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:
- Case 1: If characters lies between ‘A’ and ‘Z’ then the current character is encrypted as:
new_character = ( (current_character - 65 + N) % 26 ) + 65;
- Case 2: If characters lies between ‘a’ and ‘z’ then the current character is encrypted as:
new_character = ( (current_character - 97 + N) % 26 ) + 97;
- 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:
// 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; } |
KiiojsvKiiowKeikw163
Time Complexity: O(N + M), where N and M are the length of the two given string.
