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Math Coding Problems: 213 Questions with Solutions

213 math coding problems — 108 easy · 80 medium · 25 hard — with solutions in 13 languages. Plus a step-by-step walkthrough and a 8-day plan.

  • Problems: 213
  • By difficulty: 108 easy · 80 medium · 25 hard
  • Languages: JavaScript, TypeScript, Python, Java, C++, C, C#, Go, Kotlin, Swift, Rust, PHP and Ruby
  • Cost: Free on every plan; sign in to run and submit

Problems whose solution is a fact about numbers rather than a data structure: digit sums, divisibility, parity, arithmetic sequences, geometry on a grid, and the integer overflow and truncating division that trip a correct idea. The habit these build is to look for a closed form or an invariant before reaching for a loop.

How math works, step by step

xrev1231234x = 123rev = 1234palindrome(1234321) = true4 of 7 digits peeled
Palindrome number without a string, by peeling digits with % 10 and // 10. Example: x = 1234321
  1. Is 1234321 a palindrome? Rather than make a string, peel digits off the end: x % 10 is the last digit and x // 10 drops it. Each peeled digit is appended to rev as rev × 10 + digit, so rev grows into the right half read backwards.
  2. The last digit of 1234321 is 1234321 % 10 = 1; it moves onto rev, which becomes 0 × 10 + 1 = 1, and x becomes 1234321 // 10 = 123432. x is still the larger, so the two halves have not met yet.
  3. The last digit of 123432 is 123432 % 10 = 2; it moves onto rev, which becomes 1 × 10 + 2 = 12, and x becomes 123432 // 10 = 12343. x is still the larger, so the two halves have not met yet.
  4. The last digit of 12343 is 12343 % 10 = 3; it moves onto rev, which becomes 12 × 10 + 3 = 123, and x becomes 12343 // 10 = 1234. x is still the larger, so the two halves have not met yet.
  5. The last digit of 1234 is 1234 % 10 = 4; it moves onto rev, which becomes 123 × 10 + 4 = 1234, and x becomes 1234 // 10 = 123. Now x (123) is no larger than rev (1234): rev holds the right half, so the peeling stops.
  6. 7 digits is odd, so rev also took the middle digit 4; rev // 10 = 123 drops it. x = 123 and rev // 10 = 123 are equal, so the left half matches the right half read backwards.
  7. 1234321 is a palindrome. Only half its digits were peeled, so the test is O(log n) time and O(1) space, with no string and no full reverse to overflow; negative numbers and numbers ending in 0 are ruled out before the loop.

Math study plan

14 of the 213 Math problems (4 easy, 7 medium and 3 hard) over 8 days, about 8 h 10 min in all — the pattern first, then easiest to hardest. After that, the other 199 in the full list below are practice at your own pace. Then move on to Two Pointers.

Day 1

Learn the pattern: read the essentials and step through the walkthrough above, then solve these 3.

Day 2

Medium problems: the same pattern with one twist each. Name the twist before you code.

Day 3

More mediums. Before coding each one, write down what state the pattern keeps and when it changes.

Day 4

More mediums. Before coding each one, write down what state the pattern keeps and when it changes.

Day 5

More mediums. Before coding each one, write down what state the pattern keeps and when it changes.

Day 6

Hard problems: the pattern combined with a second idea. Give each a full attempt before reading the editorial.

Day 7

Another hard one. If it beats you after a real attempt, read the editorial, then solve it again tomorrow from memory.

Day 8

Another hard one. If it beats you after a real attempt, read the editorial, then solve it again tomorrow from memory.

Next topic: Two Pointers

Math: the essentials

When to reach for it

Numbers too large to loop up to (n up to 10⁹ or beyond), questions about digits, remainders or parity, and an operation repeated so often that it must settle into a pattern. If the brute force is "simulate a billion steps", the intended answer is a formula, a cycle or an invariant.

The pattern

Work the small cases by hand and tabulate them; the rule often shows within the first half-dozen. Take digits off an integer with n % 10 and n // 10 rather than converting to a string, which keeps sign and overflow handling explicit. Check a result against the type's limit before the operation that could exceed it, not after.

def reverse(x):                     # 32-bit signed result, or 0
    LIMIT = 2**31 - 1
    sign, x, out = (-1 if x < 0 else 1), abs(x), 0
    while x:
        x, d = divmod(x, 10)
        if out > (LIMIT - d) // 10:  # out * 10 + d would overflow
            return 0
        out = out * 10 + d
    return sign * out

Cost

A digit loop is O(log n), trial division up to √n is O(√n), and a closed form is O(1). That gap is the point: for n = 10¹² a loop to n never finishes, while a loop to √n is a million steps.

Common mistakes

  • Overflow in a 32-bit int: a * b or a + b can pass 2³¹ − 1 long before the final answer does. Widen to 64 bits first.
  • Division and remainder with negatives: in Java and C++ -7 / 2 is -3 and -7 % 2 is -1 (JavaScript's % agrees); in Python -7 // 2 is -4 and -7 % 2 is 1.
  • Using floating point for an integer question: sqrt or pow can return 2.9999… for an exact square, so confirm with integer arithmetic.
  • Forgetting 0, 1 and negative inputs, which most formulas treat specially.

Start with

All math problems

Easy (108)

Medium (80)

Hard (25)

Companies that ask math problems

Next topic: Two Pointers