What is the time complexity of Integer to English Words?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Integer to English Words?

This problem uses Math, String, Recursion. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Integer to English Words?

Insight 1: The problem can be broken down into smaller subproblems by processing the number in chunks of three digits (hundreds, tens, and ones). Insight 2: Using lookup tables (arrays or lists) to store the English words for numbers below 20, tens (20, 30, ..., 90), and powers of 1000 (Thousand, Million, Billion) simplifies the conversion process.

Which companies ask LeetCode 273. Integer to English Words in interviews?

This problem is frequently asked at Attentive, Avalara, Block, Delhivery, Gusto and 8 more companies.

The problem requires converting a non-negative integer into its English word representation. The input is an integer `num` between 0 and 2 31 - 1. The output is a string representing the English words for the given number. For example, 123 should be converted to "One Hundred Twenty Three".

Solution in python3. Solution in java. Solution in cpp. Solution in c

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Integer to English Words - LeetCode 273 Solution

Integer to English Words - Complete Solution Guide

Integer to English Words is LeetCode problem 273, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 <= 2 31 - 1

Detailed Explanation

The problem requires converting a non-negative integer into its English word representation. The input is an integer `num` between 0 and 2<sup>31</sup> - 1. The output is a string representing the English words for the given number. For example, 123 should be converted to "One Hundred Twenty Three".

Solution Approach

The solution uses a divide-and-conquer approach. It first handles the base case where the number is 0. Then, it defines lookup tables for numbers less than 20, tens, and thousands. A helper function `to_words_helper` converts a number less than 1000 into words. The main function iterates through the input number in chunks of 1000, converting each chunk to words using the helper function and appending the appropriate scale (Thousand, Million, Billion). Finally, it joins the words with spaces to form the final string.

Step-by-Step Algorithm

Step 1: Handle the base case: If the number is 0, return "Zero".
Step 2: Define lookup tables: Create arrays or lists containing English words for numbers less than 20, tens, and powers of 1000.
Step 3: Implement a helper function `to_words_helper(n)`: This function converts a number `n` less than 1000 into words. It handles numbers less than 20 directly. For numbers between 20 and 99, it uses the tens array and the ones array. For numbers between 100 and 999, it adds the "Hundred" suffix and recursively calls the helper for the remaining two digits.
Step 4: Iterate through the input number in chunks of 1000: Use a `while` loop to process the input number. In each iteration, extract the last three digits (`num % 1000`).
Step 5: Convert the chunk to words using `to_words_helper`: If the chunk is non-zero, convert it to words and append the appropriate scale (Thousand, Million, Billion) based on the current iteration.
Step 6: Prepend the chunk words to the result: Add the chunk words (and the scale) to the beginning of the result list.
Step 7: Update the number and scale index: Divide the number by 1000 (`num //= 1000`) and increment the scale index.
Step 8: Join the words in the result list with spaces: Create the final string by joining the words in the result list with spaces.

Key Insights

Insight 1: The problem can be broken down into smaller subproblems by processing the number in chunks of three digits (hundreds, tens, and ones).
Insight 2: Using lookup tables (arrays or lists) to store the English words for numbers below 20, tens (20, 30, ..., 90), and powers of 1000 (Thousand, Million, Billion) simplifies the conversion process.
Insight 3: A recursive or iterative helper function can be used to convert a three-digit number into its English word representation. Handling the edge case of zero is also essential.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Recursion.

Companies

Asked at: Attentive, Avalara, Block, Delhivery, Gusto, Nordstrom, Palantir Technologies, Roblox, Snowflake, Warnermedia, Yext, Zoho, eBay.

Integer to English Words | LeetCode 273

Hard

Math String Recursion

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public String numberToWords(int num) {
        if (num == 0) {
            return "Zero";
        }

        String[] below_20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
        String[] tens = {"", "", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
        String[] thousands = {"", "Thousand", "Million", "Billion"};

        
        
        return String.join(" ", toWordsHelper(num, below_20, tens, thousands));
    }

    private static java.util.List<String> toWordsHelper(int n, String[] below_20, String[] tens, String[] thousands) {
            java.util.List<String> res = new java.util.ArrayList<>();
            if (n < 20) {
                res.add(below_20[n]);
                return res;
            }
            if (n < 100) {
                res.add(tens[n / 10]);
                if (n % 10 != 0) {
                    res.add(below_20[n % 10]);
                }
                return res;
            }

            res.add(below_20[n / 100]);
            res.add("Hundred");
            if (n % 100 != 0) {
                res.addAll(toWordsHelper(n % 100, below_20, tens, thousands));
            }
            return res;
        }
    
}

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Problem Info

Description

Problem 273: Integer to English Words - Detailed explanation and solution approaches.

Topics

Math String Recursion

Companies

AttentiveAvalaraBlockDelhiveryGustoNordstromPalantir TechnologiesRoblox

Stats

Difficulty:Hard

Languages:4

Companies:13

Problem #:273

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Handle the base case: If the number is 0, return "Zero".
2
Step 2: Define lookup tables: Create arrays or lists containing English words for numbers less than 20, tens, and powers of 1000.
3
Step 3: Implement a helper function `to_words_helper(n)`: This function converts a number `n` less than 1000 into words. It handles numbers less than 20 directly. For numbers between 20 and 99, it uses the tens array and the ones array. For numbers between 100 and 999, it adds the "Hundred" suffix and recursively calls the helper for the remaining two digits.
4
Step 4: Iterate through the input number in chunks of 1000: Use a `while` loop to process the input number. In each iteration, extract the last three digits (`num % 1000`).
5
Step 5: Convert the chunk to words using `to_words_helper`: If the chunk is non-zero, convert it to words and append the appropriate scale (Thousand, Million, Billion) based on the current iteration.
6
Step 6: Prepend the chunk words to the result: Add the chunk words (and the scale) to the beginning of the result list.
7
Step 7: Update the number and scale index: Divide the number by 1000 (`num //= 1000`) and increment the scale index.
8
Step 8: Join the words in the result list with spaces: Create the final string by joining the words in the result list with spaces.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The number of operations is bounded by the maximum possible input (2<sup>31</sup> - 1), which has a fixed number of digits. Thus, the algorithm performs a fixed number of operations regardless of the specific input value. The space complexity is also O(1) because the lookup tables have a fixed size and don't scale with the input number. The intermediate list to hold the words will also be bound by the input size, which is limited to 2^31-1, making it constant.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Approach with a Different Structure

Instead of prepending words to the result list, you can append them. This might require more careful handling of leading and trailing spaces.

Time: O(1)Space: O(1)

Trade-offs: Might be slightly less readable due to more complex space handling, but performance is similar.

✨

Key Insights

1Insight 1: The problem can be broken down into smaller subproblems by processing the number in chunks of three digits (hundreds, tens, and ones).
2Insight 2: Using lookup tables (arrays or lists) to store the English words for numbers below 20, tens (20, 30, ..., 90), and powers of 1000 (Thousand, Million, Billion) simplifies the conversion process.
3Insight 3: A recursive or iterative helper function can be used to convert a three-digit number into its English word representation. Handling the edge case of zero is also essential.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Input is 0. The solution handles this by returning "Zero".
Edge case 2: Numbers less than 20. The solution uses the `below_20` array to handle these numbers directly.
Edge case 3: Numbers between 20 and 99. The solution uses the `tens` and `below_20` arrays to combine the tens and ones digits.
Edge case 4: Numbers ending in 00 (e.g., 100, 2000). The solution handles them correctly by not adding an extra "and".

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the input is 0.

✗Common mistake 2: Incorrectly handling the "Hundred" suffix. It should only be added if there are other digits.

✗Common mistake 3: Forgetting to add the appropriate scale (Thousand, Million, Billion) after each chunk.

✗Common mistake 4: Incorrectly handling numbers between 11 and 19 (e.g., mistaking 12 for "Ten Two").

✗Common mistake 5: Not removing leading/trailing whitespace in the output string.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Converting numerical data into human-readable text in financial reports, documents, or applications.

🏦Real-world scenario 2: Speech synthesis and text-to-speech systems need to convert numbers into their spoken form.

🎮Real-world scenario 3: Generating invoices or checks where the amount needs to be written out in words to prevent fraud or errors.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Integer to English Words"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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