3871. Count Commas in Range II

Difficulty: Medium | Published on 2026-09-08

Same problem as Part I, word for word — except INLINECODE0 can now be as large as INLINECODE1 instead of INLINECODE2. That single constraint change is the entire story here: the simple "loop from 1 to n" brute force is no longer just slow, it's completely impossible to run in time. Let's confirm why, and reuse the grouped approach from Part I — this time as a requirement, not a bonus technique.

1. Problem in Very Simple Language

Identical to Part I: write out every integer from INLINECODE3 to INLINECODE4 using standard comma formatting (a comma after every group of 3 digits from the right; numbers under 1000 get no commas). Count the total number of commas used across all of them.

What's given: a single integer INLINECODE5 (up to INLINECODE6). What's different: INLINECODE7 can now be up to INLINECODE8 (a quadrillion) — a number with 16 digits. What to return: the total comma count, as a 64-bit integer (INLINECODE9 / INLINECODE10).

2. Real-Life Analogy

Same bank-receipt analogy as Part I — except now you're asked to print a receipt for every dollar amount from INLINECODE11 up to one quadrillion. There is no printer, and no amount of time, that could physically print a quadrillion individual receipts one at a time. You need a method that answers the question without ever "touching" each individual number — exactly the grouped, digit-length-based approach from Part I.

3. Important Programming Concepts I Need First

Everything from Part I carries over directly:

Comma formula: a number with INLINECODE12 digits gets exactly INLINECODE13 commas (integer division). Grouping by digit length: all numbers with the same digit count behave identically, so we can process entire ranges of numbers at once instead of one at a time.

New concept: INLINECODE14 instead of INLINECODE15
Concept: Java's INLINECODE16 type can only hold values up to about INLINECODE17 before overflowing (wrapping around to nonsense values). Since INLINECODE18 here can be up to INLINECODE19 — far beyond that — we must use Java's INLINECODE20 type instead, which comfortably holds values up into the quintillions. Why we need it: The method signature itself has changed from INLINECODE21 to INLINECODE22 — a direct signal that both the input and the output (total comma count, which can also be huge) need this wider type. Every intermediate calculation (group boundaries, counts, sums) must also use INLINECODE23, or we risk silent overflow bugs.
Why brute force is now truly impossible (not just slow)
Concept: A loop that runs INLINECODE24 times, for INLINECODE25 (one quadrillion), would take... even at an extremely generous one billion iterations per second, that's INLINECODE26 seconds — roughly 11.5 days of continuous computation. This isn't "slow," it's simply not going to finish within any reasonable time limit (typically a couple of seconds). Why we need it: This makes Part I's grouped, digit-length-based approach mandatory — there's no way around it.

4. Understand the Input

Same structure as Part I — a single integer INLINECODE27. The only difference: it can now have up to 16 digits (since INLINECODE28 has 16 digits: INLINECODE29).

For Example 1 (INLINECODE30), nothing about the meaning changes from Part I — see Part I Sections 4-6 for the full manual walkthrough. What does change is that our algorithm must gracefully handle inputs vastly larger than this example, up to 16 digits long.

5. Understand the Output

Same as Part I: INLINECODE31, INLINECODE32. The formula and reasoning are identical; only the required scale of correctness (and data type) has grown.

6. Solve the Example Manually

Identical manual process to Part I, Section 6 — no changes here. The insight (group by digit length, cap each group's range at INLINECODE33, multiply group size by that group's fixed comma count) applies exactly the same way, whether INLINECODE34 has 4 digits or 16 digits.

7. Think Like a Programmer

What do I know? The exact same digit-length grouping insight from Part I: INLINECODE35, and each digit-length group spans INLINECODE36 to INLINECODE37. What's different this time? With INLINECODE38 up to INLINECODE39, a digit-length group can now go all the way up to digit length 16. There are at most about 16 groups to process — still an extremely small, fixed number, completely independent of how large INLINECODE40's actual value is. What must I watch out for? Every number involved in the calculation — group boundaries, counts, the running total — must use INLINECODE41, since values like INLINECODE42 and beyond would overflow a regular INLINECODE43. Is there anything conceptually new to solve? No — this is purely a "port Part I's optimized solution to use INLINECODE44 throughout" exercise. The moment you tried the brute-force loop from Part I here, it would time out instantly (well, eventually — after failing to finish for days), immediately signaling that the grouped approach isn't optional this time.

8. Why the Part I Brute Force Now Fails

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Why it fails: The loop itself would need to run up to one quadrillion times. Even if each iteration took only a single nanosecond (physically unrealistic for something involving string conversion), that's still over 11 days of runtime. This isn't a "might time out on an unlucky test case" situation — it is guaranteed to time out on any reasonably large input, every single time.

11. The Fix: Reuse Part I's Grouped Approach, with INLINECODE45

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⭐ Key Insight

Before the insight
It might seem like this "Part II" needs a fundamentally new algorithm, since it's labeled "Medium" instead of "Easy" and the constraint looks dramatically scarier.
The problem
Nothing about the underlying math changed at all — the comma formula and digit-length grouping are identical to Part I. What changed is purely the scale, which makes any INLINECODE46 approach (even a very simple one) computationally impossible, not just "less than ideal."
The insight
The grouped, digit-length-based approach from Part I was already INLINECODE47 — completely independent of INLINECODE48's actual value, only dependent on how many digits INLINECODE49 has. Since even INLINECODE50 only has 16 digits, this approach handles the new constraint just as easily as it handled Part I's tiny one — we just need to switch from INLINECODE51 to INLINECODE52 everywhere to safely hold the larger numbers involved.
After the insight
We reuse Part I's exact grouped algorithm, unchanged in structure, just with INLINECODE53 types throughout.

13. Dry Run

Identical mechanics to Part I's Section 13 dry runs (Examples 1 and 2 give INLINECODE54 and INLINECODE55 respectively, following the exact same group-by-group arithmetic). The only thing to double check: every value manipulated (INLINECODE56, INLINECODE57, INLINECODE58, INLINECODE59, INLINECODE60) is now a INLINECODE61, so that even if we later fed in INLINECODE62, none of these intermediate values would overflow.

Let's specifically verify the largest group that could ever come up, to make sure our approach scales correctly: for INLINECODE63, the relevant digit-length groups go up to INLINECODE64 (since INLINECODE65 itself is a 16-digit number: INLINECODE66 — one followed by fifteen zeros, sixteen digits total). The group for INLINECODE67 would naturally span INLINECODE68 to INLINECODE69, capped at INLINECODE70 (just the single number INLINECODE71 itself, if INLINECODE72 equals exactly that). INLINECODE73 — meaning the number INLINECODE74, written out fully, is INLINECODE75, which indeed has exactly 5 commas. This all checks out using plain INLINECODE76 arithmetic with no overflow.

14. Optimized Code

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This is line-for-line the same algorithm as Part I's optimized solution — the only changes are: the method signature now takes and returns INLINECODE77 (or INLINECODE78 in JS), INLINECODE79 is explicitly declared INLINECODE80 (a INLINECODE81 literal) to guarantee all its arithmetic stays in INLINECODE82 from the very first operation, and INLINECODE83 is declared as INLINECODE84 rather than INLINECODE85 for consistency.

15. Explain the Code Line by Line

INLINECODE86 — 64-bit accumulator for storing the large total comma count. INLINECODE87 — starts at 1, representing the lowest 1-digit number. INLINECODE88 — iterates up to 16 times as INLINECODE89 scales by $10\times$ per loop (INLINECODE90). INLINECODE91 — computes group ceiling without overflow in 64-bit integer limits (INLINECODE92). INLINECODE93 — caps the range at INLINECODE94. INLINECODE95 — counts elements inside the clamped digit bracket. INLINECODE96 — integer division yields the exact comma count per item. INLINECODE97 — batches the computation for up to hundreds of trillions of numbers in a single addition. INLINECODE98 — advances to the next digit order of magnitude. INLINECODE99 — returns the completed sum.

16. Test With Multiple Examples

Example 1 — Normal Case
INLINECODE100 → Output: INLINECODE101 ✔️ (identical to Part I)
Example 2 — Different Case
INLINECODE102 → Output: INLINECODE103 ✔️ (identical to Part I)
Example 3 — The case that actually distinguishes Part I from Part II
INLINECODE104 ($10^{15}$, the maximum allowed). Our loop runs through digit lengths INLINECODE105 through INLINECODE106 — just 16 iterations total — each doing a handful of INLINECODE107 arithmetic operations, finishing in a tiny fraction of a millisecond. The brute-force loop from Part I, by contrast, would need on the order of 10^15 iterations, which would take over 11 days to complete.

17. Edge Cases

All the same edge cases from Part I apply (n=1, boundaries at powers of 10, n fully under 1000). The genuinely new edge case here:

Maximum-size input (INLINECODE108) → our grouped approach handles this in essentially constant time (at most ~16 loop iterations), while the naive brute force would never finish. This is the entire point of the "II" sequel. INLINECODE109 overflow risk in intermediate steps → specifically, INLINECODE110 at the largest group (INLINECODE111 around INLINECODE112, multiplied by 10 gives INLINECODE113) is still comfortably within INLINECODE114's range (which extends past INLINECODE115), so there's no overflow risk even at the problem's maximum constraint.

18. Time Complexity

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Why this matters so much more now than in Part I: In Part I, INLINECODE116 meant even the "bad" INLINECODE117 approach was only 100,000 operations — trivially fast. Here, INLINECODE118 makes that same INLINECODE119 approach represent roughly ten billion times more work than Part I's worst case — a difference that turns "instant" into "practically never." This is precisely why Part II exists: to confirm you understood why the grouped approach works, rather than just having gotten lucky that Part I's small constraints let a simpler method slide by.

19. Space Complexity

INLINECODE120 — a handful of INLINECODE121 variables, none of which grow with INLINECODE122. Identical to Part I's space usage.

20. Common Mistakes Beginners Make

❌ "I'll just change INLINECODE123 to INLINECODE124 in my Part I brute-force loop and submit that." ✅ Changing the data type fixes potential overflow issues but does nothing about the fundamental speed problem — a loop of INLINECODE125 iterations is just as impossibly slow whether it's counting with INLINECODE126 or INLINECODE127. You must switch to the grouped INLINECODE128 algorithm entirely, not just patch the data types in the slow one. ❌ "Since INLINECODE129 fits in a INLINECODE130, all my intermediate variables can stay as INLINECODE131." ✅ Any variable that could hold a value derived from INLINECODE132 (like INLINECODE133, which can reach up to INLINECODE134 or beyond during the loop) must also be INLINECODE135 — mixing INLINECODE136 and INLINECODE137 carelessly risks silent overflow the moment an INLINECODE138 variable is asked to hold a value bigger than about 2.1 billion. ❌ "This must need some fancy new mathematical trick since it's rated 'Medium' now instead of 'Easy'." ✅ The difficulty bump reflects exactly one thing: whether you reach for the INLINECODE139 or the INLINECODE140 approach. The math (comma formula, digit-length grouping) is unchanged from Part I. ❌ "INLINECODE141 might overflow for very large INLINECODE142, so I need some special overflow-safe multiplication." ✅ For this problem's specific bound (INLINECODE143), plain INLINECODE144 arithmetic is entirely sufficient — INLINECODE145 safely holds values up past INLINECODE146, far beyond anything this problem could ever produce (INLINECODE147, at most, appears in intermediate group-boundary calculations).

21. How to Recognize This Pattern in Other Problems

Same meta-lesson as the other "Part II" sequels in this series:

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22. Interview Thinking

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23. Mini Challenge

Try these before checking the answers:

1. How many digit-length groups would need to be processed for INLINECODE148 exactly, and why? 2. If you forgot to change INLINECODE149 to a INLINECODE150-compatible calculation somewhere, but INLINECODE151 itself was correctly declared INLINECODE152, where specifically might a bug sneak in? 3. Why does the number of loop iterations in the grouped approach stay small even if INLINECODE153 were hypothetically increased to INLINECODE154 (close to INLINECODE155's actual maximum)?

Answer to Mini Challenge

1. INLINECODE156 is written as INLINECODE157 followed by 15 zeros — that's 16 digits total. So digit-length groups from INLINECODE158 through INLINECODE159 would all potentially be processed (16 groups), with the 16th group being just the single number INLINECODE160 itself (capped by INLINECODE161). 2. If, for example, INLINECODE162 were computed via something like INLINECODE163 where INLINECODE164 itself was accidentally still typed as INLINECODE165 somewhere in a refactor, or if a group-boundary calculation like INLINECODE166 were performed using INLINECODE167 arithmetic instead of INLINECODE168, the multiplication could silently overflow and wrap around to a nonsensical (often negative) value well before reaching realistic-sized groups — this is exactly the kind of subtle bug that "looks correct" on Part I's small INLINECODE169 but breaks silently on Part II's larger range. 3. Because the number of digit-length groups is tied to how many digits INLINECODE170 has, not INLINECODE171's raw value — and even INLINECODE172 only has 19 digits. Going from INLINECODE173 (16 digits) to INLINECODE174 (19 digits) only adds 3 more loop iterations, a negligible increase, illustrating just how powerfully "logarithmic in n" scales compared to "linear in n."

24. Final Revision

🧠 Problem in One Sentence
Identical to Part I: count total commas across standard-formatted integers from 1 to INLINECODE175 — but now INLINECODE176 can be up to INLINECODE177, making the INLINECODE178 grouped approach (and INLINECODE179 arithmetic) mandatory rather than optional.
🔑 Main Idea
Part I's digit-length-grouping insight was already independent of INLINECODE180's magnitude (only dependent on its digit count) — so the fix is simply reusing that exact algorithm with INLINECODE181 types throughout, instead of the now-infeasible INLINECODE182 brute-force loop.
⚙️ Algorithm
1. For each digit length INLINECODE183, compute the group's natural range INLINECODE184. 2. Cap the group's end at INLINECODE185. 3. Add INLINECODE186 to the running total. 4. Advance to the next digit length; stop once the group's start exceeds INLINECODE187.
⏱️ Complexity
Time: INLINECODE188 (at most ~16 iterations for INLINECODE189 up to INLINECODE190)

Space: INLINECODE191

🎯 Pattern to Remember

When a "Part II" sequel keeps the same logic but pushes INLINECODE192 from a small bound to an astronomically larger one, check whether your optimized Part I solution was already independent of INLINECODE193's raw magnitude — if it was INLINECODE194 (or similar) already, the fix is often just a data-type upgrade, not new logic.

25. Beginner Quiz

1. (Understanding) What is the only conceptual difference between Part I and Part II of this problem? 2. (Basic concept) Roughly how many seconds/days would a brute-force INLINECODE195 loop take to process INLINECODE196, assuming a generous one billion operations per second? 3. (Logic) Why does switching from INLINECODE197 to INLINECODE198 alone NOT fix the brute-force approach's fundamental problem? 4. (Dry run) For INLINECODE199 (fifteen 9s, a 15-digit number, one less than 10^15), determine how many digit-length groups are processed and what the final digit-length-15 group's comma-per-number value is. 5. (Complexity/pattern) Why does the grouped approach's runtime stay almost unchanged whether INLINECODE200 is INLINECODE201, INLINECODE202, or even INLINECODE203, and what specific aspect of the algorithm (mentioned in the Key Insight) explains this remarkable scalability?