LeetCode 3233: Find the Count of Numbers Which Are Not Special Solution

Master LeetCode problem 3233 (Find the Count of Numbers Which Are Not Special), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3233. Find the Count of Numbers Which Are Not Special

Medium

Array Math Number Theory

Problem Explanation

Explanation

To solve this problem, we need to count the numbers in the range [l, r] that are not special. A number is special if it has exactly 2 proper divisors. Therefore, we can iterate over the numbers in the range [l, r] and check if each number has exactly 2 proper divisors. We can count the numbers that do not meet this criteria and return the count.

Algorithm:
1. Define a helper function to calculate the number of divisors for a given number.
2. Iterate over the range [l, r] and for each number, check if it has exactly 2 proper divisors.
3. Increment a counter for numbers that do not have exactly 2 proper divisors.
4. Return the count of numbers that are not special.
Time Complexity: O((r-l) * sqrt(r))
Space Complexity: O(1)

Solution Code

class Solution {
    public int countNonSpecialNumbers(int l, int r) {
        int count = 0;
        for (int num = l; num <= r; num++) {
            if (divisorCount(num) != 2) {
                count++;
            }
        }
        return count;
    }

    private int divisorCount(int num) {
        int count = 0;
        for (int i = 1; i <= Math.sqrt(num); i++) {
            if (num % i == 0) {
                if (num / i == i) {
                    count++;
                } else {
                    count += 2;
                }
            }
        }
        return count;
    }
}

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