VC Pairs - HackerEarth Problem Solution

Max has a string S with length N. He needs to find the number of indices i (1≤i≤N−11≤i≤N−1) such that the i-th character of this string is a consonant and the i+1th character is a vowel. However,she is busy, so she asks for your help.
Note: The letters 'a', 'e', 'i', 'o', 'u' are vowels; all other lowercase English letters are consonants.
Input

• The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
• The first line of each test case contains a single integer N.
• The second line contains a single string S with length N.

Output

For each test case, print a single line containing one integer ― the number of occurrences of a vowel immediately after a consonant

SAMPLE INPUT

3
6
bazeci
3
abu
1
o

SAMPLE OUTPUT

3
1
0

In C -

1. #include<stdio.h>
2.  
3. int main()
4. {
5. int test,n,j,i,m,l1=0,l2=0,flag=0;
6. char str[1000000];
7. char c[30]={'b','c','d','f','g','h','j','k','l',
8. 'm','n','p','q','r','s','t','v','w','x','y','z'};
9. char v[5]={'a','e','i','o','u'};
10. scanf("%d\n",&test);
11. for(j=0;j<test;j++)
12. {
13. flag=0;
14. scanf("%d\n",&n);
15. scanf("%s",&str);
16. for(i=0;i<n;i++)
17. {
18. l1=0,l2=0;
19. for(m=0;m<21;m++)
20. {
21. if(str[i]==c[m])
22. l1=1;
23. else
24. continue;
25. }
26. for(m=0;m<5;m++)
27. {
28. if(str[i+1]==v[m])
29. l2=1;
30. else
31. continue;
32. }
34. if(l1==1 && l2==1)
35. flag+=1;
36. }
37. printf("%d\n",flag);
38. }
39. }

      

                                 by - md irfan 

In java

1. import java.io.*;
2. public class VCPairs {
3. public static void main(String[] args) throws IOException {
4. BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
5. int t = Integer.parseInt(br.readLine());
6. while(t>0) {
7. int len = Integer.parseInt(br.readLine());
8. int count = 0;
9. char[] arr = br.readLine().toCharArray();
10. for (int i = 0; i<len-1; i++) {
11. if ((arr[i+1] == 'a' || arr[i+1] == 'e' || arr[i+1] == 'i' ||
12. arr[i+1] == 'o' || arr[i+1] == 'u') &&
13. (arr[i] != 'a' && arr[i] != 'e' && arr[i] != 'i' && arr[i] != 'o' && arr[i] != 'u')
14. ) {
15. count++;
16. }
17. }
18. System.out.println(count);
19. t--;
20. }
21. br.close();
22. }
23. }

                                                                   By -  Abhinav Mukherjee

In Python -

1. import re
2. T = int(input())
3. for _ in range(T) :
4. n = int(input())
5. s = input()
6. print(len(re.findall(r'([^aeiou])([aeiou])',s)))

By - Yash Agrawal

In C++ -

1. #include<iostream>
2. using namespace std;
3. int main()
4. {
5. int t;
6. cin>>t;
7. while(t--)
8. {
9. int n;
10. cin>>n;
11. string s;
12. cin>>s;
13. int l=0;
14. for(int i=0;i<n-1;i++)
15. {
16. if((s[i]!='a'&& s[i]!='e'&&s[i]!='i'&&s[i]!='o'&&s[i]!='u')&&(s[i+1]
17. =='a'||s[i+1]=='e'||s[i+1]=='i'||s[i+1]=='o'||s[i+1]=='u'))
19. l++;
21.  
22. }
23. cout<<l<<endl;
24. }
25. return 0;
26. }

By -  Tanmay Nunhariya