Simple Cipher Amazon OA 2023

Contents

1 Simple Cipher Amazon OA 2023 Solution
- 1.1 SOLUTION
2 Amazon OA 2023 Questions with Solution

Simple Cipher Amazon OA 2023 Solution

As part of a Day 1 Challenge, your new team at Amazon has created a basic alphabet-based encryption and has asked members to test the cipher. A simple cipher is built on the alphabet wheel which has uppercase english letters[‘A’-‘Z’] written on it:

Given an encrypted string consisting of english letters[‘A’-‘Z’] only, decrypt the string by replacing each character with the kth character away on the wheel in counter clockwise direction. Counter-clockwise is the opposite direction is which the hands on a clock usually move.

Input

encryped: a string
k: an integer

Output

the decrypted string

Examples

Example 1:

Input:

1encryped = VTAOG

2k = 2

Output: TRYME

Explanation:

Looking back 2 from ‘V’ returns ‘T’, from ‘T’ returns ‘R’ and so on. The decrypted string is ‘TRYME’.

SOLUTION

Program: Simple Cipher Amazon OA Solution in C++

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int main()
{   
    string s; cin>>s;
    int k; cin>>k;
    int n=s.length();
    string ans="";
    k=k%26;
    for(int i=0;i<n;i++){
        int pos=s[i]-'A';
        if(k<=pos) ans+=s[i]-k;
        else ans+=s[i]+(26-k);
    }
    cout<<ans;
    return 0;
}

Program: Simple Cipher Amazon OA Solution in C++

A-Z letters have ascii range 65-90. We can convert char <-> int easily by primitive casting so, we need to take care of edge cases for character to be A or B to replace with Y & Z.

#include <bits/stdc++.h>
using namespace std;
int main() {
//code
vectordp={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
string s="TSNFA";
int k=10;
for(int i=0;i<s.size();i++){
    char it=s[i];
    if(it-'A'-k<0){
        s[i]=dp[it-'A'-k+26];
    }
    else{
        s[i]=dp[it-'A'-k];
    }
    
}
for(char it:s){
    cout<<it;
}
return 0;
}

Program: Simple Cipher Amazon OA Solution in Java

public static String simpleCipher(String encrypted, int k){
        char[] _encrypted = encrypted.toCharArray();
        for(int i=0; i < encrypted.length(); i++){
            char x = _encrypted[i];
            // if the previous kth element is greater than 'A'
            if(x-k>=65){
                _encrypted[i] = (char)(x-k);
            }
            //if ascii code of kth previous element if less than that of A add 26 to it
            else{
                _encrypted[i] = (char)(x-k+26);
            }
        }
        return new String(_encrypted);
    }

Program: Simple Cipher Amazon OA Solution in Java

This will give us time complexity = O(input string length) & space complexity ~= O(1) for using in memory replacement * another char[] since we create a new output string

String input = "VTAOG";
int k = 2;
int min = 65, max = 90;
char[] inputChars = input.toCharArray();
for(int p=0; p < inputChars.length; p++){
     int ascii = (int) inputChars[p];
    int targetAscii = ascii - k;
    if(targetAscii  < min){
     targetAscii = max - (ascii-targetAscii) - 1;   
//eg. k=3, ascii = 66 ie B, initial targetAscii = 66-3 ie 63, so final targetAscii =  90-(66-63)+1 ie 88 ie X where, '+1' ~= managing array. length-1 for integration target kind of logic 
   }
   
     inputChars[p] = (char) targetAscii;     //in memory replacement 
}
return new String(inputChars);

Program: Simple Cipher Amazon OA Solution in Python

The string is made up of uppercase English letters only. That means a total of 26 characters, so if k is > 26 we have to adjust it back to 0..25 by doing modulo.
Now for each character, I am converting into ASCII and subtracting ASCII of ‘A’ as ‘A’ is our base value and then subtracting k to go k steps backward.
If the value is < 0 e.g k = 2 and the character is ‘A’ so the value will be -2, but what we want is a value between 0..25, so adding 26 into the value.
value is between 0..25 and 0 is mapped to ‘A’, 1 is mapped to ‘B’, and so on. So, basically, we have to add ASCII of ‘A’ again into the value.
Now just convert the ASCII to the character and append it to string.

def simpleCipher(encrypted, k):
    k %= 26
    result = ""
    for char in encrypted:
        value = ord(char) - ord('A') - k
        if value < 0:
            value += 26
        value += ord('A')
        result += chr(value)
    return result

Simple Cipher Amazon OA 2023 Solution

SOLUTION

Amazon OA 2023 Questions with Solution