LeetCode #3233 — MEDIUM

Find the Count of Numbers Which Are Not Special

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given 2 positive integers l and r. For any number x, all positive divisors of x except x are called the proper divisors of x.

A number is called special if it has exactly 2 proper divisors. For example:

The number 4 is special because it has proper divisors 1 and 2.
The number 6 is not special because it has proper divisors 1, 2, and 3.

Return the count of numbers in the range [l, r] that are not special.

Example 1:

Input: l = 5, r = 7

Output: 3

Explanation:

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

Example 2:

Input: l = 4, r = 16

Output: 11

Explanation:

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

Constraints:

1 <= l <= r <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given 2 positive integers l and r. For any number x, all positive divisors of x except x are called the proper divisors of x. A number is called special if it has exactly 2 proper divisors. For example: The number 4 is special because it has proper divisors 1 and 2. The number 6 is not special because it has proper divisors 1, 2, and 3. Return the count of numbers in the range [l, r] that are not special.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

5
7

Example 2

4
16

Core Insight

What unlocks the optimal approach

A special number must be a square of a prime number.
We need to find all primes in the range <code>[sqrt(l), sqrt(r)]</code>.
Use sieve to find primes till <code>sqrt(10<sup>9</sup>)</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3233: Find the Count of Numbers Which Are Not Special
class Solution {
    static int m = 31623;
    static boolean[] primes = new boolean[m + 1];

    static {
        Arrays.fill(primes, true);
        primes[0] = primes[1] = false;
        for (int i = 2; i <= m; i++) {
            if (primes[i]) {
                for (int j = i + i; j <= m; j += i) {
                    primes[j] = false;
                }
            }
        }
    }

    public int nonSpecialCount(int l, int r) {
        int lo = (int) Math.ceil(Math.sqrt(l));
        int hi = (int) Math.floor(Math.sqrt(r));
        int cnt = 0;
        for (int i = lo; i <= hi; i++) {
            if (primes[i]) {
                cnt++;
            }
        }
        return r - l + 1 - cnt;
    }
}

// Accepted solution for LeetCode #3233: Find the Count of Numbers Which Are Not Special
const m = 31623

var primes [m + 1]bool

func init() {
	for i := range primes {
		primes[i] = true
	}
	primes[0] = false
	primes[1] = false
	for i := 2; i <= m; i++ {
		if primes[i] {
			for j := i * 2; j <= m; j += i {
				primes[j] = false
			}
		}
	}
}

func nonSpecialCount(l int, r int) int {
	lo := int(math.Ceil(math.Sqrt(float64(l))))
	hi := int(math.Floor(math.Sqrt(float64(r))))
	cnt := 0
	for i := lo; i <= hi; i++ {
		if primes[i] {
			cnt++
		}
	}
	return r - l + 1 - cnt
}

# Accepted solution for LeetCode #3233: Find the Count of Numbers Which Are Not Special
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

// Accepted solution for LeetCode #3233: Find the Count of Numbers Which Are Not Special
/**
 * [3233] Find the Count of Numbers Which Are Not Special
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn non_special_count(l: i32, r: i32) -> i32 {
        let n = (r as f64).sqrt() as usize;
        let mut v = vec![true; n + 1];
        let mut result = r - l + 1;

        let (l, r) = (l as usize, r as usize);
        for i in 2..=n {
            if v[i] {
                if i.pow(2) >= l && i.pow(2) <= r {
                    result -= 1;
                }

                for j in ((i * 2)..=n).step_by(i) {
                    v[j] = false;
                }
            }
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3233() {
        assert_eq!(19, Solution::non_special_count(5, 25));
        assert_eq!(3, Solution::non_special_count(5, 7));
        assert_eq!(11, Solution::non_special_count(4, 16));
    }
}

// Accepted solution for LeetCode #3233: Find the Count of Numbers Which Are Not Special
const m = 31623;
const primes: boolean[] = Array(m + 1).fill(true);

(() => {
    primes[0] = primes[1] = false;
    for (let i = 2; i <= m; ++i) {
        if (primes[i]) {
            for (let j = i * 2; j <= m; j += i) {
                primes[j] = false;
            }
        }
    }
})();

function nonSpecialCount(l: number, r: number): number {
    const lo = Math.ceil(Math.sqrt(l));
    const hi = Math.floor(Math.sqrt(r));
    let cnt = 0;
    for (let i = lo; i <= hi; ++i) {
        if (primes[i]) {
            ++cnt;
        }
    }
    return r - l + 1 - cnt;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(sqrtm)

Space

O(sqrtm)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.