Balanced Brackets


Problem Statement :


A bracket is considered to be any one of the following characters: (, ), {, }, [, or ].

Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and ().

A matching pair of brackets is not balanced if the set of brackets it encloses are not matched. For example, {[(])} is not balanced because the contents in between { and } are not balanced. The pair of square brackets encloses a single, unbalanced opening bracket, (, and the pair of parentheses encloses a single, unbalanced closing square bracket, ].

By this logic, we say a sequence of brackets is balanced if the following conditions are met:

It contains no unmatched brackets.
The subset of brackets enclosed within the confines of a matched pair of brackets is also a matched pair of brackets.

Given n strings of brackets, determine whether each sequence of brackets is balanced. If a string is balanced, return YES. Otherwise, return NO.

Function Description

Complete the function isBalanced in the editor below. It must return a string: YES if the sequence is balanced or NO if it is not.

isBalanced has the following parameter(s):

s: a string of brackets

Input Format

The first line contains a single integer n, the number of strings.
Each of the next n lines contains a single string s, a sequence of brackets.


Output Format

For each string, return YES or NO.


Solution :



title-img


                            Solution in C :

In C ++ :




#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<stack>
using namespace std;
#include<map>


int main() {
    map<char,char>lol;
    lol[')']='(';
    lol[']']='[';
    lol['}']='{';
    int t;
    cin >> t;
    while(t--)
        {
        string s;
        stack<char>st;
        cin >> s;
        int flag=0;
        for(int i=0;s[i]!='\0';i++)
            {
            if(s[i]=='(' || s[i]=='{' || s[i]=='[')
                st.push(s[i]);
            else{
                if(!st.empty() and lol[s[i]]==st.top())
                 st.pop();
                else
                {
                    flag=1;
                    break;
                }
            }
        }
        
        if(!flag and st.empty())
        cout << "YES" << endl;
        else
        cout << "NO " << endl;
    }
    return 0;
}







In Java :




import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
       
        Scanner sc = new Scanner(System.in);
        try{
          int numTest = Integer.parseInt(sc.nextLine());
          for(int i = 0; i < numTest; i++){
              
           if(isBalanced(sc.nextLine())){
               System.out.println("YES");
           } else {
               System.out.println("NO");
           }
          }
        } catch(Exception e){
            System.out.println("Invalid Input");
        }
    }
    public static boolean isBalanced(String s){
        if(s == null || s.length() % 2 != 0) return false;
        Stack<Character> stack = new Stack<Character>();
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(c == '(' || c == '{' || c == '['){
                stack.push(c);
            } else if(c == ')' || c == '}' || c == ']'){
                if(!stack.isEmpty()){
                    char latestOpenP = stack.pop();
                    if(latestOpenP == '(' && c != ')'){
                        return false;
                    } else if(latestOpenP == '{' && c != '}'){
                        return false;
                    } else if(latestOpenP == '[' && c != ']'){
                        return false;
                    }
                } else {
                    return false;
                }
            }
        }
    
        return stack.isEmpty();
    }
}







In C :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int same(char a,char b)
    {
    if(a=='['&& b==']')
        return 1;
    if(a=='{'&& b=='}')
        return 1;
    if(a=='('&& b==')')
        return 1;
    return 0;
}
int check(char *a)
    {
    char stack[1001],top=-1;
        for(int j=0;j<strlen(a);j++)
            {
         if(a[j]=='['||a[j]=='{'||a[j]=='(')
              stack[++top]=a[j]; 
           if(a[j]==']'||a[j]=='}'||a[j]==')')
               {
               if(top==-1)
                   {
               return 0;
               }
               else
                   {
                   if(!same(stack[top--],a[j]))
                       {
               return 0;
               }
           }
        } 
}
    if(top!=-1)
                {
               return 0;
            }
    return 1;
}
int main() {
char a[1001];
    int n,valid;
    scanf("%d",&n);
     for(int i=0;i<n;i++)
        {
     scanf("%s",a);
         valid = check(a);
         if(valid==1)
             printf("YES\n");
    else
        printf("NO\n");
     }
    return 0;
}








In Python3 :





#! /usr/bin/python3

def check():
    stack = []
    s = input()
    for c in s:
        #print(c)
        if c == '(':
            stack.append(0);
        elif c == ')':
            if len(stack) > 0 and stack[-1] == 0:
                stack.pop()
            else:
                return -1
        elif c == '[':
            stack.append(2)
        elif c == ']':
            if len(stack) > 0 and stack[-1] == 2:
                stack.pop()
            else:
                return -1
        if c == '{':
            stack.append(4)
        elif c == '}':
            if len(stack) > 0 and stack[-1] == 4:
                stack.pop()
            else:
                return -1
    
    if len(stack) == 0:
        return 0
    else:
        return -1

def solve():
    t = int(input())
    for i in range(0,t):
        if check() == 0:
            print("YES")
        else:
            print("NO")
solve()
                        




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