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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
}

Explain:

nope.

Complexity:

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

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