UVA – 673 – Parentheses Balance

Problem Link: https://onlinejudge.org/external/6/673.pdf

Note:
This can be solved using different data structures and logic. I have used Stack to approach the problem.

Solution:

/*
Author: Arunima Mandal
Updated Date: 06.16.2020
*/
#include <bits/stdc++.h>
#define pf printf
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define sffff(a,b,c,d) scanf("%d %d %d %d",&a,&b,&c,&d)
#define sffffl(a,b,c,d) scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sfffff(a,b,c,d,e) scanf("%d %d %d %d %d",&a,&b,&c,&d,&e)
#define sfffffl(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)
#define sfc(a) scanf("%c",&a)
#define ms(a,b) memset(a,b,sizeof(a))
#define pb(a) push_back(a)
#define db double
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pll pair<long,long>
#define ff first
#define ss second
#define sz(x) (int)x.size()
#define all(x) x.begin(),x.end()
#define CIN ios_base::sync_with_stdio(0); cin.tie(0)
#define max3(a, b, c) max(a, b) > max(b, c) ? max(a, b) : max(b, c)
#define min3(a, b, c) min(a, b) < min(b, c) ? min(a, b) : min(b, c)
#define for0(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define forrev(i,n) for(int i=n-1; i>=0; i--)
#define forab(i,a,b) for(int i=a;i<=b;i++)
#define forba(i,b,a) for(int i=b;i>=a;i--)
#define stlloop(x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
#define case1(z) cout<<"Case "<<z<<": "
#define case2(z) printf("Case %d: ",z)
#define PI acos(-1) //3.14159265358979323846264338328
#define valid(tx,ty) tx>=0 && tx<row && ty>=0 && ty<col
#define intlim 2147483648
#define MAX 1000006
#define inf 100000008
ll power(int n,int p)
{
ll ans=1;
for1(i,p) ans*=n;
return ans;
}
ll m=1000001;
ll bigmod(ll n,ll p)
{
if(p==0) return 1;
else if(p%2==0)
{
ll x=(bigmod(n,p/2))%m;
return (x*x)%m;
}
else return ((n%m)*(bigmod(n,p-1)%m))%m;
}
/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*---------------------------------------------------------------------*/
using namespace std;
stack<char> st;
string s;
int main()
{
// CIN; //if getline() is present, it can't be used.
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int t,d;
sf(t);
getchar();
for0(z,t)
{
getline(cin,s);
d=0;
for0(i,sz(s))
{
if(s[i]=='(' || s[i]=='{' || s[i]=='[')
{
st.push(s[i]);
}
else if(s[i]==' ') continue;
else
{
if(!st.empty())
{
char ch=st.top();
if((ch=='(' && s[i]==')') || (ch=='{' && s[i]=='}') || (ch=='[' && s[i]==']'))
{
st.pop();
}
else break;
}
else {d=1; break;} //there is nothing left in stack to compare with.
}
}
if(!d && st.empty()) pf("Yes\n");
else
{
pf("No\n");
while(!st.empty()) st.pop();
}
}
return 0;
}

Leave a comment