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

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)