LeetCode 3870 - Count Commas in Range

The problem asks us to count how many commas appear when writing every integer from 1 through n using standard number formatting. In standard formatting, commas are inserted every three digits from the right side of the number. For example: - 999 contains 0 commas.

leetcodeeasymath

LeetCode Problem 3870

Difficulty: 🟢 Easy
Topics: Math

Solution

LeetCode 3870 - Count Commas in Range

Problem Understanding

The problem asks us to count how many commas appear when writing every integer from 1 through n using standard number formatting.

In standard formatting, commas are inserted every three digits from the right side of the number. For example:

  • 999 contains 0 commas.
  • 1,000 contains 1 comma.
  • 12,345 contains 1 comma.
  • 1,000,000 contains 2 commas.

We are given a single integer n, and we must compute the total number of commas appearing across all formatted numbers in the inclusive range [1, n].

For example, if n = 1002, the numbers that contain commas are:

  • 1,000
  • 1,001
  • 1,002

Each contributes one comma, so the answer is 3.

The constraint is:

  • 1 <= n <= 10^5

This means the largest possible value is only 100,000. Even a straightforward solution that examines every number individually would be feasible. However, the problem is categorized under Math, which suggests there is a simpler counting observation that avoids processing every number one by one.

Some important edge cases include:

  • Very small values such as n = 1, where no number contains a comma.
  • Values just below a formatting boundary, such as 999, where the answer is still zero.
  • Values exactly at a boundary, such as 1000, where the first comma appears.
  • Values near the upper limit, such as 100000, where many numbers contain commas.

The problem guarantees that n is always positive, so we never need to handle zero or negative inputs.

Approaches

Brute Force

A direct solution is to iterate through every integer from 1 to n, determine how many commas its formatted representation would contain, and accumulate the total.

One way to determine the comma count for a number is:

  • Convert it to a string.
  • Compute (len(string) - 1) // 3.

This formula works because:

  • Numbers with 1 to 3 digits produce 0.
  • Numbers with 4 to 6 digits produce 1.
  • Numbers with 7 to 9 digits produce 2.

Since we process every number individually, the time complexity is O(n).

Although this is perfectly acceptable for n ≤ 100000, it does more work than necessary.

Key Insight

For the given constraint, n never exceeds 100,000, which means no number can contain more than one comma.

Consider the formatting ranges:

  • 1 to 999 contain 0 commas.
  • 1,000 to 999,999 contain 1 comma.

Since 100,000 lies within the second range, every number from 1000 through n contributes exactly one comma.

Therefore:

  • If n < 1000, the answer is 0.
  • Otherwise, every integer in [1000, n] contributes one comma.

The count of such integers is:

n - 1000 + 1 = n - 999

This gives an immediate mathematical solution.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force O(n) O(1) Check every number individually
Optimal O(1) O(1) Count how many numbers are at least 1000

Algorithm Walkthrough

  1. Check whether n is less than 1000.

If it is, then every number has at most three digits, so none of them contain commas. Return 0. 2. Otherwise, identify the first number that contains a comma.

The first such number is 1000. 3. Every integer from 1000 through n contributes exactly one comma.

Since the maximum value of n is 100000, there are no numbers with two commas. 4. Count how many integers lie in the range [1000, n].

The size of an inclusive range is:

n - 1000 + 1 5. Return the result:

n - 999

Why it works

Every number below 1000 contains zero commas. Every number from 1000 through 100000 contains exactly one comma because they have between four and six digits. Since the problem guarantees n ≤ 100000, no number can contain two commas. Therefore, the total comma count is exactly the number of integers in the interval [1000, n], which is n - 999.

Python Solution

class Solution:
    def countCommas(self, n: int) -> int:
        if n < 1000:
            return 0
        return n - 999

The implementation directly follows the mathematical observation.

The first condition handles all values below 1000. Since none of those numbers require comma separators, the answer is immediately 0.

For larger values, every number from 1000 onward contributes one comma. The count of numbers in that inclusive range is n - 1000 + 1, which simplifies to n - 999.

Because no iteration is required, the solution executes in constant time.

Go Solution

func countCommas(n int) int {
    if n < 1000 {
        return 0
    }
    return n - 999
}

The Go implementation is identical in logic to the Python version.

No special handling for integer overflow is required because the maximum value of n is only 100000. The calculation n - 999 safely fits within Go's int type on all supported platforms.

Worked Examples

Example 1

Input: n = 1002

Numbers containing commas:

  • 1,000
  • 1,001
  • 1,002

Algorithm trace:

Step Value
n 1002
n < 1000? No
Result 1002 - 999 = 3

Answer: 3

Example 2

Input: n = 998

Algorithm trace:

Step Value
n 998
n < 1000? Yes
Result 0

Answer: 0

Additional Example

Input: n = 1000

Algorithm trace:

Step Value
n 1000
n < 1000? No
Result 1000 - 999 = 1

Only the number 1,000 contains a comma, so the answer is 1.

Additional Example

Input: n = 100000

Algorithm trace:

Step Value
n 100000
n < 1000? No
Result 100000 - 999 = 99001

There are exactly 99,001 integers from 1000 through 100000, and each contributes one comma.

Complexity Analysis

Measure Complexity Explanation
Time O(1) Uses only a constant number of arithmetic operations
Space O(1) No extra data structures are allocated

The solution relies entirely on a mathematical counting observation. No loops, recursion, or auxiliary storage are required, so both time and space complexity remain constant regardless of the value of n.

Test Cases

sol = Solution()

assert sol.countCommas(1) == 0          # minimum input
assert sol.countCommas(9) == 0          # single-digit numbers
assert sol.countCommas(99) == 0         # two-digit boundary
assert sol.countCommas(999) == 0        # largest value without commas
assert sol.countCommas(1000) == 1       # first number with a comma
assert sol.countCommas(1001) == 2       # two comma-containing numbers
assert sol.countCommas(1002) == 3       # example 1
assert sol.countCommas(998) == 0        # example 2
assert sol.countCommas(1234) == 235     # numbers 1000..1234
assert sol.countCommas(9999) == 9000    # all numbers from 1000..9999
assert sol.countCommas(10000) == 9001   # crossing five digits
assert sol.countCommas(100000) == 99001 # maximum constraint

Test Summary

Test Why
n = 1 Minimum valid input
n = 9 Single-digit range
n = 99 Two-digit boundary
n = 999 Largest value with no commas
n = 1000 First comma-producing number
n = 1001 Small range above threshold
n = 1002 Official example
n = 998 Official example with zero result
n = 1234 General case
n = 9999 Large four-digit range
n = 10000 Five-digit boundary
n = 100000 Maximum constraint

Edge Cases

Numbers Below 1000

Values such as 1, 10, 100, and 999 contain at most three digits. Since commas are only inserted when a number has at least four digits, the correct answer is always zero. The implementation handles this through the initial n < 1000 check.

Exactly 1000

This is the first value where a comma appears. A common off-by-one mistake is forgetting that 1000 itself should be counted. The formula uses the inclusive range size n - 1000 + 1, ensuring that 1000 contributes one comma and the answer becomes 1.

Maximum Constraint

At n = 100000, every number from 1000 through 100000 contributes exactly one comma. A potential mistake is assuming larger numbers might contain two commas, but the constraint guarantees that n never reaches 1,000,000. Therefore, counting one comma per qualifying number is always correct, and the formula remains valid throughout the entire input range.