HackerRank Alternating Characters Interview preparation kit solution

In this HackerRank Alternating Characters Interview preparation kit problem 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.

Problem solution in Python programming.

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()))

Problem solution in Java Programming.

import java.io.*;
import java.util.*;
import java.math.*;


public class Solution {
	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;
	
	
	static void solve() throws Exception {
		int t = nextInt();
		for (int tt = 0; tt < t; ++tt) {
			String s = next();
			int res = 0;
			for (int i = 0; i < s.length(); ++i) {
				if (i - 1 >= 0 && s.charAt(i - 1) == s.charAt(i)) {
					++res;
				}
			}
			out.println(res);
		}
	}

	static int sqr(int x) {
		return x*x;
	}
	
	static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	static long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	static BigInteger nextBigInteger() throws IOException {
		return new BigInteger(next());
	}
	
	static String next() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}
	
	static String nextLine() throws IOException {
		tok = new StringTokenizer("");
		return in.readLine();
	}

	static boolean hasNext() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			String s = in.readLine();
			if (s == null) {
				return false;
			}
			tok = new StringTokenizer(s);
		}
		return true;
	}

	public static void main(String args[]) {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(new OutputStreamWriter(System.out));
			//in = new BufferedReader(new FileReader("input.in"));
			//out = new PrintWriter(new FileWriter("output.out"));
			solve();
			in.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			java.lang.System.exit(1);
		}
	}
}

Problem solution in C++ programming.

#include <iostream>
#include <string>

using namespace std ;

int main()
{
    int test_cases;
    cin >> test_cases ;
    string str;
    int count  ;
    int j ;
    char prev ;
    for ( int i =0 ; i < test_cases ;i++)
    {
        cin >> str ;
        j = 1; prev = str[0]; count = 0 ; 
        while((int)str[j]!=0)
        {
            if(str[j] == prev)
                count ++ ;
            else
                prev = str[j] ;
            j++ ;
        }
        cout << count << endl ;
    }
    return 0 ;
}

Problem solution in C programming.

#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;
}

Problem solution in JavaScript programming.

function processData(input) {
  var inputArray = input.split('\n');
  var t = parseInt(inputArray[0]);
  var pointer = 1;
  while (pointer <= t) {
    var target = inputArray[pointer];
    var count = 0;
    for (var i=1; i<target.length; i++) {
      if (target[i] == target[i-1]) {
        count += 1;
      }
    }
    console.log(count);
    pointer += 1;
  }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});