3871. Count Commas in Range II

MediumMath

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Problem Pattern Idea Example Code Complex Pitfalls

Problem Description

You are given an integer n. Your task is to count the total number of commas that appear when you write out every integer from 1 to n (inclusive) using standard number formatting.

In standard formatting, the rules for placing commas are:

A comma is inserted after every three digits, counting from the right side of the number.
Numbers with fewer than 4 digits contain no commas at all.

For example:

The number 123 has no commas because it has only 3 digits.
The number 1234 is written as 1,234, which contains 1 comma.
The number 1234567 is written as 1,234,567, which contains 2 commas.

So for a given n, you need to add up the commas used across all numbers in the range [1, n] and return that total count.

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How We Pick the Algorithm

Why Math / Bit Manipulation?

This problem maps to Math / Bit Manipulation through a short path in the full flowchart.

Sums n - x + 1 for each power-of-1000 threshold x <= n, using O(log n) loop over exponentially growing thresholds.

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Intuition

Instead of writing out every single number and counting its commas one by one, let's think about how many commas each number contributes based on its size.

A number gains one comma for every group of three digits beyond the first three. Let's break this down by ranges:

Numbers from 1 to 999 have at most 3 digits, so they contribute 0 commas each.
Numbers from 1,000 to 999,999 have 4 to 6 digits, so they contribute 1 comma each.
Numbers from 1,000,000 to 999,999,999 have 7 to 9 digits, so they contribute 2 commas each.
And this pattern continues, with each new range adding one more comma.

Notice the key insight: a number gets an additional comma every time it reaches a new power of 1000. So a number ≥ 1000 gets its first comma, a number ≥ 1000000 gets a second comma, and so on.

This means we can count the total commas by counting, for each threshold x (where x = 1000, 1000000, 1000000000, ...), how many numbers in [1, n] are at least x. Each such number contributes exactly one comma for crossing that threshold.

For a given threshold x, the count of numbers from x to n is n - x + 1. By summing up n - x + 1 for every threshold x that does not exceed n, we accumulate the total comma count. We start at x = 1000 and keep multiplying by 1000 until x becomes larger than n.

Pattern Learn more about Math patterns.

Solution Approach

We use a mathematical approach based on the pattern we identified earlier.

Recall the observation:

Numbers in the range [1, 999] contain no commas;
Numbers in the range [1,000, 999,999] contain one comma;
Numbers in the range [1,000,000, 999,999,999] contain two commas;
And so on.

The key idea is that a number gains one additional comma each time it reaches a new power of 1000. So instead of counting commas per number, we count how many numbers cross each threshold x, where x takes the values 1000, 1000000, 1000000000, ....

Here is how we implement it step by step:

Initialize the answer ans = 0 to accumulate the total comma count.
Start with the first threshold x = 1000. This is the smallest number that requires a comma.
Loop while x <= n:
- For the current threshold x, every number in the range [x, n] gains one comma for crossing this threshold. The count of such numbers is n - x + 1.
- Add this count to the answer: ans += n - x + 1.
- Move to the next threshold by multiplying x by 1000: x *= 1000. This advances to the next group boundary where another comma is added.
Once x exceeds n, no more numbers can cross any further threshold, so we stop and return ans.

The reasoning behind ans += n - x + 1 is that any number ≥ x (and ≤ n) earns exactly one comma at this threshold level. Numbers that are large enough to pass multiple thresholds will be counted once at each threshold they exceed, which correctly sums up to their total comma count.

For example, when n = 1,500,000:

At x = 1000: numbers from 1000 to 1500000 get a comma, contributing 1500000 - 1000 + 1 = 1499001 commas.
At x = 1000000: numbers from 1000000 to 1500000 get a second comma, contributing 1500000 - 1000000 + 1 = 500001 commas.
At x = 1000000000: since x > n, the loop stops.
Total: 1499001 + 500001 = 1999002.

Complexity Analysis:

Time complexity: O(log n), because the threshold x is multiplied by 1000 each iteration, so the number of iterations grows logarithmically with n.
Space complexity: O(1), since we only use a constant number of variables.

Example Walkthrough

Let's trace through the solution with a small example: n = 2500.

We want to count the total number of commas written across all numbers from 1 to 2500.

Setting up:

Initialize ans = 0.
Start with the first threshold x = 1000.

Iteration 1: x = 1000

Check the loop condition: is 1000 <= 2500? Yes, so we proceed.
Every number in the range [1000, 2500] gains one comma for crossing this threshold (these are the 4-digit numbers like 1,000, 1,001, ..., 2,500).
Count of such numbers: n - x + 1 = 2500 - 1000 + 1 = 1501.
Update the answer: ans = 0 + 1501 = 1501.
Advance the threshold: x = 1000 * 1000 = 1000000.

Iteration 2: x = 1000000

Check the loop condition: is 1000000 <= 2500? No.
The loop stops.

Result:

Return ans = 1501.

Verifying the logic:

Numbers 1 through 999 have at most 3 digits, so they contribute 0 commas. ✓
Numbers 1000 through 2500 are written as 1,000 through 2,500, each containing exactly 1 comma. There are 2500 - 1000 + 1 = 1501 such numbers, contributing 1501 commas. ✓
No number in [1, 2500] reaches 1,000,000, so no second comma is ever added. ✓

The total of 1501 commas matches our manual reasoning, confirming the approach works correctly.

Solution Implementation

1class Solution:
2    def countCommas(self, n: int) -> int:
3        # Total number of comma separators across all integers from 1 to n
4        total_commas = 0
5
6        # Each power of 1000 marks where an additional comma is required:
7        # numbers >= 1000 need 1+ comma, numbers >= 1_000_000 need 2+ commas, etc.
8        power = 1000
9
10        # Continue while the current power threshold does not exceed n
11        while power <= n:
12            # Count how many integers in [power, n] cross this threshold;
13            # each such integer contributes one comma at this level
14            total_commas += n - power + 1
15
16            # Move to the next thousands grouping (next comma level)
17            power *= 1000
18
19        return total_commas
20

1class Solution {
2    /**
3     * Counts the total number of thousands-separator commas needed
4     * when formatting every integer from 1 to n.
5     *
6     * A number gets one comma for every power of 1000 it reaches or exceeds:
7     *   >= 1,000        -> 1 comma
8     *   >= 1,000,000    -> 2 commas
9     *   >= 1,000,000,000 -> 3 commas, and so on.
10     *
11     * For each power-of-1000 threshold "threshold", the count of integers
12     * in [1, n] that are >= threshold is exactly (n - threshold + 1).
13     * Summing this over all thresholds <= n yields the total comma count.
14     *
15     * @param n the upper bound of the range [1, n]
16     * @return the total number of commas across all numbers in the range
17     */
18    public long countCommas(long n) {
19        long totalCommas = 0;
20
21        // Iterate over each thousands threshold: 1e3, 1e6, 1e9, ...
22        // Multiply by 1000 each step to move to the next comma boundary.
23        for (long threshold = 1000; threshold <= n; threshold *= 1000) {
24            // Add the count of numbers in [threshold, n], each of which
25            // contributes one comma at this threshold level.
26            totalCommas += n - threshold + 1;
27        }
28
29        return totalCommas;
30    }
31}
32

1class Solution {
2public:
3    long long countCommas(long long n) {
4        // Accumulates the total number of comma separators needed
5        // when formatting every integer from 1 to n.
6        long long totalCommas = 0;
7
8        // Each 'boundary' marks a position where a new comma is introduced
9        // (1000 -> 1 comma, 1,000,000 -> 2 commas, etc.).
10        // Multiply by 1000 each iteration to advance to the next comma group.
11        for (long long boundary = 1000; boundary <= n; boundary *= 1000) {
12            // Every number in the range [boundary, n] requires one extra comma
13            // at this boundary level. There are (n - boundary + 1) such numbers.
14            totalCommas += n - boundary + 1;
15        }
16
17        return totalCommas;
18    }
19};
20

1/**
2 * Counts the total number of comma separators required when writing
3 * every integer from 1 to n using standard thousands grouping
4 * (e.g. 1,000 has one comma; 1,000,000 has two commas).
5 *
6 * @param n - The upper bound of the range [1, n].
7 * @returns The total count of commas across all numbers in the range.
8 */
9function countCommas(n: number): number {
10  // Accumulator for the total number of commas.
11  let count = 0;
12
13  // Iterate over each thousands boundary: 1000, 1000^2, 1000^3, ...
14  // Stop once the threshold exceeds n, since no number can reach it.
15  for (let threshold = 1000; threshold <= n; threshold *= 1000) {
16    // Every integer in [threshold, n] gains one comma at this boundary.
17    // There are (n - threshold + 1) such integers.
18    count += n - threshold + 1;
19  }
20
21  return count;
22}
23

Time and Space Complexity

Time Complexity: O(log n)

The core of the algorithm is the while loop where the variable x starts at 1000 and is multiplied by 1000 in each iteration (x *= 1000). The loop continues as long as x <= n. Since x grows exponentially (specifically, after k iterations x = 1000^(k+1)), the number of iterations is bounded by the value k for which 1000^k <= n. This gives k <= log_1000(n), so the loop executes approximately log_1000(n) times. Therefore, the time complexity is O(log n).
Space Complexity: O(1)

The algorithm only uses a constant number of variables (ans, x), regardless of the size of the input n. No additional data structures that scale with the input are allocated. Therefore, the space complexity is O(1).

Pattern Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall: Overflow when computing the next threshold power *= 1000

In languages with fixed-width integers (such as Java, C++, or C#), a subtle but dangerous bug occurs when advancing the threshold with power *= 1000. Even though the loop condition power <= n correctly stops processing, the multiplication power *= 1000 is executed before the condition is re-checked. If power is already large (for example, near the maximum representable value), multiplying it by 1000 can overflow and wrap around to a negative or small number, causing the loop to either run forever or produce a wildly incorrect result.

Consider n close to the upper bound of a 64-bit integer. Suppose power reaches 10^18. The next multiplication power *= 1000 yields 10^21, which exceeds the 64-bit signed range (about 9.2 × 10^18). Instead of becoming a huge value that safely fails the power <= n check, it overflows into a smaller (possibly negative) number that passes the check, breaking the algorithm.

In Python this pitfall does not occur, because Python integers have arbitrary precision and grow as needed. That is why the given Python code is correct as written. However, the same logic ported to a fixed-width language is vulnerable.

Solution: Guard the multiplication or restructure the loop to avoid overflow

The safest fix is to check whether multiplying would overflow before performing it, or to restructure the loop so the threshold is compared against n using division rather than multiplying it past the limit.

Approach 1 — divide instead of multiply to test the bound:

class Solution:
    def countCommas(self, n: int) -> int:
        total_commas = 0
        power = 1000
        while power <= n:
            total_commas += n - power + 1
            # Stop before overflowing: only multiply if it's safe.
            # In a fixed-width language, check `power > n // 1000` to break early.
            if power > n // 1000:
                break
            power *= 1000
        return total_commas

Here power > n // 1000 is equivalent to power * 1000 > n but performed via division, which never overflows. If the next threshold would exceed n anyway, there is no need to compute it, so we break out safely.

Approach 2 — In a language like Java, use an explicit overflow-safe guard before the multiplication:

class Solution {
    public long countCommas(long n) {
        long total = 0;
        long power = 1000;
        while (power <= n) {
            total += n - power + 1;
            if (power > n / 1000) break;  // next step would exceed n
            power *= 1000;
        }
        return total;
    }
}

Additional related pitfall: integer overflow in the accumulated answer

For very large n, the running total total_commas can itself grow large (on the order of n times the number of thresholds). In fixed-width languages, ensure the accumulator uses a 64-bit type (long) rather than a 32-bit int, otherwise the sum silently overflows even when the loop logic is correct. Python users are immune to this, but it is essential to keep in mind when translating the solution.

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