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发布于 2024-06-17 01:03:05 字数 5140 浏览 0 评论 0 收藏 0

2485. Find the Pivot Integer

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Description

Given a positive integer n, find the pivot integer x such that:

  • The sum of all elements between 1 and x inclusively equals the sum of all elements between x and n inclusively.

Return _the pivot integer _x. If no such integer exists, return -1. It is guaranteed that there will be at most one pivot index for the given input.

 

Example 1:

Input: n = 8
Output: 6
Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21.

Example 2:

Input: n = 1
Output: 1
Explanation: 1 is the pivot integer since: 1 = 1.

Example 3:

Input: n = 4
Output: -1
Explanation: It can be proved that no such integer exist.

 

Constraints:

  • 1 <= n <= 1000

Solutions

Solution 1: Enumeration

We can directly enumerate $x$ in the range of $[1,..n]$, and check whether the following equation holds. If it holds, then $x$ is the pivot integer, and we can directly return $x$.

$$ (1 + x) \times x = (x + n) \times (n - x + 1) $$

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

class Solution:
  def pivotInteger(self, n: int) -> int:
    for x in range(1, n + 1):
      if (1 + x) * x == (x + n) * (n - x + 1):
        return x
    return -1
class Solution {
  public int pivotInteger(int n) {
    for (int x = 1; x <= n; ++x) {
      if ((1 + x) * x == (x + n) * (n - x + 1)) {
        return x;
      }
    }
    return -1;
  }
}
class Solution {
public:
  int pivotInteger(int n) {
    for (int x = 1; x <= n; ++x) {
      if ((1 + x) * x == (x + n) * (n - x + 1)) {
        return x;
      }
    }
    return -1;
  }
};
func pivotInteger(n int) int {
  for x := 1; x <= n; x++ {
    if (1+x)*x == (x+n)*(n-x+1) {
      return x
    }
  }
  return -1
}
function pivotInteger(n: number): number {
  for (let x = 1; x <= n; ++x) {
    if ((1 + x) * x === (x + n) * (n - x + 1)) {
      return x;
    }
  }
  return -1;
}
impl Solution {
  pub fn pivot_integer(n: i32) -> i32 {
    let y = (n * (n + 1)) / 2;
    let x = (y as f64).sqrt() as i32;

    if x * x == y {
      return x;
    }

    -1
  }
}
class Solution {
  /**
   * @param Integer $n
   * @return Integer
   */
  function pivotInteger($n) {
    $sum = ($n * ($n + 1)) / 2;
    $pre = 0;
    for ($i = 1; $i <= $n; $i++) {
      if ($pre + $i === $sum - $pre) {
        return $i;
      }
      $pre += $i;
    }
    return -1;
  }
}

Solution 2: Mathematics

We can transform the above equation to get:

$$ n \times (n + 1) = 2 \times x^2 $$

That is:

$$ x = \sqrt{\frac{n \times (n + 1)}{2}} $$

If $x$ is an integer, then $x$ is the pivot integer, otherwise there is no pivot integer.

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

class Solution:
  def pivotInteger(self, n: int) -> int:
    y = n * (n + 1) // 2
    x = int(sqrt(y))
    return x if x * x == y else -1
class Solution {
  public int pivotInteger(int n) {
    int y = n * (n + 1) / 2;
    int x = (int) Math.sqrt(y);
    return x * x == y ? x : -1;
  }
}
class Solution {
public:
  int pivotInteger(int n) {
    int y = n * (n + 1) / 2;
    int x = sqrt(y);
    return x * x == y ? x : -1;
  }
};
func pivotInteger(n int) int {
  y := n * (n + 1) / 2
  x := int(math.Sqrt(float64(y)))
  if x*x == y {
    return x
  }
  return -1
}
function pivotInteger(n: number): number {
  const y = Math.floor((n * (n + 1)) / 2);
  const x = Math.floor(Math.sqrt(y));
  return x * x === y ? x : -1;
}

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