LeetCode 1952: Three Divisors Solution
Master LeetCode problem 1952 (Three Divisors), a easy 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.
1952. Three Divisors
Problem Explanation
Explanation
To solve this problem, we need to check if the given integer n has exactly three positive divisors. We can determine the number of divisors by iterating from 1 to the square root of n and counting the divisors. If the count reaches 3, we return true. The time complexity of this approach is O(sqrt(n)).
Solution Code
class Solution {
public boolean isThree(int n) {
int count = 0;
for (int i = 1; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
count++;
if (n / i != i) {
count++;
}
}
if (count > 3) {
return false;
}
}
return count == 3;
}
}Try It Yourself
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