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HackerRank Alternating Characters Problem Solution

HackerRank Alternating Characters Solution

Posted on April 26, 2023April 28, 2023 By Yashwant Parihar No Comments on HackerRank Alternating Characters Solution

In this post, we will solve HackerRank Alternating Characters 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.
Example
8 = AABAAB
Remove an A at positions 0 and 3 to make s = ABAB in 2 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.

Sample Input

5
AAAA
BBBBB
ABABABAB
BABABA
AAABBB

Sample Output

3
4
0
0
4

Explanation

The characters marked red are the ones that can be deleted so that the string does not have matching adjacent characters.

HackerRank Alternating Characters Problem Solution
HackerRank Alternating Characters Problem Solution

Table of Contents

  • Alternating Characters C Solution
  • Alternating Characters C++ Solution
  • Alternating Characters C Sharp Solution
  • Alternating Characters Java Solution
  • Alternating Characters JavaScript Solution
  • Alternating Characters Python Solution

Alternating Characters C Solution

#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#define _CRT_DISABLE_PERFCRIT_LOCKS
#define getchar_unlocked() getchar()
#define putchar_unlocked(c) putchar(c)
#endif

#include <stdio.h>
#include <stdlib.h>

#define mygc(c) (c)=getchar_unlocked()
#define mypc(c) putchar_unlocked(c)

void reader(int *x){ int k, m = 0; *x = 0; for (;;){ mygc(k); if (k == '-'){ m = 1; break; }if ('0' <= k&&k <= '9'){ *x = k - '0'; break; } }for (;;){ mygc(k); if (k<'0' || k>'9')break; *x = (*x) * 10 + k - '0'; }if (m)(*x) = -(*x); }
void readln(char *x){ int i = 1, k;  for (;;){ mygc(k); if ('-' <= k&&k <= 'z'){ x[0] = k; break; } }for (;;){ mygc(k); if (k<'-' || k>'z') { x[i] = 0; break; } x[i] = k, i++; } }
void writer(int x, char c){ int sz = 0, m = 0; char buf[10]; if (x<0)m = 1, x = -x; while (x)buf[sz++] = x % 10, x /= 10; if (!sz)buf[sz++] = 0; if (m)mypc('-'); while (sz--)mypc(buf[sz] + '0'); mypc(c); }

int main() {
	int T, ans;
	char cur, pre;
	char buf[100002];

	reader(&T);

	for (int i = 0; i < T; i++) {
		ans = 0, cur = -1;
		readln(buf);
		int j = 0;
		while (buf[j]) {
			pre = cur;
			cur = buf[j];
			if (cur == pre) { ans++; }
			j++;
		}
		writer(ans, '\n');
	}

	getchar();
	return 0;
}

Alternating Characters C++ Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

int main() {
    std::string input;
    std::getline(std::cin, input);
    int tests = std::stoi(input);
    
    for (int i = 0; i < tests; i++) {
        std::string s;
        std::getline(std::cin, s);
        
        int deletions = 0;
        char current = 0;
        for (auto c: s) {
            if (c == current) {
                deletions++;
            } else {
                current = c;
            }
        }
        
        std::cout << deletions << std::endl;
    }
    
    return 0;
}

Alternating Characters C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
class Solution {
    static void Main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
        int count = Convert.ToInt32(Console.ReadLine());
        string[] strings = new string[count];
        for (int i = 0; i < count; i++)
        {
            strings[i] = Console.ReadLine();
        }
        for (int i = 0; i < count; i++)
        {
            string s = strings[i];
            int res = FindMinDeletion(s);
            Console.WriteLine(res);
        }
    }
    
    static int FindMinDeletion(string s)
    {
        char[] chars = s.ToCharArray();
        char pervious = chars[0];
        int deletion = 0;
        for (int i = 1; i < chars.Length; i++)
        {
            char next = chars[i];
            if (pervious == next)
                deletion += 1;
            else
                pervious = next;
        }
        return deletion;
    }
}

Alternating Characters Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'alternatingCharacters' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts STRING s as parameter.
     */

    public static int alternatingCharacters(String s) {
    // Write your code here
      int currentChar = -1;
      int toReturn = 0;
      char[] chars = s.toCharArray();
      for(int i = 0; i <chars.length; i++) {
        if(currentChar == chars[i] && currentChar != -1) {
            toReturn++;
        } else {
            currentChar = chars[i];
        }
            
      }
      
      return toReturn;
       
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = Integer.parseInt(bufferedReader.readLine().trim());

        IntStream.range(0, q).forEach(qItr -> {
            try {
                String s = bufferedReader.readLine();

                int result = Result.alternatingCharacters(s);

                bufferedWriter.write(String.valueOf(result));
                bufferedWriter.newLine();
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        });

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Alternating Characters JavaScript Solution

function processData(input) {
    var inputArr = input.split('\n');
    var T = parseInt(inputArr.shift());
    if(T<1 || T>10)
        return;
    
    for(var j=0;j<T; j++){
        var i=0, noOfDels=0, k=0;
        var str = inputArr[j], len=str.length;
        if(len>100000)
            continue;
        
        while(i<len){
            k=i+1;
            while(str[i]==str[k]){
                ++noOfDels;
                k++;
            }
            i=k;
        }
        
        console.log(noOfDels);
    }
} 

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 Python Solution

N = int(input())

for i in range(N):
    s = input()
    lastchar = s[0]
    cnt = 0
    for char in s[1:]:
        if(char == lastchar):
            cnt += 1
        else:
            lastchar = char
    print(cnt)

Other Solutions

  • HackerRank The Full Counting Sort Solution
  • HackerRank Beautiful Binary String Solution
c, C#, C++, HackerRank Solutions, java, javascript, python Tags:C, cpp, CSharp, Hackerrank Solutions, java, javascript, python

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