LeetCode 1945: Sum of Digits of String After Convert Solution

Master LeetCode problem 1945 (Sum of Digits of String After Convert), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1945. Sum of Digits of String After Convert

Easy

String Simulation

Problem Explanation

Explanation:

To solve this problem, we will follow these steps:

Convert each letter of the string to its corresponding position in the alphabet.
Sum the digits of the resulting integer.
Repeat the above step k times.

Algorithm:

Iterate over the characters in the string and convert each character to its position in the alphabet.
Calculate the sum of the digits of the resulting integer.
Repeat step 2 for k times.
Return the final sum.

Time Complexity:

The time complexity of this algorithm is O(n * k), where n is the length of the input string and k is the number of iterations.

Space Complexity:

The space complexity of this algorithm is O(1) as we are not using any extra space proportional to the input size.

Solution Code

class Solution {
    public int getLucky(String s, int k) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            int digit = c - 'a' + 1;
            sb.append(digit);
        }
        
        String numStr = sb.toString();
        int sum = 0;
        for (int i = 0; i < k; i++) {
            for (char c : numStr.toCharArray()) {
                sum += c - '0';
            }
            numStr = String.valueOf(sum);
            sum = 0;
        }
        
        return Integer.parseInt(numStr);
    }
}

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What is the time complexity of LeetCode 1945 (Sum of Digits of String After Convert)?

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