Generate string by incrementing character of given string by number present at corresponding index of second string

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Given two strings S[] and N[] of the same size, the task is to update string S[] by adding the digit of string N[] of respective indices.

Examples:

Input: S = “sun”, N = “966”
Output: bat

Input: S = “apple”, N = “12580”
Output: brute

Approach: The idea is to traverse the string S[] from left to right. Get the ASCII value of string N[] and add it to the ASCII value of string S[]. If the value exceeds 122, which is the ASCII value of the last alphabet ‘z’. Then subtract the value by 26, which is the total count of English alphabets. Update string S with the character of ASCII value obtained. Follow the steps below to solve the problem:

Iterate over the range [0, S.size()) using the variable i and perform the following tasks:
- Initialize the variables a and b as the integer and ascii value of N[i] and S[i].
- If b is greater than 122 then subtract 26 from b.
- Set S[i] as char(b).
After performing the above steps, print the value of S[] as the answer.

Below is the implementation of the above approach.

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to update string
string updateStr(string S, string N)
{
    for (int i = 0; i < S.size(); i++) {
 
        // Get ASCII value
        int a = int(N[i]) - '0';
        int b = int(S[i]) + a;
 
        if (b > 122)
            b -= 26;
 
        S[i] = char(b);
    }
    return S;
}
 
// Driver Code
int main()
{
    string S = "sun";
    string N = "966";
    cout << updateStr(S, N);
    return 0;
}

Java

// Java code to implement above approach
import java.util.*;
public class GFG {
 
  // Function to update string
  static String updateStr(String S, String N)
  {
    String t = "";
    for (int i = 0; i < S.length(); i++) {
 
      // Get ASCII value
      int a = (int)(N.charAt(i) - '0');
      int b = (int)(S.charAt(i) + a);
 
      if (b > 122)
        b -= 26;
 
      char x = (char)b;
      t +=x;
    }
    return t;
  }
 
  // Driver code
  public static void main(String args[])
  {
    String S = "sun";
    String N = "966";
    System.out.println(updateStr(S, N));
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to update string
def updateStr(S, N):
    S = list(S)
    for i in range(len(S)):
 
        # Get ASCII value
        a = ord(N[i]) - ord('0')
        b = ord(S[i]) + a
 
        if (b > 122):
            b -= 26
 
        S[i] = chr(b)
    return "".join(S)
 
# Driver Code
 
S = "sun"
N = "966"
print(updateStr(S, N))
 
# This code is contributed by Saurabh Jaiswal

C#

// C# code to implement above approach
using System;
public class GFG {
 
  // Function to update string
  static String updateStr(String S, String N)
  {
    String t = "";
    for (int i = 0; i < S.Length; i++) {
 
      // Get ASCII value
      int a = (int)(N[i] - '0');
      int b = (int)(S[i] + a);
 
      if (b > 122)
        b -= 26;
 
      char x = (char)b;
      t +=x;
    }
    return t;
  }
 
  // Driver code
  public static void Main(String []args)
  {
    String S = "sun";
    String N = "966";
    Console.WriteLine(updateStr(S, N));
 
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

<script>
   // JavaScript code for the above approach
 
   // Function to update string
   function updateStr(S, N) {
     S = S.split('')
     for (let i = 0; i < S.length; i++) {
 
       // Get ASCII value
       let a = (N[i].charCodeAt(0) - '0'.charCodeAt(0));
       let b = (S[i].charCodeAt(0)) + a;
 
       if (b > 122)
         b -= 26;
 
       S[i] = String.fromCharCode(b);
     }
     return S.join('');
   }
 
   // Driver Code
 
   let S = "sun";
   let N = "966";
   document.write(updateStr(S, N));
 
 // This code is contributed by Potta Lokesh
 </script>

Output

bat

Time Complexity: O(|S|)
Auxiliary Space: O(1)

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