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Tested Python solution for LeetCode 2523 with 22 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 2523 # by problem number
lcpy gen -s closest_prime_numbers_in_range # by problem name
left and right, find the two integers num1 and num2 such that:
left <= num1 < num2 <= right.num1 and num2 are prime numbers.num2 - num1 is the minimum amongst all other pairs satisfying the above conditions.ans = [num1, num2]. If there are multiple pairs satisfying these conditions, return the one with the smallest num1 value. If no such numbers exist, return [-1, -1].
Input: left = 10, right = 19
Output: [11,13]
Explanation: The prime numbers between 10 and 19 are 11, 13, 17, and 19.
The closest gap between any pair is 2, which can be achieved by [11,13] or [17,19].
Since 11 is smaller than 17, we return the first pair.
Input: left = 4, right = 6
Output: [-1,-1]
Explanation: There exists only one prime number in the given range, so the conditions cannot be satisfied.
class Solution:
# Time: O(right * log(log(right)))
# Space: O(right)
def closest_primes(self, left: int, right: int) -> list[int]:
if right < 2:
return [-1, -1]
sieve = bytearray([1]) * (right + 1)
sieve[0] = sieve[1] = 0
i = 2
while i * i <= right:
if sieve[i]:
sieve[i * i : right + 1 : i] = bytearray(len(sieve[i * i : right + 1 : i]))
i += 1
best: list[int] = [-1, -1]
prev = -1
for num in range(max(left, 2), right + 1):
if not sieve[num]:
continue
if prev != -1 and (best[0] == -1 or num - prev < best[1] - best[0]):
best = [prev, num]
prev = num
return best
| Time | Space |
|---|---|
| O(right * log(log(right))) | O(right) |