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2485. Find the Pivot Integer

Description

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

Solutions

Solution 1: Enumeration

We can directly enumerate \(x\) in the range of \([1,..n]\), and check whether the following equation holds. If it holds, then \(x\) is the pivot integer, and we can directly return \(x\).

\[ (1 + x) \times x = (x + n) \times (n - x + 1) \]

The time complexity is \(O(n)\), where \(n\) is the given positive integer \(n\). The space complexity is \(O(1)\).

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class Solution:
    def pivotInteger(self, n: int) -> int:
        for x in range(1, n + 1):
            if (1 + x) * x == (x + n) * (n - x + 1):
                return x
        return -1

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class Solution {
    public int pivotInteger(int n) {
        for (int x = 1; x <= n; ++x) {
            if ((1 + x) * x == (x + n) * (n - x + 1)) {
                return x;
            }
        }
        return -1;
    }
}

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class Solution {
public:
    int pivotInteger(int n) {
        for (int x = 1; x <= n; ++x) {
            if ((1 + x) * x == (x + n) * (n - x + 1)) {
                return x;
            }
        }
        return -1;
    }
};

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func pivotInteger(n int) int {
    for x := 1; x <= n; x++ {
        if (1+x)*x == (x+n)*(n-x+1) {
            return x
        }
    }
    return -1
}

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function pivotInteger(n: number): number {
    for (let x = 1; x <= n; ++x) {
        if ((1 + x) * x === (x + n) * (n - x + 1)) {
            return x;
        }
    }
    return -1;
}

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impl Solution {
    pub fn pivot_integer(n: i32) -> i32 {
        let y = (n * (n + 1)) / 2;
        let x = (y as f64).sqrt() as i32;

        if x * x == y {
            return x;
        }

        -1
    }
}

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class Solution {
    /**
     * @param Integer $n
     * @return Integer
     */
    function pivotInteger($n) {
        $sum = ($n * ($n + 1)) / 2;
        $pre = 0;
        for ($i = 1; $i <= $n; $i++) {
            if ($pre + $i === $sum - $pre) {
                return $i;
            }
            $pre += $i;
        }
        return -1;
    }
}

Solution 2: Mathematics

We can transform the above equation to get:

\[ n \times (n + 1) = 2 \times x^2 \]

That is:

\[ x = \sqrt{\frac{n \times (n + 1)}{2}} \]

If \(x\) is an integer, then \(x\) is the pivot integer, otherwise there is no pivot integer.

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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class Solution:
    def pivotInteger(self, n: int) -> int:
        y = n * (n + 1) // 2
        x = int(sqrt(y))
        return x if x * x == y else -1

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

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class Solution {
public:
    int pivotInteger(int n) {
        int y = n * (n + 1) / 2;
        int x = sqrt(y);
        return x * x == y ? x : -1;
    }
};

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

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function pivotInteger(n: number): number {
    const y = Math.floor((n * (n + 1)) / 2);
    const x = Math.floor(Math.sqrt(y));
    return x * x === y ? x : -1;
}

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