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Formatted question description: https://leetcode.ca/all/1952.html

# 1952. Three Divisors

Easy

## Description

Given an integer n, return true if n has exactly three positive divisors. Otherwise, return false.

An integer m is a divisor of n if there exists an integer k such that n = k * m.

Example 1:

Input: n = 2

Output: false

Explantion: 2 has only two divisors: 1 and 2.

Example 2:

Input: n = 4

Output: true

Explantion: 4 has three divisors: 1, 2, and 4.

Constraints:

• 1 <= n <= 10^4

## Solution

If n has three divisors, then n must be a perfect square number. Calculate the square root of n, and loop over all integers from 1 to the square root minus 1. For each i such that n is divisible by i, add 2 to the number of divisors. The final divisor is the square root of n. Return true if the number of divisors equals 3.

• class Solution {
public boolean isThree(int n) {
int sqrt = (int) Math.sqrt(n);
if (sqrt * sqrt != n)
return false;
int divisors = 1;
for (int i = 1; i < sqrt; i++) {
if (n % i == 0)
divisors += 2;
}
return divisors == 3;
}
}

############

class Solution {
public boolean isThree(int n) {
int cnt = 0;
for (int i = 1; i <= n / i; ++i) {
if (n % i == 0) {
cnt += n / i == i ? 1 : 2;
}
}
return cnt == 3;
}
}

• // OJ: https://leetcode.com/problems/three-divisors/
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool isThree(int n) {
int ans = 0;
for (int i = 1; i <= n && ans <= 3; ++i) {
if (n % i == 0) ++ans;
}
return ans == 3;
}
};

• class Solution:
def isThree(self, n: int) -> bool:
return sum(n % i == 0 for i in range(2, n)) == 1

############

# 1952. Three Divisors
# https://leetcode.com/problems/three-divisors/

class Solution:
def isThree(self, n: int) -> bool:
res = 0

for k in range(1, n + 1):
if n % k == 0: res += 1

if res > 3: return False

return res == 3


• func isThree(n int) bool {
cnt := 0
for i := 1; i <= n/i; i++ {
if n%i == 0 {
if n/i == i {
cnt++
} else {
cnt += 2
}
}
}
return cnt == 3
}

• /**
* @param {number} n
* @return {boolean}
*/
var isThree = function (n) {
let cnt = 0;
for (let i = 1; i <= n / i; ++i) {
if (n % i == 0) {
cnt += ~~(n / i) == i ? 1 : 2;
}
}
return cnt == 3;
};