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Caesar Concatenation
  • Last Updated : 07 May, 2020

Given two strings str1 and str2 containing alpha-numeric characters and a number N. The task is to form a new encrypted string which contains the string str1 with a Ceaser Encryption of N characters and the string str2 with a Ceaser Encryption of N characters at odd indices.

Example:

Input: str1 = “GeekforGeeks”, str2 = “Geeks123”, N = 4
Output: KiiojsvKiiowKeikw163
Explanation:
Caesar Text for string str1 with a shift of 4 is “KiiojsvKiiow”
Caesar Text for string str2 with a shift of 4 at all even indexes is “Keikw163”
Resultant string is “KiiojsvKiiow” + “Keikw163” = “KiiojsvKiiowKeikw163”

Input: str1 = “ABcdE23”, str2 = “efda2w”, N = 9
Output: JKlmN12nfma1w
Explanation:
Caesar Text for string str1 with a shift of 9 is “JKlmN12”
Caesar Text for string str2 with a shift of 9 at all even indexes is “nfma1w”
Resultant string is “JKlmN12” + “nfma1w” = “JKlmN12nfma1w”

Approach:
This problem is an application of Caesar Cipher in Cryptography. Below are the steps:
The idea is to traverse the given string str1 and str2 and convert all the characters at every index of str1 and at even indexes of str2 by a shift of N on the basis of below 3 cases:



  1. Case 1: If characters lies between ‘A’ and ‘Z’ then the current character is encrypted as:
    new_character = ( (current_character - 65 + N) % 26 ) + 65;
    
  2. Case 2: If characters lies between ‘a’ and ‘z’ then the current character is encrypted as:
    new_character = ( (current_character - 97 + N) % 26 ) + 97;
    
  3. Case 3: If characters lies between ‘A’ and ‘Z’ then the current character is encrypted as:
    new_character = ( (current_character - 48 + N) % 10 ) + 48;
    

Below is the implementation of the above approach:

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// C++ implementation of the above
// approach
#include <bits/stdc++.h>
using namespace std;
  
void printCaesarText(string str1,
                     string str2, int N)
{
  
    // Traverse the string str1
    for (int i = 0; str1[i]; i++) {
  
        // Current character
        char ch = str1[i];
  
        // Case 1:
        if (ch >= 'A' && ch <= 'Z') {
            str1[i] = (ch - 65 + N) % 26 + 65;
        }
  
        // Case 2:
        else if (ch >= 'a' && ch <= 'z') {
            str1[i] = (ch - 97 + N) % 26 + 97;
        }
  
        // Case 3:
        else if (ch >= '0' && ch <= '9') {
            str1[i] = (ch - 48 + N) % 10 + 48;
        }
    }
  
    for (int i = 0; str2[i]; i++) {
  
        // If current index is odd, then
        // do nothing
        if (i & 1)
            continue;
  
        // Current character
        char ch = str2[i];
  
        // Case 1:
        if (ch >= 'A' && ch <= 'Z') {
            str2[i] = (ch - 65 + N) % 26 + 65;
        }
  
        // Case 2:
        else if (ch >= 'a' && ch <= 'z') {
            str2[i] = (ch - 97 + N) % 26 + 97;
        }
  
        // Case 3:
        else if (ch >= '0' && ch <= '9') {
            str2[i] = (ch - 48 + N) % 10 + 48;
        }
    }
  
    // Print the concatenated strings
    // str1 + str2
    cout << str1 + str2;
}
  
// Driver Code
int main()
{
  
    string str1 = "GeekforGeeks";
    string str2 = "Geeks123";
    int N = 4;
  
    printCaesarText(str1, str2, N);
  
    return 0;
}

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Output:

KiiojsvKiiowKeikw163

Time Complexity: O(N + M), where N and M are the length of the two given string.

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