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 <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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