HackerRank Alternating Characters problem solution

In this HackerRank Alternating Characters Interview preparation kit problem solution, 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

HackerRank Alternating Characters problem solution in Python.

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()))

Alternating Characters problem solution in Java.

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("%un",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);
});