What is the Time Complexity of LeetCode 273. Integer to English Words?

Question

I am struggling to understand the time complexity of this solution. This question is about converting number to English words.

For Example, Input: num = 1234567891 Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

StringBuilder insert() has O(n) time Complexity.
I suspect the time Complexity is O(n^2). But I am not sure. Where n is the number of digits. Link to question: englishToWords

There is my code:

Code:

class Solution {
    private final String[] THOUSANDS = {"", "Thousand", "Million", "Billion"};
    
    private final String[] LESS_THAN_TWENTY = {"", "One", "Two", "Three", "Four", "Five", 
    "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen",
     "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
    
    private final String[] TENS = {"", "", "Twenty", "Thirty", "Forty", "Fifty",
     "Sixty", "Seventy", "Eighty", "Ninety"};
    //O(n) solution
    public String numberToWords(int num) {
        if (num == 0){
          return "Zero";  
        } 
        StringBuilder sb = new StringBuilder();
        int index = 0;
        //this contributes towards time complexity
        while (num > 0) {
            if (num % 1000 > 0) {
                StringBuilder tmp = new StringBuilder();
                helper(tmp, num % 1000);
                System.out.println(index);
                System.out.println("tmp: "+ tmp);
                tmp.append(THOUSANDS[index]).append(" ");
                //I suspect the time complexity will increase because of this to O(n^2)
                sb.insert(0, tmp);
            }
            index++;
            num = num / 1000;
        }
        return sb.toString().trim();
    }
    
    private void helper(StringBuilder tmp, int num) {
        if (num == 0) {
            return;
        } else if (num < 20) {
            tmp.append(LESS_THAN_TWENTY[num]).append(" ");
            return;
        } else if (num < 100) {
            tmp.append(TENS[num / 10]).append(" ");
            helper(tmp, num % 10);
        } else {
            tmp.append(LESS_THAN_TWENTY[num / 100]).append(" Hundred ");
            helper(tmp, num % 100);
        }
    }
}

Answer 1

Time Complexity is O(n) where n is the number of digits in the given number. We are traversing the number once.

I confirmed this from my few friends working in Cisco and Google.