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 :
Solution in C :
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;
}
Solution in C++ :
In C ++ :
#include <cmath>
#include <cstdio>
#include <stack>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int t,i;
string s;
stack<char>a;
cin>>t;
while(t--){
cin>>s;
for(i=0;i<s.size();i++){
if(s[i]=='(')
a.push(s[i]);
else if(s[i]=='{')
a.push(s[i]);
else if(s[i]=='['){
a.push(s[i]);
}
else if(s[i]==')'){
if(!a.empty()){
if(a.top()=='(')
a.pop();
else
break;
}
else
break;
}
else if(s[i]=='}'){
if(!a.empty()){
if(a.top()=='{')
a.pop();
else
break;
}
else
break;
}
else if(s[i]==']'){
if(!a.empty()){
if(a.top()=='[')
a.pop();
else
break;
}
else
break;
}
}
if(a.empty()&&i==s.size())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
while(!a.empty()){
a.pop();
}
}
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
//int arr[] = new int[n];
for(int arr_i=0; arr_i < n; arr_i++){
Stack st = new Stack();
//arr[arr_i] = in.nextInt();
String ipSeq = in.next();
//System.out.println(ipSeq);
boolean match = true;
for(int ind=0; ind<ipSeq.length(); ind++){
char ch = ipSeq.charAt(ind);
if(ch=='(' || ch=='{' || ch=='['){
st.push(ch);
}else if(st.isEmpty()){
match = false;
break;
}else if(ch==')'){
if('('!=(char)st.pop()){
match = false;
break;
}
}else if(ch=='}'){
if('{'!=(char)st.pop()){
match = false;
break;
}
}else if(ch==']'){
if('['!=(char)st.pop()){
match = false;
break;
}
}
}
if(match){
if(!st.isEmpty()){
System.out.println("NO");
}else{
System.out.println("YES");
}
}else{
System.out.println("NO");
}
}
}
}
Solution in Python :
In Python3 :
t = int(input())
while t:
ar = ['e']
s = input()
for i in s:
if i == '(':
ar.append('(')
elif i == '[':
ar.append('[')
elif i == '{':
ar.append('{')
elif i == ')':
k = ar.pop()
if k != '(':
ar.append('k')
break
elif i == ']':
k = ar.pop()
if k != '[':
ar.append('k')
break
elif i == '}':
k = ar.pop()
if k != '{':
ar.append('k')
break
#print(ar)
if len(ar) == 0 or ar[len(ar)-1] != 'e':
print('NO')
else:
#print(len(ar), ar[len(ar)-1])
print('YES')
t-=1
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