LeetCode Problem Workspace

Find the Count of Numbers Which Are Not Special

Count the numbers between two integers that are not special, where special numbers are squares of primes.

category

3

Topics

code_blocks

5

Code langs

hub

3

Related

Practice Focus

Medium · Array plus Math

bolt

Answer-first summary

Count the numbers between two integers that are not special, where special numbers are squares of primes.

Interview AiBox logo

Ace coding interviews with Interview AiBox guidance for Array plus Math

Try AiBox Copilotarrow_forward

This problem requires counting numbers in the range [l, r] that are not 'special'. A special number is defined as a square of a prime. Identifying special numbers and efficiently excluding them from the count forms the core of the solution.

Problem Statement

You are given two integers, l and r, where l <= r. A special number is defined as a number that has exactly two proper divisors. In other words, the number must be the square of a prime number. Your task is to find how many numbers in the range [l, r] are not special.

For example, consider the range [5, 7]. The numbers 5, 6, and 7 are not special since they are not squares of primes. However, for the range [4, 16], the numbers 4 and 9 are special, being squares of the primes 2 and 3, respectively.

Examples

Example 1

Input: l = 5, r = 7

Output: 3

There are no special numbers in the range [5, 7] .

Example 2

Input: l = 4, r = 16

Output: 11

The special numbers in the range [4, 16] are 4 and 9.

Constraints

  • 1 <= l <= r <= 109

Solution Approach

Prime Checking for Squares

The first step is to identify numbers that are squares of primes. To do this efficiently, check each number in the range to determine if its square root is prime. This will allow identifying special numbers.

Efficient Square Identification

Iterate through the numbers in the range and check if their square roots are integers and primes. If so, mark the number as special and exclude it from the count of non-special numbers.

Final Count Computation

Once the special numbers are identified, simply subtract the count of special numbers from the total count of numbers in the range to get the number of non-special numbers.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time complexity depends on the method used for prime number generation and checking. A sieve approach for prime generation combined with efficient square root checks can make the solution feasible for large ranges. The space complexity depends on the storage of primes and squares of primes within the range [l, r].

What Interviewers Usually Probe

  • Check if the candidate can efficiently handle the identification of prime numbers.
  • Look for understanding of number theory concepts like divisors and prime squares.
  • Assess the candidate's ability to optimize the solution for large ranges of numbers.

Common Pitfalls or Variants

Common pitfalls

  • Misidentifying numbers as special when they are not squares of primes.
  • Inefficient prime checking methods that lead to slower performance on large ranges.
  • Failing to account for edge cases, such as very small or very large values of l and r.

Follow-up variants

  • Optimize the solution for very large ranges, e.g., when l and r are close to 109.
  • Consider variations where the definition of special numbers changes, such as requiring more than two divisors.
  • Change the problem to count the special numbers instead of the non-special ones.

FAQ

What is the definition of a special number in this problem?

A special number is a number that is the square of a prime number.

How do I efficiently find prime numbers in the range [l, r]?

You can use a sieve algorithm to generate primes up to sqrt(r) and then check if the square roots of numbers in [l, r] are prime.

What is the pattern used in this problem?

This problem combines Array and Math patterns, particularly focusing on identifying squares of prime numbers.

How can I optimize my solution for large inputs?

By using a sieve to precompute primes up to sqrt(r) and checking each number's square root efficiently, you can handle larger inputs effectively.

What is the expected time complexity for this problem?

The time complexity depends on the method used for prime checking and the size of the range [l, r]. A sieve-based approach can make it feasible for large inputs.

terminal

Solution

Solution 1: Mathematics

According to the problem description, we can observe that only the squares of prime numbers are special numbers. Therefore, we can first preprocess all prime numbers less than or equal to $\sqrt{10^9}$, and then iterate through the interval $[\lceil\sqrt{l}\rceil, \lfloor\sqrt{r}\rfloor]$, counting the number of primes $\textit{cnt}$ in the interval. Finally, we return $r - l + 1 - \textit{cnt}$.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
m = 31623
primes = [True] * (m + 1)
primes[0] = primes[1] = False
for i in range(2, m + 1):
    if primes[i]:
        for j in range(i + i, m + 1, i):
            primes[j] = False


class Solution:
    def nonSpecialCount(self, l: int, r: int) -> int:
        lo = ceil(sqrt(l))
        hi = floor(sqrt(r))
        cnt = sum(primes[i] for i in range(lo, hi + 1))
        return r - l + 1 - cnt
Find the Count of Numbers Which Are Not Special Solution: Array plus Math | LeetCode #3233 Medium