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: batInput: 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 stringstring 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 Codeint main(){ string S = "sun"; string N = "966"; cout << updateStr(S, N); return 0;} |
Java
// Java code to implement above approachimport 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 stringdef 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 CodeS = "sun"N = "966"print(updateStr(S, N))# This code is contributed by Saurabh Jaiswal |
C#
// C# code to implement above approachusing 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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