HackerRank Two Characters problem solution

HackerRank Two Characters problem solution – In this HackerRank Two Characters problem, we have given a string, remove characters until the string is made up of any two alternating characters. When you choose a character to remove, all instances of that character must be removed. Determine the longest string possible that contains just two alternating letters.

Function Description

Complete the alternate function in the editor below.

alternate has the following parameter(s):

• string s: a string

Returns.

• int: the length of the longest valid string, or 0 if there are none

Input Format

The first line contains a single integer that denotes the length of s.
The second line contains string s.

HackerRank Two Characters problem solution in Python.

#!/bin/python3

import sys

def valid(s):
    if len(s) <= 1:
        return False
    if s[0] == s[1]:
        return False
    if len(set(s)) > 2:
        return False
    for i in range(2, len(s)):
        if i%2 == 0:
            if s[i] != s[0]:
                return False
        else:
            if s[i] != s[1]:
                return False
    return True
    

n = int(input().strip())
s = input().strip()

r = 0
alp = 'abcdefghijklmnopqrstuvwxyz'
for k in alp:
    for l in alp:
        if k >= l:
            continue
        f = list(filter(lambda x: x == k or x == l, s))
        if valid(f):
            r = max(r, len(f))
print(r)

Two Characters problem solution in Java.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.util.stream.*;

public class Solution {
    static boolean isAlt(int[] str){
        return str[0]!=str[1] && IntStream.range(2,str.length).allMatch(i->str[i]==str[i%2]);
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int len = in.nextInt();
        String s = in.next();
        int[] nums=s.chars().distinct().toArray();
        int maxLength=0;
        for(int i=0; i<nums.length; i++){
            for(int j=i+1; j<nums.length; j++){
                int a=nums[i]; int b=nums[j];
                int [] removed=s.chars().filter(c->c==a||c==b).toArray();
                if(isAlt(removed) && removed.length>maxLength)
                    maxLength=removed.length;
            }
        }
        System.out.println(maxLength);
            
    }
}

Two Characters problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

bool alternating(string &s) {
    if (s.size() < 2) return 0;
    char a = s[0];
    char b = s[1];
    if (a == b) return 0;
    
    for (int i=2; i<s.size(); i++) {
        if (i & 1) {
            if (s[i] != b) return 0;
        } else if (s[i] != a) return 0;
    }
    
    return 1;
}

int main() {
    string s;
    cin >> s;
    cin >> s;
    
    int ans = 0;
    for (char a = 'a'; a <= 'z'; a++) {
        for (char b = 'a'; b <= 'z'; b++) {
            if (a == b) continue;
            string t;
            for (int i=0; i<s.size(); i++) {
                if (s[i] == a || s[i] == b) t += s[i];
            }
            
            if (alternating(t)) ans = max(ans, (int) t.size());
        }
    }
    cout << ans << endl;
    return 0;
}

Problem solution in C programming.

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

int alternating_length(char *s, char x, char y) {
    int len = 0, which = 0;
    for (int i = 0; s[i]; i++) {
        if (s[i] == x) {
            if (which == 0) {
                len++;
                which = 1;
            } else {
                return 0;
            }
        } else if (s[i] == y) {
            if (which == 1) {
                len++;
                which = 0;
            } else {
                return 0;
            }
        }
    }
    return len;
}

int main(void) {
    int n;
    scanf("%d", &n);
    char s[1002];
    scanf("%s", s);
    int maxLen = 0;
    for (char x = 'a'; x <= 'z'; x++) {
        for (char y = 'a'; y <= 'z'; y++) {
            if (x == y) continue;
            int len = alternating_length(s, x, y);
            if (len > maxLen) maxLen = len;
        }
    }
    if (maxLen == 1) maxLen = 0;
    printf("%dn", maxLen);
    return 0;
}

Problem solution in JavaScript programming.

process.stdin.resume();
process.stdin.setEncoding('ascii');

var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;

process.stdin.on('data', function (data) {
    input_stdin += data;
});

process.stdin.on('end', function () {
    input_stdin_array = input_stdin.split("n");
    main();    
});

function readLine() {
    return input_stdin_array[input_currentline++];
}

/////////////// ignore above this line ////////////////////

function main() {
    var len = parseInt(readLine());
    var s = readLine();
    var m = {};
    for (let i = 0, len = s.length; i < len; i++) {
        m[s[i]] ? m[s[i]]++ : m[s[i]] = 1;
    }
    const sorted = Object.keys(m).sort((a, b) => m[a] > m[b]);
    let pairs = [];
    for (let i = 0, len = sorted.length; i < len; i++) {
        for (let j = i + 1; j < len; j++) {
            Math.abs(m[sorted[i]] - m[sorted[j]]) <= 1 ? pairs.push([sorted[i], sorted[j]]) : null;
        }
    }
    let maxCount = 0;
    pairs.some(pair => {
        let isFirst = null;
        let count = 0;
        for (let i = 0, len = s.length; i < len; i++) {
            if (s[i] === pair[0]) {
                if (isFirst == null || !isFirst) {
                    isFirst = true;
                    count++;
                } else {
                    break;
                }
            }
            else if (s[i] === pair[1]) {
                if (isFirst == null || isFirst) {
                    isFirst = false;
                    count++;
                } else {
                    break;
                }
            }
            if (i === s.length - 1 && count > maxCount) {
                maxCount = count;
            }
        }
    });
    console.log(maxCount);
}