How do you solve Integer to English Words on LeetCode?

Integer to English Words (LeetCode #273) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Breaking down the number into chunks of three digits helps in managing the conversion.

What is the brute force solution for Integer to English Words?

The brute force approach for Integer to English Words is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves converting each segment of the number into words without any optimization. We will break down the number into groups of three digits and convert each group separately, which can lead to repetitive calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Integer to English Words?

Integer to English Words uses the following concepts: Math, String, Recursion. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Integer to English Words?

Explain your thought process clearly while coding. Test your code with different inputs, especially edge cases. Be prepared to discuss the time and space complexities of your solution.

What are common mistakes when solving Integer to English Words?

Not handling edge cases like zero properly. Failing to manage spaces between words correctly.

#273

Integer to English Words

Hard

MathStringRecursionHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach builds on the brute force method but optimizes the way we handle the conversion of numbers into words. By using a helper function for numbers less than 1000 and processing the number in chunks, we reduce redundancy and improve efficiency.

⚙️

Algorithm

3 steps

1Step 1: Create a helper function to convert numbers less than 1000 to words.
2Step 2: Use a loop to process the input number in chunks of three digits (thousands).
3Step 3: Concatenate the results from the helper function with the appropriate place values.

solution.py25 lines

1# Full working Python code
2class Solution:
3    def numberToWords(self, num: int) -> str:
4        if num == 0:
5            return 'Zero'
6        below_20 = ['', 'One', 'Two', 'Three', 'Four', 'Five', 'Six', 'Seven', 'Eight', 'Nine', 'Ten', 'Eleven', 'Twelve', 'Thirteen', 'Fourteen', 'Fifteen', 'Sixteen', 'Seventeen', 'Eighteen', 'Nineteen']
7        tens = ['', '', 'Twenty', 'Thirty', 'Forty', 'Fifty', 'Sixty', 'Seventy', 'Eighty', 'Ninety']
8        thousands = ['', 'Thousand', 'Million', 'Billion']
9
10        def helper(n):
11            if n == 0:
12                return ''
13            elif n < 20:
14                return below_20[n] + ' '
15            elif n < 100:
16                return tens[n // 10] + ' ' + helper(n % 10)
17            else:
18                return below_20[n // 100] + ' Hundred ' + helper(n % 100)
19
20        res = ''
21        for i in range(len(thousands)):
22            if num % 1000 != 0:
23                res = helper(num % 1000) + thousands[i] + ' ' + res
24            num //= 1000
25        return res.strip()

ℹ

Complexity note: The time complexity is O(n) because we process each digit of the number only once. The space complexity is O(n) due to the storage of the resulting string.

1Breaking down the number into chunks of three digits helps in managing the conversion.
2Using a helper function for numbers less than 1000 avoids redundancy.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.