LeetCode #273 — HARD

Integer to English Words

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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"

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Example 3

Core Insight

What unlocks the optimal approach

Did you see a pattern in dividing the number into chunk of words? For example, 123 and 123000.
Group the number by thousands (3 digits). You can write a helper function that takes a number less than 1000 and convert just that chunk to words.
There are many edge cases. What are some good test cases? Does your code work with input such as 0? Or 1000010? (middle chunk is zero and should not be printed out)

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #273: Integer to English Words
class Solution {
    private String[] lt20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight",
        "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen",
        "Seventeen", "Eighteen", "Nineteen"};
    private String[] tens
        = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
    private String[] thousands = {"Billion", "Million", "Thousand", ""};

    public String numberToWords(int num) {
        if (num == 0) {
            return "Zero";
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 1000000000, j = 0; i > 0; i /= 1000, ++j) {
            if (num / i == 0) {
                continue;
            }
            sb.append(transfer(num / i)).append(thousands[j]).append(' ');
            num %= i;
        }
        return sb.toString().trim();
    }

    private String transfer(int num) {
        if (num == 0) {
            return "";
        }
        if (num < 20) {
            return lt20[num] + " ";
        }
        if (num < 100) {
            return tens[num / 10] + " " + transfer(num % 10);
        }
        return lt20[num / 100] + " Hundred " + transfer(num % 100);
    }
}

// Accepted solution for LeetCode #273: Integer to English Words
func numberToWords(num int) string {
	if num == 0 {
		return "Zero"
	}

	lt20 := []string{
		"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight",
		"Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen",
		"Sixteen", "Seventeen", "Eighteen", "Nineteen",
	}
	tens := []string{
		"", "Ten", "Twenty", "Thirty", "Forty", "Fifty",
		"Sixty", "Seventy", "Eighty", "Ninety",
	}
	thousands := []string{"Billion", "Million", "Thousand", ""}

	var transfer func(int) string
	transfer = func(num int) string {
		if num == 0 {
			return ""
		}
		if num < 20 {
			return lt20[num]
		}
		if num < 100 {
			if num%10 == 0 {
				return tens[num/10]
			}
			return tens[num/10] + " " + transfer(num%10)
		}
		if num%100 == 0 {
			return lt20[num/100] + " Hundred"
		}
		return lt20[num/100] + " Hundred " + transfer(num%100)
	}

	res := ""
	for i, j := 1000000000, 0; i > 0; i, j = i/1000, j+1 {
		cur := num / i
		if cur == 0 {
			continue
		}
		if res != "" {
			res += " "
		}
		res += transfer(cur)
		if thousands[j] != "" {
			res += " " + thousands[j]
		}
		num %= i
	}
	return res
}

# Accepted solution for LeetCode #273: Integer to English Words
class Solution:
    def numberToWords(self, num: int) -> str:
        if num == 0:
            return 'Zero'

        lt20 = [
            '',
            'One',
            'Two',
            'Three',
            'Four',
            'Five',
            'Six',
            'Seven',
            'Eight',
            'Nine',
            'Ten',
            'Eleven',
            'Twelve',
            'Thirteen',
            'Fourteen',
            'Fifteen',
            'Sixteen',
            'Seventeen',
            'Eighteen',
            'Nineteen',
        ]
        tens = [
            '',
            'Ten',
            'Twenty',
            'Thirty',
            'Forty',
            'Fifty',
            'Sixty',
            'Seventy',
            'Eighty',
            'Ninety',
        ]
        thousands = ['Billion', 'Million', 'Thousand', '']

        def transfer(num):
            if num == 0:
                return ''
            if num < 20:
                return lt20[num] + ' '
            if num < 100:
                return tens[num // 10] + ' ' + transfer(num % 10)
            return lt20[num // 100] + ' Hundred ' + transfer(num % 100)

        res = []
        i, j = 1000000000, 0
        while i > 0:
            if num // i != 0:
                res.append(transfer(num // i))
                res.append(thousands[j])
                res.append(' ')
                num %= i
            j += 1
            i //= 1000
        return ''.join(res).strip()

// Accepted solution for LeetCode #273: Integer to English Words
struct Solution;

impl Solution {
    fn number_to_words(num: i32) -> String {
        if num == 0 {
            return "Zero".to_string();
        }
        let nineteen: &'static str = "One Two Three Four Five Six Seven Eight Nine Ten Eleven Twelve Thirteen Fourteen Fifteen Sixteen Seventeen Eighteen Nineteen";
        let nineteen: Vec<&'static str> = nineteen.split_whitespace().collect();
        let tens: &'static str = "Twenty Thirty Forty Fifty Sixty Seventy Eighty Ninety";
        let tens: Vec<&'static str> = tens.split_whitespace().collect();
        let units: &'static str = "Hundred Thousand Million Billion";
        let units: Vec<&'static str> = units.split_whitespace().collect();
        Self::words(num as usize, &nineteen, &tens, &units).join(" ")
    }

    fn words(
        num: usize,
        nineteen: &[&'static str],
        tens: &[&'static str],
        units: &[&'static str],
    ) -> Vec<&'static str> {
        if num < 20 {
            if num > 0 {
                vec![nineteen[num - 1]]
            } else {
                vec![]
            }
        } else if num < 100 {
            let d = num / 10;
            vec![
                if d > 1 { vec![tens[d - 2]] } else { vec![] },
                Self::words(num % 10, nineteen, tens, units),
            ]
            .concat()
        } else if num < 1000 {
            vec![
                Self::words(num / 100, nineteen, tens, units),
                vec![units[0]],
                Self::words(num % 100, nineteen, tens, units),
            ]
            .concat()
        } else if num < 1_000_000 {
            vec![
                Self::words(num / 1000, nineteen, tens, units),
                vec![units[1]],
                Self::words(num % 1000, nineteen, tens, units),
            ]
            .concat()
        } else if num < 1_000_000_000 {
            vec![
                Self::words(num / 1_000_000, nineteen, tens, units),
                vec![units[2]],
                Self::words(num % 1_000_000, nineteen, tens, units),
            ]
            .concat()
        } else {
            vec![
                Self::words(num / 1_000_000_000, nineteen, tens, units),
                vec![units[3]],
                Self::words(num % 1_000_000_000, nineteen, tens, units),
            ]
            .concat()
        }
    }
}

#[test]
fn test() {
    let num = 123;
    let res = "One Hundred Twenty Three".to_string();
    assert_eq!(Solution::number_to_words(num), res);
    let num = 12345;
    let res = "Twelve Thousand Three Hundred Forty Five".to_string();
    assert_eq!(Solution::number_to_words(num), res);
    let num = 1_234_567;
    let res = "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven".to_string();
    assert_eq!(Solution::number_to_words(num), res);
    let num = 1_234_567_891;
    let res = "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One".to_string();
    assert_eq!(Solution::number_to_words(num), res);
    let num = 20;
    let res = "Twenty".to_string();
    assert_eq!(Solution::number_to_words(num), res);
}

// Accepted solution for LeetCode #273: Integer to English Words
function numberToWords(num: number): string {
    if (num === 0) {
        return 'Zero';
    }

    const lt20 = [
        '',
        'One',
        'Two',
        'Three',
        'Four',
        'Five',
        'Six',
        'Seven',
        'Eight',
        'Nine',
        'Ten',
        'Eleven',
        'Twelve',
        'Thirteen',
        'Fourteen',
        'Fifteen',
        'Sixteen',
        'Seventeen',
        'Eighteen',
        'Nineteen',
    ];

    const tens = [
        '',
        'Ten',
        'Twenty',
        'Thirty',
        'Forty',
        'Fifty',
        'Sixty',
        'Seventy',
        'Eighty',
        'Ninety',
    ];

    const thousands = ['Billion', 'Million', 'Thousand', ''];

    const transfer = (n: number): string => {
        if (n === 0) {
            return '';
        }
        if (n < 20) {
            return lt20[n];
        }
        if (n < 100) {
            if (n % 10 === 0) {
                return tens[Math.floor(n / 10)];
            }
            return tens[Math.floor(n / 10)] + ' ' + transfer(n % 10);
        }
        if (n % 100 === 0) {
            return lt20[Math.floor(n / 100)] + ' Hundred';
        }
        return lt20[Math.floor(n / 100)] + ' Hundred ' + transfer(n % 100);
    };

    let res = '';
    for (let i = 1_000_000_000, j = 0; i > 0; i = Math.floor(i / 1000), ++j) {
        const cur = Math.floor(num / i);
        if (cur === 0) {
            continue;
        }
        if (res !== '') {
            res += ' ';
        }
        res += transfer(cur);
        if (thousands[j] !== '') {
            res += ' ' + thousands[j];
        }
        num %= i;
    }
    return res;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.