LeetCode 3918: Sum of Primes Between Number and Its Reverse ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2026-05-03

Submissions: Wrong Answer → Accepted

Approach
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With r the reverse of n, the range is [min(n, r), max(n, r)]. It is small, so trial-divide each number and sum the primes. (For larger ranges, a sieve computes them all at once.)

My Attempts
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Attempt 1’s primality test:

tested divisibility with n // i == n / i, relying on floating point, and
tried range(2, ceil(√n) + 1), so for n = 2 it tried i = 2, where 2 // 2 == 2 / 2, judging 2 composite.

For n = 2, the reverse is still 2 and the answer is 2; it returned 0. Handling 2 first, trying only odd divisors, and using % passed.

Solution
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from math import isqrt

class Solution:
    def sumOfPrimesInRange(self, n: int) -> int:
        def is_prime(x: int) -> bool:
            if x < 2:
                return False
            if x % 2 == 0:
                return x == 2
            for d in range(3, isqrt(x) + 1, 2):
                if x % d == 0:
                    return False
            return True

        r = int(str(n)[::-1])
        return sum(x for x in range(min(n, r), max(n, r) + 1) if is_prime(x))

Complexity
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Time O(R √R) with R the range’s upper end; space O(1).

Mistakes to Avoid
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Testing divisibility with floating-point division.
Using ceil(√n) as the limit lets small numbers try themselves; use isqrt and stop at √x.

Interview Explanation
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Reversing n gives the range; trial division tests each number and the primes are summed. Trial division stops at √x, divisibility uses %, and 2 is handled first. For a bigger range, switch to a sieve.