Find the Pivot Integer - LeetCode Daily Challenge

Problem Statement

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.

Try here before watching the video.

Video Explanation

Java Code

class Solution {
    public int pivotInteger(int n) {
        int sumOfN = (n * (n + 1)) / 2;
        int sqrtOfSum = (int)Math.sqrt(sumOfN);
        return (sqrtOfSum * sqrtOfSum) == sumOfN ? sqrtOfSum : -1;
    }
}

C++ Code

class Solution {
public:
    int pivotInteger(int n) {
        int sumOfN = (n * (n + 1)) / 2;
        int sqrtOfSum = static_cast<int>(sqrt(sumOfN));
        return (sqrtOfSum * sqrtOfSum) == sumOfN ? sqrtOfSum : -1;
    }
};

Python Code

class Solution:
    def pivotInteger(self, n: int) -> int:
        sum_of_n = (n * (n + 1)) // 2
        sqrt_of_sum = int(math.sqrt(sum_of_n))
        return sqrt_of_sum if sqrt_of_sum * sqrt_of_sum == sum_of_n else -1

Javascript Code

var pivotInteger = function(n) {
    let sumOfN = (n * (n + 1)) / 2;
    let sqrtOfSum = Math.floor(Math.sqrt(sumOfN));
    return sqrtOfSum * sqrtOfSum === sumOfN ? sqrtOfSum : -1;
};

Go Code

func pivotInteger(n int) int {
	sumOfN := (n * (n + 1)) / 2
	sqrtOfSum := int(math.Sqrt(float64(sumOfN)))
	if sqrtOfSum*sqrtOfSum == sumOfN {
		return sqrtOfSum
	}
	return -1
}

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(1)