LeetCode-in-Java

2485. Find the Pivot Integer

Easy

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

public class Solution {
    public int pivotInteger(int n) {
        if (n == 0 || n == 1) {
            return n;
        }
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += i;
        }
        int ad = 0;
        for (int i = 1; i <= n; i++) {
            ad += i - 1;
            sum -= i;
            if (sum == ad) {
                return i;
            }
        }
        return -1;
    }
}

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