Hello Programmers, In this post, you will learn how to solve HackerRank Alternating Characters Solution. This problem is a part of the HackerRank Algorithms Series.HackerRank Alternating Characters SolutionOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.HackerRank Alternating Characters SolutionTaskYou 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.Examples = AABAABRemove an A at positions 0 and 3 to make s = ABAB in 2 deletions.Function DescriptionComplete the alternatingCharacters function in the editor below.alternatingCharacters has the following parameter(s):string s: a stringReturnsint: the minimum number of deletions requiredInput FormatThe first line contains an integer q, the number of queries.The next q lines each contain a string s to analyze.Constraints1 <= q <= 101 <= length of s <= 105Each string s will consist only of characters A and B.Sample Input5AAAABBBBBABABABABBABABAAAABBBSample Output34004ExplanationThe characters marked red are the ones that can be deleted so that the string does not have matching adjacent characters.HackerRank Alternating Characters SolutionsAlternating Characters Solution 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; }Alternating Characters Solution in Cpp#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; }Alternating Characters Solution in Javaimport java.util.Scanner; public class Solution { /** * @param args */ 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); } } }Alternating Characters Solution in Pythont = int(raw_input()) for j in range(t): res=0 str = raw_input() if 'A' not in str or 'B' not in str: res = len(str)-1 print res continue i=0 while i<(len(str)-1): if str[i+1]==str[i]: res+=1 i+=1 print resAlternating Characters Solution using JavaScriptfunction 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); });Alternating Characters Solution in Scalaobject Solution extends App { for (T <- 1 to readInt) { val s = readLine var i = 0 var ans = 0 while (i < s.length) { var j = i var cnt = 0 while (j < s.length && s(i) == s(j)) { j += 1 cnt += 1 } ans += cnt - 1 i = j } println(ans) } }Alternating Characters Solution in Pascal(* Enter your code here. Read input from STDIN. Print output to STDOUT *) program abab; var T : integer; word : ansistring; result : array [1..10] of longint; i : integer; j : longint; begin readln(T); for i := 1 to T do begin result[i] := 0; readln(word); for j := 1 to (length(word)-1) do if word[j] = word[j+1] then result[i] := result[i] + 1; end; for i := 1 to T do writeln(result[i]); end.Disclaimer: This problem (Alternating Characters) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: HackerRank Modified Kaprekar Numbers Solution Post navigationHackerRank Gemstones Solution HackerRank Beautiful Binary String Solution