Leetcode 2485: Find the Pivot Integer

Input: A positive integer n representing the upper bound of the sequence.

Example: n = 8

Constraints:

• 1 <= n <= 1000

Output: The pivot integer x such that the sum of numbers from 1 to x equals the sum of numbers from x to n, or -1 if no such x exists.

Example: Output: 6

Constraints:

• The returned value will be an integer between 1 and n or -1.

Goal: Find the pivot integer x that satisfies the sum condition or return -1 if no such integer exists.

Steps:

• 1. Calculate the total sum of all integers from 1 to n.

• 2. Check if there exists an integer x where the sum of integers from 1 to x equals the sum of integers from x to n.

• 3. If such x exists, return it; otherwise, return -1.

Goal: The input integer n is guaranteed to be between 1 and 1000.

Steps:

• 1 <= n <= 1000

Assumptions:

• The solution assumes the sum formula for the first n integers is correct.

• Input: n = 8

• Explanation: For n = 8, the total sum of numbers from 1 to 8 is 36. The sum from 1 to 6 is 21, and the sum from 6 to 8 is also 21. Hence, 6 is the pivot integer.

• Input: n = 1

• Explanation: For n = 1, the sum of numbers from 1 to 1 is 1. The sum from 1 to 1 is also 1, so 1 is the pivot integer.

• Input: n = 4

• Explanation: For n = 4, the total sum of numbers from 1 to 4 is 10. There is no integer x where the sum of numbers from 1 to x equals the sum from x to 4, so the output is -1.

Approach: To solve this problem, we can first compute the total sum of the integers from 1 to n. We then check for a pivot integer x such that the sum of integers from 1 to x equals the sum from x to n. We use the fact that the sum of the first n integers is given by the formula n * (n + 1) / 2.

Observations:

• We need to find an integer x where the sum of numbers from 1 to x is equal to the sum of numbers from x to n.

• This can be simplified using the formula for the sum of the first n integers. By calculating the total sum and checking if the square root of that sum is an integer, we can determine the pivot integer.

Steps:

• 1. Calculate the sum of integers from 1 to n using the formula n * (n + 1) / 2.

• 2. Find the square root of the total sum and check if it is an integer.

• 3. If the square root is an integer, return it as the pivot integer; otherwise, return -1.

Empty Inputs:

• No empty inputs are allowed as per the constraints.

Large Inputs:

• For large values of n (up to 1000), the solution must efficiently compute the sum and check for a valid pivot integer.

Special Values:

• If n is a small value like 1, the solution must correctly handle this edge case.

Constraints:

• Ensure that the solution works for all n between 1 and 1000.

int pivotInteger(int n) {
    int sum = n * (n + 1) / 2, x = sqrt(sum);
    return x * x == sum ? x : -1;
}

1 : Function Declaration

int pivotInteger(int n) {

This line defines the function `pivotInteger`, which takes an integer `n` and returns the pivot integer `x` or -1 if no such pivot exists.

2 : Sum Calculation

    int sum = n * (n + 1) / 2, x = sqrt(sum);

Here, `sum` calculates the total sum of integers from 1 to `n`. `x` is the square root of this sum, representing the potential pivot integer.

3 : Conditional Check

    return x * x == sum ? x : -1;

This line checks if the square of `x` equals `sum`. If true, it returns `x` as the pivot integer; otherwise, it returns -1 to indicate no such pivot exists.

Best Case: O(1)

Average Case: O(1)

Worst Case: O(1)

Description: The solution computes the sum in constant time and checks if the square root of the sum is an integer, which takes constant time.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is constant because we only store a few variables.

Leetcode 2485: Find the Pivot Integer

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