Welcome to Subscribe On Youtube

2485. Find the Pivot Integer

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 $O(1)$.

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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Php
• RenderScript

• 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;
      }
  };
  

• 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
  
  

• 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;
  }
  
  

• 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;
      }
  }
  
  

• 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
      }
  }
  
  

All Problems

All Solutions