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Formatted question description: https://leetcode.ca/all/2485.html

# 2485. Find the Pivot Integer

• Difficulty: Easy.
• Related Topics: Math, Prefix Sum.
• Similar Questions: Bulb Switcher.

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public int pivotInteger(int n) {
for (int x = 1; x < 1000; ++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 < 1000; ++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, 1000):
if (1 + x) * x == (x + n) * (n - x + 1):
return x
return -1


• func pivotInteger(n int) int {
for x := 1; x < 1000; x++ {
if (1+x)*x == (x+n)*(n-x+1) {
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;
}


• 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
}
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).