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 :



title-img 


                            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;
}







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);	
		}
	}

}







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;
}








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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