2485. Find the Pivot Integer

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)

class Solution {
    public int pivotInteger(int n) {
        int ans = (n * n + n ) /2;
        int sq = (int)Math.sqrt(ans);
        if(sq * sq == ans)return sq;
        else return -1;
    }
}

Explain:

nope.

Complexity:

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