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2485. Find the Pivot Integer

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:

class Solution {
    public int pivotInteger(int n) {
        int result = -1;
        for (int i = 1; i <=n; i++){
            int sum = (1+i)*i/2;
            int remain = (i + n)* (n-i +1)/2;
            if (sum == remain){
                result = i;
                break;
            }
        }
        return result;
    }
}

// TC: O(n)
// SC: O(1)