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  1. Practice
  2. Data Structures
  3. Stacks
  4. Balanced Brackets
  5. Discussions

Balanced Brackets

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  • alexYer + 10 comments

    C++ solution.

    #include <stack>
#include <string>
#include <iostream>

using namespace std;

string isBalanced(string s) {
    stack<char> st;  
    
    for (auto c: s) {
        switch (c) {
            case '{':
            case '(':
            case '[':
                st.push(c);
                break;
            case '}':
                if (st.empty() || (st.top() != '{')) {
                    return "NO";
                }
                st.pop();
                break;
            case ')':
                if (st.empty() || (st.top() != '(')) {
                    return "NO";
                }
                st.pop();
                break;
            case ']':
                if (st.empty() || (st.top() != '[')) {
                    return "NO";
                }
                st.pop();
                break;
        }
    }
    
    return st.empty() ? "YES" : "NO";
}


int main(){
    int t;
    cin >> t;
    for(int a0 = 0; a0 < t; a0++){
        string s;
        cin >> s;
        cout << isBalanced(s) << endl;
    }
    return 0;
}
    • vishy12697 + 2 comments

      What is auto in it??

      • alexYer + 1 comment

        http://en.cppreference.com/w/cpp/language/auto

        • benedict_ + 3 comments
          string isBalanced(string s) {
    // Complete this function
    std::stack<char> st;
    st.push('@');
    
    for(int i=0;i<s.length();i++)
    {
        if(s[i]=='('||s[i]=='['||s[i]=='{')
            st.push(s[i]);
        else
        {
            while(!st.empty()&&((s[i]==')'&&st.top()=='(')||(s[i]==']'&&st.top()=='[')||(s[i]=='}'&&st.top()=='{')))
                  {
                      st.pop();
                  }
        }
    }
                  if(st.top()=='@')
                  return "YES";
                  else
                  return "NO";
}

          why this code fail for two of the testcases.

          • miguel_hdezp21 + 0 comments

            I dont know if are the same cases like me , but "))" one of that cases was i do not contemplate when only have close cases.

          • vgautam99 + 0 comments

            @benedict_ I think if string is "))" or any string starts with closing brackets your code will fail.

            try this,it might help you

            #include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

struct Stack
{
    int data;
    struct Stack *next;
};

typedef struct Stack s;
s *top=NULL,*neu;

void push(int data)
{
    neu = (s *)malloc(sizeof(s));
    neu->data = data;
    neu->next = NULL;
    
    if(!top)
    {
        top = neu;
    }
    else
    {
        neu->next = top;
        top = neu;
    }
}

void pop()
{
    s *temp = top;
    top = top->next;
    free(temp);
}


char* isBalanced(char* s) {
    // Complete this function
    int i,len,flag=1;
    
    len = strlen(s);
    for(i=0;i<len;i++)
    {
        if((s[i]==')' || s[i]=='}' || s[i]==']') && (!top))
        {
            push(7);
            break;
        }
        else
        {
            if(s[i]=='(')
            {
                push(1);
            }
            if(s[i]=='{')
            {
                push(2);
            }
            if(s[i]=='[')
            {
                push(3);
            }
            if(s[i]==')' && top->data==1)
            {
                pop();
            }
            if(s[i]=='}' && top->data==2)
            {
                pop();
            }
            if(s[i]==']' && top->data==3)
            {
                pop();
            }
        }    
        
    }
    if(top )
    {
        while(top)
        {
            struct Stack *temp;
            temp = top;
            top = top->next;
            free(temp);
        }
        return "NO";
    }
    else
    {
        return "YES";
    }
}

int main() {
    int t; 
    scanf("%i", &t);
    for(int a0 = 0; a0 < t; a0++){
        char* s = (char *)malloc(512000 * sizeof(char));
        scanf("%s", s);
        int result_size;
        char* result = isBalanced(s);
        printf("%s\n", result);
    }
    return 0;
}
          • anujagrazzel + 0 comments

            @benedict_ your testcases are failing, when a string starts from close bracket, so just put a check on first element of string, both the testcases will pass.
            if(s[0]=='}'||s[0]==']'||s[0]==')') { st.push(s[0]); break; }

      • naphstor + 0 comments

        It is a new feature in C++ that automatically applies the datatype to a variable according to the value stored in the variable. It is similar to "var" keyword in some of the other languages.

    • messi655 + 3 comments

      your code will failed with test case: )), ]], }}

      • Snigdha_Sanat + 3 comments

        try this:

        import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;

        public class Solution { /use a stack to put the opening brackets, if a closing bracket comes, compare it with the top element, and if it is the corresponding opening bracket, return false/

        boolean check_bal(String s,int length)
    {
    int i,top;
    char [] stack=new char [100000];
    top=-1;
    for(i=0;i<length;i++)
        {
        if((s.charAt(i)=='[') ||  (s.charAt(i)=='{') ||   (s.charAt(i)=='(' ) )
            {stack[++top]=s.charAt(i);}//if opening bracket, push
        if(s.charAt(i)==']')
            {
            if(top>=0 && stack[top]=='[') top-=1;//check if top>=0, since starting element cannot be a closing bracket
            else return false;
        }// if

        if(s.charAt(i)=='}')
            {
            if(top>=0 && stack[top]=='{') top-=1;
            else return false;
        }// if

        if(s.charAt(i)==')')
            {
            if(top>=0 && stack[top]=='(') top-=1;
            else return false;
        }// if

    }//for loop
    //System.out.println("Top:"+top);
    if(top==-1)return true ;
    else return false;
}//end of function

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int iter,len;
    boolean res;
    int t = in.nextInt();
    for(iter = 0; iter < t; iter++){
        String s = in.next();
        len=s.length();
        if(len%2!=0) System.out.println("NO");//if string is of odd length
        else{
        //System.out.println(len);
        res=new Solution().check_bal(s,len);
         if (res==true) System.out.println("YES");
         else System.out.println("NO");
        }//else
    }//for
}//main

        }//Solution

        I am trying to make this code more and more robust. Hence any test case that this code fails ,are very welcome :)

        • learner_codes + 1 comment
          [deleted]
        • mk9440 + 2 comments

          my java sol :

          public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
         int t = in.nextInt();
        for(int a0 = 0; a0 < t; a0++){
            String s = in.next();
            Stack<Integer> st =new Stack<Integer>();
            char a[]=s.toCharArray();
            int comp=0;
            for(int i=0;i<s.length();i++){
            	if(i>0 && !st.isEmpty()){comp=st.peek();}
            	st.push((int) a[i]);
            	if(comp!=0 && !st.isEmpty() && st.size()>1){
            	if((comp==91 && st.peek()==93)||(comp==123 && st.peek()==125)||(comp==40 && st.peek()==41)){
            	st.pop();st.pop();}
            	}
            	}
            if(st.isEmpty()){System.out.println("YES");}
            else{System.out.println("NO");}
        }
    }
}
          • iamtodor + 4 comments

            I have improved and cleaned up your code. Here the effect:

            private static String isBalancedBrackets(String value) {
        Stack<Character> stack = new Stack<>();
        char upperElement = 0;
        for (int i = 0; i < value.length(); i++) {
            if (!stack.isEmpty()) {
                upperElement = stack.peek();
            }
            stack.push(value.charAt(i));
            if (!stack.isEmpty() && stack.size() > 1) {
                if ((upperElement == '[' && stack.peek() == ']') ||
                        (upperElement == '{' && stack.peek() == '}') ||
                        (upperElement == '(' && stack.peek() == ')')) {
                    stack.pop();
                    stack.pop();
                }
            }
        }
        return stack.isEmpty() ? "YES" : "NO";
            • sowmi546 + 3 comments

              My Java Solution 

              import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static boolean isBalanced(String s) {
   int len=s.length();
    if(len==0 || s==null) return true;
      Stack<Character> stack = new Stack<Character>();
    for(int i=0;i<s.length();i++)
    {
        if(s.charAt(i)=='(' || s.charAt(i)=='[' || s.charAt(i)=='{')  stack.push(s.charAt(i));
        else if(s.charAt(i)==')' && !stack.empty() && stack.peek()=='(') stack.pop();
        else if(s.charAt(i)==']' && !stack.empty() && stack.peek()=='[') stack.pop();

        else if(s.charAt(i)=='}' && !stack.empty() && stack.peek()=='{') stack.pop();
        else return false;


    }
    return stack.empty();
    }

    public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    for (int a0 = 0; a0 < t; a0++) {
        String expression = in.next();
        System.out.println( (isBalanced(expression)) ? "YES" : "NO" );
    }
}
}
              • saif01md + 1 comment
                else if(str.charAt(i)==']' && s.peek()=='[' && !s.empty())

                Please reply .. If we write the statement on either side of && as the order shown above ( peek before and empty after) , it is showing runtime error . Is the order of statement also important in writing these expressions ?

                • claytonsibanda60 + 0 comments

                  Yes it is. If !s.empty() is placed first and the stack is empty, s.peek()=='[' will not be executed, hence no runtime error.This is due to the short circuit evaluation nature of if statements. However the order you gave above gives runtime error because s.peek()=='[' is executed first and this may cause errors if the stack is empty.

              • saif01md + 2 comments

                Another questionisthat why !s.empty() is even required in the ?

                • TusharSh + 0 comments

                  I think You can't pop is stack is empty

                • aleksei_maide + 0 comments

                  If stack is not empty... then there is an extra unbalanced parenthesis on the stack...

              • b_vsbrk + 1 comment

                My code looks exactly like your code but still I'm not able to pass 3 test cases...

                import java.util.Scanner;
import java.util.Stack;

/**
 * Created by BK on 06-08-2017.
 */

public class BalancedParanthesis {
    public static void main(String... strings) {
        Scanner sc = new Scanner(System.in);
        int tc = sc.nextInt();
        for (int i = 0; i < tc; i++) {
            printAnswer(sc.next());
        }
    }

    private static void printAnswer(String input) {
        Stack<Character> stack = new Stack<>();
        boolean isValid=true;
        for (int i = 0; i < input.length(); i++) {
            char currentChar = input.charAt(i);
            if (currentChar == '{' || currentChar == '(' || currentChar == '[') stack.push(currentChar);
            else if (currentChar == '}') {
                if (stack.isEmpty())isValid=false;
                else {
                    if (stack.pop() != '{') {
                        isValid=false;
                    }
                }
            } else if (currentChar == ')') {
                if (stack.isEmpty())isValid=false;
                else {
                    if (stack.pop() != '(') {
                        isValid=false;
                    }
                }
            } else if (currentChar == ']') {
                if (stack.isEmpty())isValid=false;
                else {
                    if (stack.pop() != '[') {
                        isValid=false;
                    }
                }
            }
        }
        System.out.println(isValid?"YES":"NO");
    }
}
                • aishik_swarez + 0 comments

                  import java.util.Scanner; import java.util.Stack;

                  public class Solution {

                  static String isBalanced(String s) {
    Stack stack = new Stack();
    stack.push(s.charAt(0));
    char peek;
    char charAt;

    for (int i = 1; i < s.length(); i++) {
        try {
            peek = (char) stack.peek();
            charAt = s.charAt(i);
            if(!stack.empty()) {
            if ((charAt == '}' && peek == '{') || (charAt == ')' && peek == '(')
                    || (charAt == ']' && peek == '[')) {
                stack.pop();
            } else
                stack.push(charAt);
            }
            else
                stack.push(i);
        } catch (Exception e) {
            //i++;
            charAt = s.charAt(i);
            stack.push(charAt);
        }
    }
    if (stack.empty())
        return "YES";
    else
        return "NO";

}

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    in.nextLine();
    String result[] = new String[t];
    for (int a0 = 0; a0 < t; a0++) {
        String s = in.next();
        result[a0] = isBalanced(s);
    }
    in.close();
    for (int i = 0; i < t; i++)
        System.out.println(result[i]);
}

                  }

            • nadoo1110 + 0 comments
              [deleted]
            • tambekarv + 0 comments

              Nice Solution! No need for !stack.isEmpty() in second if statement

            • tambekarv + 0 comments

              Nice Solution! No need of !stack.isEmpty() in second if statement.

          • piyushmishra + 1 comment

            what are magic numbers like 91 ,93 etc?

            • mk9440 + 0 comments

              [ has ascii code 91 , ] has ascii code 93

              ascii code table

        • shashisuman + 2 comments
          import java.util.*;

public class Solution {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();

    for(int a0 = 0; a0 < t; a0++){
        String s = in.next();
        int l = s.length();
        while(true){
            s = s.replaceAll("\\(\\)","");s = s.replaceAll("\\{\\}","");s = s.replaceAll("\\[]","");s=s.trim();
            if(s.length()==l)
                break;
            else if(s.length()==0)
                break;
            l = s.length();
        }
        if(s.length()==0)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

          }

          • himanshub010 + 0 comments

            can u expalain what is s.replaceAll("","");

          • anishdgr8111 + 0 comments

            Smart solution :)

      • AshishSinha5 + 0 comments

        my c++ solution taking care of special cases

        #include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    while(n!=0){
        string a;
    cin>>a;
    stack <char> s;
    int i,flag=0;
    for(i=0;i<a.length();i++){
        switch(a[i]){
        case '{':
        case '[':
        case '(':
            s.push(a[i]);
            break;
        case '}':
            if(s.empty()||s.top()!='{'){
                cout<<"NO"<<endl;
                flag=1;
                goto exit;
            }
            s.pop();
            break;
        case ']':
            if(s.empty()||s.top()!='['){
                cout<<"NO"<<endl;
                flag=1;
                goto exit;
            }
            s.pop();
            break;
        case ')':
            if(s.empty()||s.top()!='('){
                cout<<"NO"<<endl;
                flag=1;
                goto exit;
            }
            s.pop();
            break;
        }
    }
    exit: if(s.empty()&&!flag)
                cout<<"YES"<<endl;
          if(!flag&&!s.empty())
                cout<<"NO"<<endl;

    n--;
    }

    return 0;
}
      • sahilshelangia + 0 comments

        for each test case intialise top=-1

    • sophiekakker1998 + 1 comment

      for (auto c: s) what is the meaning of this statement?

      • schardong_gg + 0 comments

        The same as "for each" in other languages. For each char c in s, do ...

        The "auto" keyword is used to cast the type to "char" automatically.

    • bachhal + 1 comment

      {[[}[] }(]}){

      tell me how this is balanced

      • charan_sai + 0 comments

        It's UnBalanced!!

    • _silent_ + 0 comments

      i like this code

    • fheil0815 + 0 comments

      c#:

      static string isBalanced(string s) {
        // Complete this function
        if (s.Length%2==1) return "NO";
        Char[] bc = s.ToCharArray();
        int[] b = new int[bc.Length];
        for(int i=0;i<b.Length;i++) b[i]= Convert.ToInt32(bc[i]);
        Stack<int> sta = new Stack<int>();
        int c;
        for (int i=0;i<b.Length;i++){
            if (sta.Count==0){
            if (b[i]==125 || b[i]==93 || b[i]==41) return "NO";
            sta.Push(b[i]);
            continue;}
            c = sta.Peek();
            switch(b[i]){
                case 125: case 93:
                    if (c!=b[i]-2) return "NO";
                    sta.Pop();
                    break;
                case 41:
                    if (c!=40) return "NO";
                    sta.Pop();
                    break;
                default: sta.Push(b[i]); break;
            }     
        }
        return (sta.Count>0) ? "NO": "YES";
    }
    • bibhas_abhishek + 0 comments

      Nice use of switch :)

    • neerajnitjsr20 + 0 comments
      #include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

char* isBalanced(char* s) {
    int i=0,j=0;
    char c[strlen(s)-1];


  for(i=0;i<=strlen(s)-1;i++)
  {
      if(s[i]=='('||s[i]=='{'||s[i]=='[')
         {
          c[j++]=s[i];
          //printf("%d%d%c%c",i,j,s[i],c[--j]);
         }

      else
          {
              if(s[i]==')'&&c[--j]=='(')
              ;
              else{
                    if(s[i]=='}'&&c[--j]=='{')
                    ;
                    else
                        {
                         if(s[i]==']'&&c[--j]=='[')
                         ;
                         else
                            return "NO";

                         }
                  }
           }
      }
    if(j==0)
        return "YES";
    else
        return "NO";


}

int main() {
    int t,a0;
    scanf("%i", &t);
    for( a0 = 0; a0 < t; a0++){
        char* s = (char *)malloc(512000 * sizeof(char));
        scanf("%s", s);
        char* result = isBalanced(s);
        printf("%s\n", result);
    }
    return 0;
}
    • rwan7727 + 0 comments
      [deleted]
    • bennattj + 0 comments

      The fact that you rewrite if(st.empty() || ...) over and over again, suggests there's a better way to write this.

  • w00tles + 7 comments

    Python solution

    lefts = '{[('
rights = '}])'
closes = { a:b for a,b in zip(rights,lefts)}

def valid(s):
    stack = []
    for c in s:
        if c in lefts:
            stack.append(c)
        elif c in rights:
            if not stack or stack.pop() != closes[c]:
                return False
    return not stack  # stack must be empty at the end

t = int(raw_input().strip())
for a0 in xrange(t):    
    s = raw_input().strip()    
    if valid(s):
        print 'YES'
    else:
        print 'NO'
    • bplusv + 0 comments

      Nice & clean!

    • s_h_u_bham + 1 comment

      Sexy!

      • kgkoete + 0 comments

        Very neat (pun intended)

    • DanHaggard + 1 comment

      Adding my python3 solution here for convenience:

      def check_str(string):
    stack = []
    table = {
        ")": "(",
        "}": "{",
        "]": "[",
    }
    for char in string:
        if not stack:
            stack.append(char)
        elif char not in table:
            stack.append(char)
        elif table[char] == stack[-1]:
            stack.pop()
        else:
            stack.append(char)
    if stack:
        print("NO")
    else:
        print("YES")    

N = int(input().strip())
for i in range(N):
    check_str(input())
      • jojo_lapin + 2 comments

        A slightly shorter version ;)

        table =  { ')': '(', ']':'[', '}':'{' }

for _ in range(int(input())):
    stack = []
    for x in input():
        if stack and table.get(x) == stack[-1]:
            stack.pop()
        else:
            stack.append(x)
    print("NO" if stack else "YES")
        • Aniruddha_ + 2 comments
          if stack and table.get(x) == stack[-1]:

          Please explain this line of code

          • stephen_akerson + 0 comments

            That line is saying:

            If stack exists (which equates to False if stack is empty)

            AND

            The table lookup for the current iteration of x is equivalent to the element atop stack

            It might be helpful to take a look at the .get() method: get(key[, default]) Return the value for key if key is in the dictionary, else default. If default is not given, it defaults to None, so that this method never raises a KeyError.

          • Anyone_ + 0 comments

            this line explain that firstly stack(consider it as a list) is not empty and element in table is equal to the element at last of stack(consider it as a list) is equal. [-1] is to catch the last element in list.

        • stephen_akerson + 0 comments

          It's so pretty! (Ï€_Ï€) Thanks for sharing

    • vmane858 + 0 comments

      pretty solution

    • ElizaW + 0 comments

      Thank you for reminding me that stack must be empty at the end. That's why I failed the first few times.

    • ujjl_248 + 0 comments

      What does this line do ? 

      if not stack or stack.pop() != closes[c]:

      I only know that, 

      stack.pop()
    • shuhwsudh + 0 comments

      Nice

  • sanju1 + 1 comment

    solution using stacks all test cases passed!

    public boolean isValid(String s){
        Stack<Character> stack=new Stack<Character>();
        for(char c:s.toCharArray()){
            if(c=='(')
                stack.push(')');
            else if(c=='{')
               stack.push('}');
               else if(c=='[')
               stack.push(']');
               else if(stack.isEmpty()||stack.pop()!=c)
               return false;
        }
               return stack.isEmpty();
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for(int a0 = 0; a0 < t; a0++){
            String s = in.next();
                System.out.println(isValid(s) ? "YES":"NO");
        }
    }
    • miere00 + 0 comments

      I think that was the most efficient one here. My solution was close, but not that efficient.

          
    static final List<Character> 
        OPENING = Arrays.asList( '{', '[', '(' ),
        CLOSING = Arrays.asList( '}', ']', ')' );
    
    static final String TRUE = "YES", FALSE = "NO";
    static final int NOT_FOUND = -1;

    static String isBalanced(String s)
    {
        Deque<Character> brackets = new ArrayDeque<>();
        for ( int i=0; i<s.length(); i++ )
        {
            Character ch = s.charAt(i);
            int pos = CLOSING.indexOf(ch);
            if ( pos != NOT_FOUND ) {
                if ( !OPENING.get(pos).equals( brackets.poll() ) )
                    return FALSE;
            } else
                brackets.push( ch );
        }
        return brackets.isEmpty() ? TRUE : FALSE;
    }

  • shilay + 5 comments

    ()( take care of this case also

    • developer_akhil1 + 0 comments

      After taking care of this case, all test cases are passed. Thanks :)

    • overl0rd11 + 0 comments

      thanks a lot bro

    • akwatz + 1 comment

      i took care of this case by checking whether the string length is odd but didnt workout for me,any guesses??

      • tyronep + 1 comment

        Make sure it passes ()((, too.

        • magneton + 0 comments

          Couldn't figure out why I needed to check if my stack was empty at the end. This case shows why. Thanks for the example, helped me understand the solution. Cheers.

    • bhgupta + 0 comments

      thanks. you saved my hackos :)

    • baby_groot + 0 comments

      This should be pinned. Thanks a lot.

  • anonymouse_pal + 1 comment

    try these special case:

    }}}
}{
    • akgarhwal + 0 comments

      Thanks for Special Case '}}}' @anonymouspal

  • pareit + 2 comments

    javascript: 

    function isBalanced(s) {
    var result = 'YES';
    var stack = [];
    s.split('').forEach(function(val) {
        switch(val) {
            case '{':
                stack.push('}');
                break;
            case '[':
                stack.push(']');
                break;
            case '(':
                stack.push(')');
                break;
            default:
                var test = stack.pop();
                if (val !== test) {
                    result = 'NO';
                }    
        }
    })
    if (stack.length) {
        result = 'NO';
    }
    return result;
}
    • chadwalt + 0 comments

      Awesome solution.

    • gogia_prateek + 0 comments

      This is the finest!

  • ankit_kumar_pune + 2 comments

    C++ solution

    bool isValid(string s)
{
    int i=0,flag=0;
    stack<char> s1;
    while(s[i]!='\0')
    {
        if(s[i]=='(' )
            s1.push(')');
        else if(s[i]=='{' )
            s1.push('}');
        else if(s[i]=='[' )
            s1.push(']');
        else if(s1.empty() || s1.top()!=s[i])
        {
            return false;
        }
        else
        s1.pop();
        i++;
    }
   return s1.empty();
}
    • positivedeist_07 + 0 comments

      Is it okay to push the closing brackets in the stack? It seems pretty clean but how does this legitimately works?

    • shravankumar + 0 comments

      Can you explain the last else if?

  • parasou79 + 0 comments

    Without using stack

    let n = -1;
	while (s.length != n) {
		n = s.length;
		s = s.replace('()', '');
		s = s.replace('[]', '');
		s = s.replace('{}', '');
	}
	if (s.length == 0)
		return "YES"
	else 
		return "NO"

  • marzi_golbaz + 1 comment

    Python short solution:

    def isBalanced(s):
    # Complete this function
    bracket_map = {'}': '{', ')': '(', ']': '['}
    stack = []
    for b in s:
        if b in bracket_map.values():  # openning
            stack.append(b)
        elif b in bracket_map.keys():  # closing
            if not stack or stack.pop() != bracket_map[b]:
                return "NO"
    if not stack:
        return "YES"
    else:
        return "NO"
    • hirajawed3 + 0 comments

      how would your code look like if u choose to write bracket_map = {'{': '}', '(': ')', '[': ']'} as this way? I was trying to do it but I couldn't figure out.. maybe then it would be the different approach?

  • humoyun_ahmad + 1 comment

    My solution:

    def is_balanced(string):
    b_stc=[]
    closing=[')',']','}']
    opening=['(','[','{']
    
    for ch in string:
        if ch in opening:
            b_stc.append(ch)
        else:
            if not b_stc:
                return "NO"
            last=b_stc.pop()
            if opening.index(last) != closing.index(ch):
                return "NO"
    return "NO" if b_stc else "YES" 

t = int(input().strip())
for x in range(t):
    s = input().strip()
    print(is_balanced(s))
    • vivek_keshore + 0 comments

      Though the solution is good and working, but in stacks you don't have access to indices. You only have access to TOS. I think by using the list's index method, this is violating the principles of stacks.

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