Sakib and Shamim are good friends. As Sakib is a little naughty, every night before exam he disturbs…
Click here to read the complete problem statement.
If you need help solving this problem, mention your approach and ask specific questions. Please avoid sharing your code and asking the Community to figure out “what’s wrong”.
Iam also having the same doubt
@Aritra15 @mustafizz_22 Looks like the statement of this problem wasn’t very clear in its definition of what a valid subsequence is. I have tried to make some quick changes to it to make the statement clearer.
#include <bits/stdc++.h>
using namespace std;
int main() {stack s;
string a;
while((cin>>a))
{int max=0;
int count=0;
for(int i=0;i<a.size();i++)
{char c=a[i];
if(i!=0 && s.empty()){if(count>max)max=count;
count=0;}
if(c=='{' || c=='[' || c=='(')s.push(c);
else if(c=='}' && s.top()=='{'){s.pop();
count++;}
else if (c==']' && s.top()=='['){s.pop();
count++;}
else if (c==‘)’ && s.top()==‘(’){s.pop();
count++;}
}
cout<<2*max<<endl;}
return 0;
}
Can anyone please help me to correct this code?I really can’t find my mistake.
#include <bits/stdc++.h>
#define optimize ios_base::sync_with_stdio(0) ; cin.tie(0) ; cout.tie(0) ;
#define ll long long
#define vi vector
#define pb push_back
#define lp(i,a,b) for(int i = a ; i <= b ; i ++)
#define rlp(i,a,b) for(int i = b ; i >= a ; i --)
#define a_sort(v) sort(v.begin(), v.end())
#define d_sort(v) sort(v.rbegin(), v.rend())
#define pii pair<int,int>
using namespace std ;
const int MOD = 1E9 + 7 ;
int main() {
#ifdef debug
freopen("input.txt", "r", stdin) ;
freopen("output.txt", "w", stdout) ;
#endif // debug
optimize
string s ;
while(cin >> s) {
int l = s.size() ;
if(l == 1) cout << 1 << endl ;
else {
int mx = 0 , cnt = 0 ; stack<char> st ;
lp(i, 0, l-1) {
char current = s[i] ; // cout << "current = " << current << endl ;
if(st.empty()) {
st.push(current) ; // cout << "stack empty, " << st.top() << " was pushed" << endl ;
cnt = 0 ;
}
else {
char top = st.top() ; // cout << "stack top = " << top << endl ;
if(current == ']') {
if(top == '[') {
st.pop() ; // cout << "[ was poped" << endl ;
cnt += 2 ;
mx = max(cnt, mx) ;
}
else {
st.push(current) ;
// cout << "] was pushed" << endl ;
}
}
else if(current == ')') {
if(top == '(') {
st.pop() ; // cout << "( was poped" << endl ;
cnt += 2 ;
mx = max(cnt, mx) ;
}
else {
st.push(current) ;
// cout << ") was pushed" << endl ;
}
}
else st.push(current) ; // cout << current << " was pushed" << endl ;
}
// cout << "mx = " << mx << " cnt = " << cnt << endl ;
}
cout << mx << endl ;
}
}
}
what’s wrong with this code?
What should be the answer for these inputs ?
[()()]
2 , 4 or 6
[()(]]
2 or 4
Please reply soon.
I’m confused. I tried 4 times, but wrong answer in test 2. 
([])() in here the number of valid parentheses is 4.
[ ])()[ ])()[ ])()[ ])()[()()]
ans=> 4 : [()()], [()()], [()()], [()()]
[()(]]
ans=>3 : [()(]], [()(]], [()(]]
I wrote a program that failed only on last test. I think the code is correct for this problem. can tell me, what am I missing to check?
def countValidParentheses(sp:str, vp_count:int):
if(sp=='\n' or sp==''):return 1
for i in range(0,len(sp)):
for j in range(i+1, len(sp)):
if((sp[i]== '(' and sp[j]==')')
or (sp[i]== '{' and sp[j]=='}')
or (sp[i]== '[' and sp[j]==']')):
vp_count+=1
return vp_count
from sys import stdin
strings=[]
for line in stdin:
# if(line=='\n'):break
strings.append(line.strip())
for string in strings:
result=countValidParentheses(sp=string,vp_count=0)
print(result)
I wrote a program that failed only on last test. I think the code is correct for this problem. can tell me, what am I missing to check?
def countValidParentheses(sp:str, vp_count:int):
if(sp=='\n' or sp==''):return 1
for i in range(0,len(sp)):
for j in range(i+1, len(sp)):
if((sp[i]== '(' and sp[j]==')')
or (sp[i]== '{' and sp[j]=='}')
or (sp[i]== '[' and sp[j]==']')):
vp_count+=1
return vp_count
from sys import stdin
strings=[]
for line in stdin:
# if(line=='\n'):break
strings.append(line.strip())
for string in strings:
result=countValidParentheses(sp=string,vp_count=0)
print(result)
How can the length of a valid subsequence be an odd number?
Because there are always pairs of brackets in a valid subsequence.
You have to count the number of valid parentheses not the length.