Replace every character of string by character whose ASCII value is K times more than it

Difficulty Level : Easy
Last Updated : 12 May, 2021

Given string str consisting of lowercase letters only and an integer k, the task is to replace every character of the given string with a character whose ASCII value is k times more than it. If the ASCII value exceeds ‘z’, then start checking from ‘a’ in a cyclic manner.

Examples:

Input: str = “abc”, k = 2
Output: cde
a is moved by 2 times which results in character c
b is moved by 2 times which results in character d
c is moved by 2 times which results in character e
Input: str = “abc”, k = 28
Output: cde
a is moved 25 times, z is reached. Then 26th character will be a, 27-th b and 28-th c.
b is moved 24 times, z is reached. 28-th is d.
b is moved 23 times, z is reached. 28-th is e.

Approach: Iterate for every character in the string and perform the below steps for each character:

Add k to the ASCII value of character str[i].
If it exceeds 122, then perform a modulus operation of k with 26 to reduce the number of steps, as 26 is the maximum number of shifts that can be performed in a rotation.
To find the character, add k to 96. Hence, the character with ASCII value k+96 will be a new character.

Repeat the above steps for every character of the given string.

Below is the implementation of the above approach:

C++

// CPP program to move every character
// K times ahead in a given string
#include <bits/stdc++.h>
using namespace std;
 
// Function to move string character
void encode(string s,int k){
 
    // changed string
    string newS;
 
    // iterate for every characters
    for(int i=0; i<s.length(); ++i)
    {
        // ASCII value
        int val = int(s[i]);
 
        // store the duplicate
        int dup = k;
 
        // if k-th ahead character exceed 'z'
        if(val + k > 122){
            k -= (122-val);
            k = k % 26;
            newS += char(96 + k);
        }
        else
            newS += char(val + k);
 
        k = dup;
    }
 
    // print the new string
    cout<<newS;
}
 
// driver code
int main(){
    string str = "abc";
    int k = 28;
 
    // function call
    encode(str, k);
 
    return 0;
}
 
// This code is contributed by Sanjit_Prasad

Java

// Java program to move every character
// K times ahead in a given string
 
class GFG {
 
// Function to move string character
    static void encode(String s, int k) {
 
        // changed string
        String newS = "";
 
        // iterate for every characters
        for (int i = 0; i < s.length(); ++i) {
            // ASCII value
            int val = s.charAt(i);
            // store the duplicate
            int dup = k;
 
            // if k-th ahead character exceed 'z'
            if (val + k > 122) {
                k -= (122 - val);
                k = k % 26;
                 
                newS += (char)(96 + k);
            } else {
                newS += (char)(val + k);
            }
 
            k = dup;
        }
 
        // print the new string
        System.out.println(newS);
    }
 
// Driver Code
    public static void main(String[] args) {
        String str = "abc";
        int k = 28;
 
        // function call
        encode(str, k);
    }
}
 
// This code is contributed by Rajput-JI

Python3

# Python program to move every character
# K times ahead in a given string
 
# Function to move string character
def encode(s, k):
     
    # changed string
    newS = ""
     
    # iterate for every characters
    for i in range(len(s)):
         
        # ASCII value
        val = ord(s[i])
         
        # store the duplicate
        dup = k
         
        # if k-th ahead character exceed 'z'
        if val + k>122:
            k -= (122-val)
            k = k % 26
            newS += chr(96 + k)
             
        else:
            newS += chr(val + k)
         
        k = dup
     
    # print the new string
    print (newS)
             
# driver code    
str = "abc"
k = 28
 
encode(str, k)

C#

// C# program to move every character
// K times ahead in a given string
using System;
public class GFG {
 
// Function to move string character
    static void encode(String s, int k) {
 
        // changed string
        String newS = "";
 
        // iterate for every characters
        for (int i = 0; i < s.Length; ++i) {
            // ASCII value
            int val = s[i];
            // store the duplicate
            int dup = k;
 
            // if k-th ahead character exceed 'z'
            if (val + k > 122) {
                k -= (122 - val);
                k = k % 26;
                 
                newS += (char)(96 + k);
            } else {
                newS += (char)(96 + k);
            }
 
            k = dup;
        }
 
        // print the new string
        Console.Write(newS);
    }
 
// Driver Code
    public static void Main() {
        String str = "abc";
        int k = 28;
 
        // function call
        encode(str, k);
    }
}
 
// This code is contributed by Rajput-JI

PHP

<?php
// PHP program to move every character
// K times ahead in a given string
 
// Function to move string character
function encode($s, $k)
{
 
    // changed string
    $newS = "";
 
    // iterate for every characters
    for($i = 0; $i < strlen($s); ++$i)
    {
        // ASCII value
        $val = ord($s[$i]);
 
        // store the duplicate
        $dup = $k;
 
        // if k-th ahead character exceed 'z'
        if($val + $k > 122)
        {
            $k -= (122 - $val);
            $k = $k % 26;
            $newS = $newS.chr(96 + $k);
        }
        else
            $newS = $newS.chr($val + $k);
 
        $k = $dup;
    }
 
    // print the new string
    echo $newS;
}
 
// Driver code
 
$str = "abc";
$k = 28;
 
// function call
encode($str, $k);
 
// This code is contributed by ita_c
?>

Javascript

<script>
// Javascript program to move every character
// K times ahead in a given string
 
     
// Function to move string character
 
    function encode(s,k)
    {
        // changed string
        let newS = "";
   
        // iterate for every characters
        for (let i = 0; i < s.length; ++i) {
            // ASCII value
            let val = s[i].charCodeAt(0);
            // store the duplicate
            let dup = k;
   
            // if k-th ahead character exceed 'z'
            if (val + k > 122) {
                k -= (122 - val);
                k = k % 26;
                   
                newS += String.fromCharCode(96 + k);
            } else {
                newS += String.fromCharCode(val + k);
            }
   
            k = dup;
        }
   
        // print the new string
        document.write(newS);
    }
     
    // Driver Code
    let str = "abc";
    let k = 28;
     
    // function call
    encode(str, k);
     
 
 
// This code is contributed by rag2127
</script>

Output:

cde

Time Complexity: O(N), where N is the length of the string.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes