Given two strings str1 and str2 containing alphanumeric 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;
}
Output:
KiiojsvKiiowKeikw163
Time Complexity: O(N + M), where N and M are the lengths of the two given string.
Top comments (1)
chcan be reference to avoid writingstr[i]. Instead you can writech = ch + 1. If the index is not used in computations, you may use a range-based for loop:Weird include.