Hello Programmers, In this post, you will learn how to solve HackerRank Caesar Cipher Solution. This problem is a part of the HackerRank Algorithms Series.HackerRank Caesar Cipher 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 Caesar Cipher SolutionTaskJulius Caesar protected his confidential information by encrypting it using a cipher. Caesar’s cipher shifts each letter by a number of letters. If the shift takes you past the end of the alphabet, just rotate back to the front of the alphabet. In the case of a rotation by 3, w, x, y and z would map to z, a, b and c.Examples = There’s-a-starman-waiting-in-the-skyk = 3The alphabet is rotated by 3, matching the mapping above. The encrypted string is Wkhuh’v-d-vwdupqd-zdlwlqj-lq-wkh-vnb.Note: The cipher only encrypts letters; symbols, such as -, remain unencrypted.Function DescriptionComplete the caesarCipher function in the editor below.caesarCipher has the following parameter(s):string s: cleartextint k: the alphabet rotation factorReturnsstring: the encrypted stringInput FormatThe first line contains the integer, n, the length of the unencrypted string.The second line contains the unencrypted string, s.The third line contains k, the number of letters to rotate the alphabet by.Constraints1 <= n <= 1000 <= k <= 100s is a valid ASCII string without any spaces.Sample Input11middle-Outz2Sample Outputokffng-QwvbExplanationOriginal alphabet: abcdefghijklmnopqrstuvwxyzAlphabet rotated +2: cdefghijklmnopqrstuvwxyzabm -> oi -> kd -> fd -> fl -> ne -> gO -> Qu –> wt -> vz -> bHackerRank Caesar Cipher SolutionCaesar Cipher Solution in C#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int i,j,k,g,n,m=0;scanf("%d",&n); char ch1[130]={'a'},ch2[130]={'A'},s[100];scanf("%s",s);scanf("%d",&k); while(m<130) { for(i=m,g=0;i<m+26;i++) { ch1[i]='a'+g++; }m+=26; } m=0; while(m<130) { for(i=m,g=0;i<m+26;i++) { ch2[i]='A'+g++; }m+=26; } for(i=0;i<n;i++) { for(j=0;j<26;j++) { if(s[i]==ch1[j]) {s[i]=ch1[j+k];j=26;} else if(s[i]==ch2[j]) {s[i]=ch2[j+k];j=26;} } } printf("%s",s); return 0; }Caesar Cipher Solution in Cpp#include <iostream> #include <string> using namespace std; int main() { int N = 0, K = 0; string str, dummy; cin >> N; getline(cin, dummy); getline(cin, str); cin >> K; int len = str.length(); for (int i = 0; i < len; ++i) { if (65 <= str[i] && str[i] <= 90) str[i] = char(65 + ((str[i] - 65) + K) % 26); else if (97 <= str[i] && str[i] <= 122) str[i] = char(97 + ((str[i] - 97) + K) % 26); } cout << str << endl; return 0; }Caesar Cipher Solution in Javaimport java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner s = new Scanner(System.in); int len = s.nextInt(); s.nextLine(); String str = s.nextLine(); int shift = s.nextInt(); char sarr[] = str.toCharArray(); for (int i=0; i<sarr.length; i++) { sarr[i] = cryptIt(sarr[i], shift); } System.out.println(new String(sarr)); } public static char cryptIt(char c, int shift) { if (!Character.isAlphabetic(c)) return c; char base = 'A'; if (c >= 'a') base = 'a'; return (char)(((c - base + shift) % 26) + base); } }Caesar Cipher Solution in Pythondef ceasar(s,n): coded = [] for el in s: if ord(el) >96 and ord(el) < 123: a = chr((ord(el)-97 +n) % 26+97) coded.append(a) elif ord(el) > 64 and ord(el) < 91: a = chr((ord(el)-65+n) % 26 + 65) coded.append(a) else: coded.append(el) return ''.join(coded) l = int(raw_input()) s = raw_input() n = int(raw_input()) print ceasar(s,n)Caesar Cipher Solution using JavaScriptfunction processData(input) { //Enter your code here var lines = input.split('\n'); var chars = parseInt(lines[0].trim(), 10); var text = lines[1].trim(); var shift = parseInt(lines[2].trim(), 10)%26; var output = ""; for(var i=0; i<chars; i++) { var character = text.charCodeAt(i); if (character >= 65 && character <= 90) { character += shift; if (character > 90) { character = 64 + (character - 90); } } else if (character >= 97 && character <= 122) { character += shift; if (character > 122) { character = 96 + (character - 122); } } output += String.fromCharCode(character); } process.stdout.write(output+"\n") } process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { processData(_input); });Caesar Cipher Solution in Scalaobject Solution { def main(args: Array[String]) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */ val len = readInt val s = readLine val k = readInt var result: String = "" for (i <- 0 to len - 1) { val item = s.charAt(i) if (item >= 'a' && item <= 'z') { result = result + ((item - 'a' + k) % 26 + 'a').asInstanceOf[Char].toString } else if (item >= 'A' && item <= 'Z') { result = result + ((item - 'A' + k) % 26 + 'A').asInstanceOf[Char].toString } else { result = result + item.toString } } println(result) } }Caesar Cipher Solution in Pascalprogram caesar_cipher; var big,small:set of Char; N,K,i:integer; S:string; begin small:=['a'..'z']; big:=['A'..'Z']; readln(N); ReadLn(S); ReadLn(K); for i:=1 to N do begin if S[i] in small then begin if ord(S[i])+K > 122 then S[i]:=chr(ord('a')+((ord(S[i])+K-97) mod 26)) else S[i]:=chr(ord(S[i])+K); end; if S[i] in big then begin if ord(S[i])+K > 90 then S[i]:=chr(ord('A')+((ord(S[i])+K-65) mod 26)) else S[i]:=chr(ord(S[i])+K); end; end; Writeln(S); end. Disclaimer: This problem (Caesar Cipher) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: HackerRank Two Characters Solution Post navigationHackerRank Happy Ladybugs Solution HackerRank Two Characters Solution