**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. Your task is to find the minimum number of required deletions. 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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