# Solving Leetcode Interviews in Seconds with AI: Integer to English Words


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "273" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Convert a non-negative integer num to its English words representation.   Example 1:  Input: num = 123 Output: "One Hundred Twenty Three"  Example 2:  Input: num = 12345 Output: "Twelve Thousand Three Hundred Forty Five"  Example 3:  Input: num = 1234567 Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"    Constraints:  0 <= num <= 231 - 1  

	# Explanation
	Here's a solution for converting non-negative integers to their English word representations, along with an explanation of the approach and complexity analysis.

*   **Divide and Conquer:** Break down the number into groups of three digits (hundreds, tens, and ones) starting from the right. Process each group independently.
*   **Lookup Tables:** Use dictionaries to store the English words for numbers less than 20 and for multiples of 10. This avoids complex conditional logic.
*   **Handle Place Values:** Append the appropriate place value (Thousand, Million, Billion) after processing each group of three digits.

*   **Time Complexity:** O(1), as the number of operations is bounded by the maximum possible input value (2<sup>31</sup> - 1), and the lookup tables provide constant-time access. **Space Complexity:** O(1), as the space used is determined by the size of the lookup tables, which is constant.

	
	# Code
	```python
	class Solution:
    def numberToWords(self, num: int) -> str:
        less_than_20 = ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven",
                          "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"]
        tens = ["", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"]
        thousands = ["", "Thousand", "Million", "Billion"]

        def helper(n):
            if n < 20:
                return less_than_20[n]
            elif n < 100:
                return tens[n // 10] + (" " + less_than_20[n % 10]) if n % 10 != 0 else tens[n//10]
            else:
                return less_than_20[n // 100] + " Hundred " + helper(n % 100)

        if num == 0:
            return "Zero"

        result = []
        for i in range(4):
            if num % 1000 != 0:
                result.append(helper(num % 1000) + " " + thousands[i])
            num //= 1000

        return " ".join(result[::-1]).strip()
	```