Help me in solving EZSPEAK problem

help
,
1

My issue

My code

# cook your dish here
v=["a","e","i","o","u"]
count=0
t=int(input())
for i in range(t):
    n=int(input())
    s=input()
    for k in range(n):
        if(s[k]!=v): 
            count=count+1
            if(count>=4):
                print("NO")
                break
            else: 
                print("YES")  
                break
        elif(s[k]==v):
            count=0

Problem Link: EZSPEAK Problem - CodeChef

2

@malav1115
In your code, there seems to be a few errors which i can see from a glance.

  • Counter variable placement: You have taken your counter variable outside the test case loop so it is not resetting to zero for each individual test case and only keeps increasing.

  • Wrongly checking vowels: You should have used in and not in to check if the letter is present in list of vowels, v.

  • Printing before the string has fully been checked: You should not have put the print statement while checking the counter value, as it would keep executing each time and thus mess with the desired output.

  • Breaking the loop without need: There was no need for else condition at all, you could have checked if counter has reached a particular value or not. Then you could have used break to break from checking the string if the value was reached.

The following code might help you understand the solution better.

# cook your dish here
v=['a','e','i','o','u']
for _ in range(int(input())):
    n=int(input())
    s=input()
    c=0
    for i in s:
        if i in v:
            c=0
        else:
            c+=1
            if(c==4):
                break
    if(c==4):
        print('NO')
    else:
        print('YES')
3

May this help you.
code written in c++

#include<bits/stdc++.h>
using namespace std;

int main() {
    int t;
    cin >> t;

    while(t--) { 
        int n;
        cin >> n;
        string s;
        cin >> s;
        int count = 0;
        int notor_easy_abhay = 0;

        for(int i = 0; i < n; i++) {
            if(s[i] != 'a' && s[i] != 'e' && s[i] != 'i' && s[i] != 'o' && s[i] != 'u') {
                count++;
            }

            if(s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u') {
                count = 0;
            }

            if(count >= 4) {
                notor_easy_abhay++;
                break;
            }
        }

        if(notor_easy_abhay != 0) {
            cout << "NO" << endl;
        }
        else {
            cout << "YES" << endl;
        }
    }

    return 0;
}
1 Like