Is Count Commas in Range easy or hard?

Count Commas in Range is classified as an Easy problem. The main insight is recognizing that comma placement depends only on the number of digits. Once you group numbers by digit length, the implementation becomes a straightforward math calculation.

Count Commas in Range Python/Java solution

Implement the math approach by iterating over possible digit lengths, computing the overlap of each digit interval with [low, high], and multiplying by the comma count floor((digits-1)/3). The same logic translates directly across Python, Java, C++, Go, and TypeScript since it only uses integer arithmetic.

How to solve Count Commas in Range in O(log n)?

Calculate the digit length ranges that intersect with the query interval [low, high]. For each digit length d, determine how many numbers in the range have that length. Each of those numbers contributes floor((d-1)/3) commas. Summing these contributions across all digit groups produces the final answer in logarithmic time.

What is the best approach for Count Commas in Range?

The optimal solution groups numbers by digit length and counts how many numbers of each length fall within the range. Every number with the same digit length produces the same number of commas, equal to floor((digits-1)/3). By multiplying this comma count with the number of values in each digit interval, the problem can be solved in O(log n) time and O(1) space.

Is Count Commas in Range asked at Google/Amazon/Meta?

This type of question appears as a math-style brain teaser in interviews that test number properties and range counting. Similar problems have been asked at companies like Amazon and Google where candidates are expected to replace brute force iteration with mathematical aggregation.

What data structure is used in Count Commas in Range?

No complex data structure is required. The solution relies on arithmetic calculations, digit length checks, and simple range counting. The core idea comes from math and number grouping rather than arrays, trees, or hash maps.

What is the time complexity of Count Commas in Range?

The optimized math approach runs in O(log n) time because it only iterates through digit-length boundaries such as 1–3 digits, 4–6 digits, and so on. The brute force approach takes O(n log n) time since it processes every number in the range and computes its digit length individually.

Description Hints 1 Solution Video 3:19

3870. Count Commas in Range

Easy78.0% acceptance

You are given an integer n.

Return the total number of commas used when writing all integers from [1, n] (inclusive) in standard number formatting.

In standard formatting:

A comma is inserted after every three digits from the right.
Numbers with fewer than 4 digits contain no commas.

Example 1:

Input: n = 1002

Output: 3

Explanation:

The numbers "1,000", "1,001", and "1,002" each contain one comma, giving a total of 3.

Example 2:

Input: n = 998

Output: 0

Explanation:

All numbers from 1 to 998 have fewer than four digits. Therefore, no commas are used.

Constraints:

1 <= n <= 10⁵

Topics

Math

Hints open one at a time

Hint 1

Numbers in the range [1000, 100000] have one comma.

3:19

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1class Solution:
2    def countCommas(self, n: int) -> int:
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