Question

2652. Sum Multiples

中文文档

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 $O(1)$.

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 $O(1)$.

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

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)
  }
}