Problem of The Day 30 December — Nice Arches

Nice Arches

Nikki’s latest work is writing an story of letters. However, she finds writing story so boring that, after working for three hours, she realized that all she has written are M long words consisting entirely of letters A and B. Having accepted that she will never finish the story in time, poor Nikki has decided to at least have some fun with it by counting bubbly words.

Now Nikki is connecting pairs of identical letters (A with A, B with B) by drawing arches above the word. A given word is bubbly if each letter can be connected to exactly one other letter in such a way that no two arches intersect. So here is your task. Help Nikki count how many words are bubbly.

Input :

The first line of input contains the positive integer M , the number of words written down by Nikki. Each of the following M lines contains a single word consisting of letters A and B, with length between 2 and 10⁵, inclusive. The sum of lengths of all words doesn’t exceed 10⁶.

Output :

The first and only line of output must contain the number of bubbly words.

Constraints:

1 <= M <= 100

SAMPLE INPUT

3
ABAB
AABB
ABBA

SAMPLE OUTPUT

2

Explanation

ABAB — It is not bubbly as A(indexed 1) will connect to A(indexed 3) by an arch and when we try to connect B(indexed 2) with B(indexed 4) by an arch then it will intersect with the arch b/w A and A. AABB — It is bubbly as arch b/w A and A will not intersect with the arch b/w B and B. ABBA — It is also bubbly as arches will not intersect. we can draw arches b/w A and A above the arch b/w B and B.

Problem Link: Nice Arches

The solution will be posted tomorrow.

SOLUTION

In this problem, we have to make arches of minimum length and with the nearest A or B.
So we are using a stack here. We push the element and check if top two are same then they form arches and that arch won’t intersect with any other arch.
And in end, if the stack is empty then the string is Bubbly.

CODE

#include<iostream>
using namespace std;
 
int main()
{
    int M;  cin>>M;
    int count = 0;
    while(M--) {
        string s;   cin>>s;
        int len = s.length();
        int ans = 0;
        int top = -1;
        char *arr = new char[s.length()];
        for(int i=0; i<len; i++) {
             top = top + 1; 
             arr[top] = s[i]; 
             if(arr[top-1] == arr[top] && top>0)
                top = top-2;
        }
        if(top == -1)
            ans=1;
        count+=ans;
    }
    cout<<count<<endl;
}