# Alternating Characters

### Problem Statement :

```You are given a string containing characters A and B only. Your task is to change it into a string such that there are no matching adjacent characters. To do this, you are allowed to delete zero or more characters in the string.

Function Description

Complete the alternatingCharacters function in the editor below.

alternatingCharacters has the following parameter(s):

string s: a string
Returns

int: the minimum number of deletions required
Input Format

The first line contains an integer q, the number of queries.
The next q lines each contain a string s to analyze.

Constraints

1  <=   q   <=   10
1  <=  length of s  <=  10^5
Each string s will consist only of characters A and B.

Sample Input

5
AAAA
BBBBB
ABABABAB
BABABA
AAABBB
Sample Output

3
4
0
0
4```

### Solution :

```                            ```Solution in C :

In  C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main()
{
int t;
long i=0;
unsigned int count=0;
char * c;
scanf("%d",&t);
c=(char *)malloc(sizeof(char)*(100002));
while(t--)
{
scanf("%s",c);
for(i=0; *(c+i); i++)
{
if(c[i]==c[i+1])
{
count++;
}
}
printf("%u\n",count);
count=0;
}

return 0;
}```
```

```                        ```Solution in C++ :

In C++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
int t;
cin>>t;
while(t--)
{
string s;int c=0,a=0;
cin>>s;
for(int i=1;s[i]!='\0';i++)
{
if((s[i]==65 && s[a]==66)||(s[i]==66 &&s[a]==65))
{
a=i;
// cout<<a<<endl;
}
else{
c++;
//cout<<c<<" "<<i<<endl;

}

}
cout<<c<<endl;

}
return 0;
}```
```

```                        ```Solution in Java :

In Java :

import java.util.Scanner;
public class Solution {

public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s = new Scanner(System.in);
int t = s.nextInt();
s.nextLine();
while(t-- > 0)
{
int count = 0;
String str = s.nextLine();
for(int i=1;i<str.length();i++)
{
if(str.charAt(i)==str.charAt(i-1))
{
count++;
}
}
System.out.println(count);
}
}

}```
```

```                        ```Solution in Python :

In  Python3 :

def f(s):
return sum(1 for c1, c2 in zip(s, s[1:]) if c1 == c2)

t = int(input())
for _ in range(t):
print(f(input()))```
```

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