 0, flag = 1;
    let open = 0, close = 0;

    for (let i2=0;i2< k.length;i2++)
    {

        // Check if character is '('
        if (k[i2] == '(')
        {
            // Increment depth by 1
            depth += 1;

            // Increment open by 1
            open += 1;

        // Check if character is ')'
    }else if (k[i2] == ')'){

            // Decrement depth by 1
            depth -= 1;

            // Increment close by 1
            close += 1;
        }
        else{
            if (k[i2]-'0' != depth){
                flag = 0;
                break;
            }
        }
    }

    // Check if open parentheses
    // NOT equals close parentheses
    if (open != close)
        flag = 0;

    return (flag == 1)?true:false;
}

// Driver Code
let s = "((2)((3)))";
let k = Formatted(s);
if (k == true)
    document.write("Yes");
else
    document.write("No");


// This code is contributed by patel2127
</script>

Output:

Yes

Time Complexity: O(n)
Auxiliary Space: O(n), where n is the length of the given string.