[Coding Challenge] Secure My Conversation by Encryption and Decryption
This is competitive coding question was asked in Goldman Sachs coding round. There were two coding questions which are needed to be solved in two hours.
Basically you can take one hour to solve this question.
Problem Statement: Secure My Conversation by Encryption and Decryption
Person A and B use an encryption-based system for their conversation.
Each conversation message is encrypted from the source and decrypted in the destination using a shared private positive number key known to each other.
The algorithm is illustrated with an example.
Input format with explanation:
- Operation (1 for Encryption and 2 for Decryption)
- Input message
- Input private key
Output format with explanation:
- Output message
Example 1:
Input:
1 Open 123
Output:
Oppeeen
Here, the input message characters are duplicated based on each digit in the key.
Example 2:
Input:
2 Oppeeen 123
Output:
Open
Here, the input message characters are compressed based on each digit in the key.
The conversation message and the private key need NOT be in equal length and the encoding/decoding takes place till the end is reached either conversation message or private key while retaining the rest of the conversation message.
Programming Solution
Python Code
#!/bin/python3 import math import os import random import re import sys # Complete the secureChannel function below. def secureChannel(operation, message, key): if len(message)==0 or len(key)==0: return "-1" strRet="" if operation==1: index=0 for keyVal in key: nKeyVal=int(keyVal) if index>=len(message): return strRet strRet=strRet+message[index]*nKeyVal index=index+1 if index < len(message): strRet=strRet+message[index:] elif operation==2: index=0 for keyVal in key: nKeyVal=int(keyVal) if index>=len(message): return strRet for i in range(0,nKeyVal): if (index+i)>=len(message): return "-1" if message[index] != message[index+i]: return "-1" strRet=strRet+message[index] index=index+nKeyVal if index < len(message): strRet=strRet+message[index:] else: return "-1" return strRet if __name__ == '__main__': print(secureChannel(1, 'Open', '123')) print(secureChannel(2, 'Oppeeen', '123'))
To understand this code, you can refer Python Programming tutorial.
C++ Code
(This solution in C++ is shared by the Shivani Singh.)
#include <string> #include <iostream> using namespace std; string encoding( string msg, string key) { string ans; int i=0;//for msg int j=0;//for key int keysize = key.size(); int msgsize= msg.size(); if(keysize==0 || msgsize==0) return "";//to be defined in question while(j<keysize && i<msgsize) { int count= key[j]-'0'; for(int it=0;it<count;it++) ans.push_back(msg[i]); j++; i++; } while(i< msgsize) { ans.push_back(msg[i]); i++; } return ans; } string decoding(string msg, string key) { string ans; int i=0;//for msg int j=0;//for key int keysize = key.size(); int msgsize= msg.size(); if(keysize==0 || msgsize==0) return " ";//to be defined in question while(i<msgsize && j<keysize) { ans.push_back(msg[i]); i=i+(key[j]-'0'); j++; } while(i<msgsize) { ans.push_back(msg[i]); i++; } return ans; } int main() { string msg= "open"; string key= "123"; cout<<encoding(msg,key)<<endl; cout<<decoding("oppeeen","123"); return 0; }
To understand this code, refer C/C++ programming tutorial.
Output:
Oppeeen Open
The program is self-explanatory. If you have any doubt or finding I difficult to understand the code, kindly comment your query below.
You can solve this coding question in any programming language. I choose Python as I feel more comfortable with it.
For more practice check other competitive coding challenge questions.
Comments
Shivani Singh
hey, I tried solving this question in c++.. but since I can’t test my code anywhere, feedbacks and corrections are welcome.
Aniruddha Chaudhari
Hi Shivani, I tested your code and it’s working. Awesome! I have added your code in our article.
This will help many. Thanks for submitting your code.
If you are interested to share your knowledge and contribute to our portal, kindly check here.