2652. Sum Multiples

Description

Given a positive integer n, find the sum of all integers in the range [1, n] inclusive that are divisible by 3, 5, or 7.

Return an integer denoting the sum of all numbers in the given range satisfying the constraint.

 

Example 1:

Input: n = 7
Output: 21
Explanation: Numbers in the range [1, 7] that are divisible by 3, 5, or 7 are 3, 5, 6, 7. The sum of these numbers is 21.

Example 2:

Input: n = 10
Output: 40
Explanation: Numbers in the range [1, 10] that are divisible by 3, 5, or 7 are 3, 5, 6, 7, 9, 10. The sum of these numbers is 40.

Example 3:

Input: n = 9
Output: 30
Explanation: Numbers in the range [1, 9] that are divisible by 3, 5, or 7 are 3, 5, 6, 7, 9. The sum of these numbers is 30.

 

Constraints:

• 1 <= n <= 103

Solutions

Solution 1: Enumeration

We directly enumerate every number $x$ in $[1,..n]$, and if $x$ is divisible by $3$, $5$, and $7$, we add $x$ to the answer.

After the enumeration, we return the answer.

The time complexity is $O(n)$, where $n$ is the given integer. The space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRust

class Solution:
    def sumOfMultiples(self, n: int) -> int:
        return sum(x for x in range(1, n + 1) if x % 3 == 0 or x % 5 == 0 or x % 7 == 0)

class Solution {
    public int sumOfMultiples(int n) {
        int ans = 0;
        for (int x = 1; x <= n; ++x) {
            if (x % 3 == 0 || x % 5 == 0 || x % 7 == 0) {
                ans += x;
            }
        }
        return ans;
    }
}

class Solution {
public:
    int sumOfMultiples(int n) {
        int ans = 0;
        for (int x = 1; x <= n; ++x) {
            if (x % 3 == 0 || x % 5 == 0 || x % 7 == 0) {
                ans += x;
            }
        }
        return ans;
    }
};

func sumOfMultiples(n int) (ans int) {
    for x := 1; x <= n; x++ {
        if x%3 == 0 || x%5 == 0 || x%7 == 0 {
            ans += x
        }
    }
    return
}

function sumOfMultiples(n: number): number {
    let ans = 0;
    for (let x = 1; x <= n; ++x) {
        if (x % 3 === 0 || x % 5 === 0 || x % 7 === 0) {
            ans += x;
        }
    }
    return ans;
}

impl Solution {
    pub fn sum_of_multiples(n: i32) -> i32 {
        let mut ans = 0;

        for x in 1..=n {
            if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 {
                ans += x;
            }
        }

        ans
    }
}

Solution 2: Mathematics (Inclusion-Exclusion Principle)

We define a function $f(x)$ to represent the sum of numbers in $[1,..n]$ that are divisible by $x$. There are $m = \left\lfloor \frac{n}{x} \right\rfloor$ numbers that are divisible by $x$, which are $x$, $2x$, $3x$, $\cdots$, $mx$, forming an arithmetic sequence with the first term $x$, the last term $mx$, and the number of terms $m$. Therefore, $f(x) = \frac{(x + mx) \times m}{2}$.

According to the inclusion-exclusion principle, we can obtain the answer as:

$$ f(3) + f(5) + f(7) - f(3 \times 5) - f(3 \times 7) - f(5 \times 7) + f(3 \times 5 \times 7) $$

The time complexity is $O(1)$, and the space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRust

class Solution:
    def sumOfMultiples(self, n: int) -> int:
        def f(x: int) -> int:
            m = n // x
            return (x + m * x) * m // 2

        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7)

class Solution {
    private int n;

    public int sumOfMultiples(int n) {
        this.n = n;
        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
    }

    private int f(int x) {
        int m = n / x;
        return (x + m * x) * m / 2;
    }
}

class Solution {
public:
    int sumOfMultiples(int n) {
        auto f = [&](int x) {
            int m = n / x;
            return (x + m * x) * m / 2;
        };
        return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
    }
};

func sumOfMultiples(n int) int {
    f := func(x int) int {
        m := n / x
        return (x + m*x) * m / 2
    }
    return f(3) + f(5) + f(7) - f(3*5) - f(3*7) - f(5*7) + f(3*5*7)
}

function sumOfMultiples(n: number): number {
    const f = (x: number): number => {
        const m = Math.floor(n / x);
        return ((x + m * x) * m) >> 1;
    };
    return f(3) + f(5) + f(7) - f(3 * 5) - f(3 * 7) - f(5 * 7) + f(3 * 5 * 7);
}

impl Solution {
    pub fn sum_of_multiples(n: i32) -> i32 {
        (1..=n)
            .filter(|&x| (x % 3 == 0 || x % 5 == 0 || x % 7 == 0))
            .sum()
    }
}

Solution 3

Rust

impl Solution {
    pub fn sum_of_multiples(n: i32) -> i32 {
        fn f(x: i32, n: i32) -> i32 {
            let m = n / x;
            ((x + m * x) * m) / 2
        }

        f(3, n) + f(5, n) + f(7, n) - f(3 * 5, n) - f(3 * 7, n) - f(5 * 7, n) + f(3 * 5 * 7, n)
    }
}

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