Given a positive integer n, find the pivot integer x such that:
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 <= 1000class Solution {
public:
int pivotInteger(int n) {
float x = sqrt(n*(n+1)/2);
return x==floor(x) ? x : -1;
}
};