Find the character made by adding all the characters of the given string

Given a string str consisting of lowercase English alphabets. The task is to add all the character values i.e. ‘a’ = 1, ‘b’ = 2, ‘c’ = 3, …, ‘z’ = 26 and output the character corresponding to the sum value. If it exceeds 26 then take sum % 26.

Examples:

Input: str = “gfg” 
Output: t 
(g + f + g) = 7 + 6 + 7 = 20 and t = 20

Input: str = “geeks” 
Output: u 

Approach:

• Initialize a variable sum to 0.
• Loop through each character of the string.
• For each character, get its ASCII value by typecasting it to an integer and subtracting 96 from it (as ‘a’ has an ASCII value of 97, and we want ‘a’ to have a value of 1).
• Add this value to the sum.
• If sum is greater than 26, take its modulo with 26.
• Convert the value obtained in step 5 to the corresponding character value by adding 96 to it and typecasting it to a character. If the value is 0, set it to ‘z’.

Below is the implementation of above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Function to return the required character
char getChar(string str)
{
    int n = str.length();
    int max_sum = 0;
    char max_char = 'a';
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = i; j < n; j++) {
            sum += (str[j] - 'a' + 1);
            if (sum > max_sum) {
                max_sum = sum;
                if (max_sum % 26 == 0)
                    max_char = 'z';
                else
                    max_char = (char)('a' + max_sum % 26 - 1);
            }
        }
    }
    return max_char;
}

// Driver code
int main()
{
    string str = "gfg";

    cout << getChar(str);

    return 0;
}

Java

import java.util.*;

public class Main {
    public static void main(String[] args) {
        String str = "gfg";
        System.out.println(getChar(str));
    }

    public static char getChar(String str) {
        int n = str.length();
        int maxSum = 0;
        char maxChar = 'a';

        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
                sum += (str.charAt(j) - 'a' + 1);
                if (sum > maxSum) {
                    maxSum = sum;
                    if (maxSum % 26 == 0)
                        maxChar = 'z';
                    else
                        maxChar = (char) ('a' + maxSum % 26 - 1);
                }
            }
        }

        return maxChar;
    }
}

Python

# Function to return the required character
def get_char(s):
    n = len(s)
    max_sum = 0
    max_char = 'a'
    for i in range(n):
        sum_val = 0
        for j in range(i, n):
            sum_val += (ord(s[j]) - ord('a') + 1)
            if sum_val > max_sum:
                max_sum = sum_val
                if max_sum % 26 == 0:
                    max_char = 'z'
                else:
                    max_char = chr(ord('a') + max_sum % 26 - 1)
    return max_char

# Driver code


def main():
    s = "gfg"
    print(get_char(s))


if __name__ == "__main__":
    main()

C#

using System;

class Program {
    // Function to return the required character
    static char GetChar(string str)
    {
        int n = str.Length;
        int maxSum = 0;
        char maxChar = 'a';

        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
                sum += (str[j] - 'a' + 1);

                if (sum > maxSum) {
                    maxSum = sum;
                    if (maxSum % 26 == 0)
                        maxChar = 'z';
                    else
                        maxChar
                            = (char)('a' + maxSum % 26 - 1);
                }
            }
        }

        return maxChar;
    }

    // Driver code
    static void Main()
    {
        string str = "gfg";
        Console.WriteLine(GetChar(str));
    }
}
// sinudp5vi

Javascript

function getChar(str) {
    let n = str.length;
    let maxSum = 0;
    let maxChar = 'a';
    for (let i = 0; i < n; i++) {
        let sum = 0;
        for (let j = i; j < n; j++) {
            sum += (str[j].charCodeAt(0) - 'a'.charCodeAt(0) + 1);
            if (sum > maxSum) {
                maxSum = sum;
                if (maxSum % 26 === 0)
                    maxChar = 'z';
                else
                    maxChar = String.fromCharCode('a'.charCodeAt(0) + (maxSum % 26) - 1);
            }
        }
    }
    return maxChar;
}

// Test the getChar function
let str = "gfg";
console.log(getChar(str));

Output

t

Time Complexity: O(n^2) where n is the length of the given string.
Space Complexity: O(1), as we are not using any extra space.

Approach:

1. Find the sum of all characters of the string and store it in a variable sum.
2. If sum % 26 = 0 then print ‘z’.
3. Else update sum = sum % 26 and print (sum + ‘a’ + 1).

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the required character
char getChar(string str)
{

    // To store the sum of the characters
    // of the given string
    int sum = 0;

    for (int i = 0; i < str.length(); i++) {

        // Add the current character to the sum
        sum += (str[i] - 'a' + 1);
    }

    // Return the required character
    if (sum % 26 == 0)
        return 'z';
    else {
        sum = sum % 26;
        return (char)('a' + sum - 1);
    }
}

// Driver code
int main()
{
    string str = "gfg";

    cout << getChar(str);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the required character
static char getChar(String str)
{

    // To store the sum of the characters
    // of the given string
    int sum = 0;

    for (int i = 0; i < str.length(); i++)
    {

        // Add the current character to the sum
        sum += (str.charAt(i) - 'a' + 1);
    }

    // Return the required character
    if (sum % 26 == 0)
        return 'z';
    else
    {
        sum = sum % 26;
        return (char)('a' + sum - 1);
    }
}

// Driver code
public static void main (String[] args)
{
    String str = "gfg";

    System.out.println(getChar(str));
}
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return the required character
def getChar(strr):

    # To store the summ of the characters
    # of the given string
    summ = 0

    for i in range(len(strr)):

        # Add the current character to the summ
        summ += (ord(strr[i]) - ord('a') + 1)

    # Return the required character
    if (summ % 26 == 0):
        return ord('z')
    else:
        summ = summ % 26
        return chr(ord('a') + summ - 1)

# Driver code
strr = "gfg"

print(getChar(strr))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the required character
static char getChar(String str)
{

    // To store the sum of the characters
    // of the given string
    int sum = 0;

    for (int i = 0; i < str.Length; i++)
    {

        // Add the current character to the sum
        sum += (str[i] - 'a' + 1);
    }

    // Return the required character
    if (sum % 26 == 0)
        return 'z';
    else
    {
        sum = sum % 26;
        return (char)('a' + sum - 1);
    }
}

// Driver code
public static void Main (String[] args)
{
    String str = "gfg";

    Console.WriteLine(getChar(str));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the required character
    function getChar(str)
    {

        // To store the sum of the characters
        // of the given string
        let sum = 0;

        for (let i = 0; i < str.length; i++)
        {

            // Add the current character to the sum
            sum += (str[i].charCodeAt() - 'a'.charCodeAt() + 1);
        }

        // Return the required character
        if (sum % 26 == 0)
            return 'z';
        else {
            sum = sum % 26;
            return String.fromCharCode('a'.charCodeAt() + sum - 1);
        }
    }

    let str = "gfg";
    document.write(getChar(str));

    // This code is contributed by divyeshrabadiya07.
</script>

Output

t

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.