# 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 $O(1)$.

• 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
}
}