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1952. Three Divisors

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 <= 104

Solutions

Solution 1

Thinking

A number has exactly three positive divisors iff it is the square of a prime. \(n\) is small, so counting divisors in \(2..n-1\) and asking whether that count is \(1\) suffices.

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

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

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class Solution {
public:
    bool isThree(int n) {
        int cnt = 0;
        for (int i = 2; i < n; ++i) {
            cnt += n % i == 0;
        }
        return cnt == 1;
    }
};

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

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

Solution 2

Thinking

A linear scan wastes time near the upper bound. Enumerating up to \(\sqrt{n}\) counts each pair once (or once for a square) and checks whether the total is \(3\).

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class Solution:
    def isThree(self, n: int) -> bool:
        cnt = 0
        i = 1
        while i <= n // i:
            if n % i == 0:
                cnt += 1 if i == n // i else 2
            i += 1
        return cnt == 3

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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;
    }
}

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class Solution {
public:
    bool 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;
    }
};

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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
}

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/**
 * @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;
};

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