#2485

Easy

LeetCode Problem Workspace

Find the Pivot Integer

Find the Pivot Integer problem challenges you to identify an integer x that balances prefix sums on both sides.

Problem Statement

Given a positive integer n, your task is to find an integer x, known as the pivot integer, such that the sum of integers from 1 to x equals the sum of integers from x to n. If such an integer exists, return it. If no integer satisfies this condition, return -1.

For example, when n = 8, the pivot integer is 6 because the sum from 1 to 6 (1 + 2 + 3 + 4 + 5 + 6) equals the sum from 6 to 8 (6 + 7 + 8). If no such integer can be found, return -1, as shown in the case for n = 4, where no pivot integer exists.

Examples

Example 1

Input: n = 8

Output: 6

6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21.

Example 2

Input: n = 1

Output: 1

1 is the pivot integer since: 1 = 1.

Example 3

Input: n = 4

Output: -1

It can be proved that no such integer exist.

Constraints

1 <= n <= 1000

Solution Approach

Math Formula

The problem revolves around the mathematical relationship between prefix sums. The key observation is that the sum of integers from 1 to x equals the sum from x to n. You can use the formula for the sum of an arithmetic series, S = n(n + 1)/2, to determine potential pivots.

Efficient Calculation

Instead of brute-forcing through all values, leverage the sum formula and solve for x. Check if the sum of integers from 1 to x equals the sum from x to n using a mathematical approach that directly compares sums.

Brute Force for Validation

If the direct mathematical formula seems tricky or for validation, you can use a brute force approach to check if any x satisfies the condition from 1 to n. This method works for smaller inputs but becomes inefficient as n grows.

Complexity Analysis

Metric	Value
Time	O(1)
Space	Since the input size (in terms of bits) is bounded by a constant multiple of

The time complexity is O(1) since the pivot integer can be calculated using a mathematical formula, not requiring iteration over all values of n. The space complexity is O(1) as no extra space is needed besides basic variables.

What Interviewers Usually Probe

Checks if the candidate understands the relation between prefix sums.
Evaluates if the candidate can apply an efficient mathematical formula.
Assesses how the candidate handles edge cases like n = 1 or no pivot integer.

Common Pitfalls or Variants

Common pitfalls

Ignoring the need for an efficient solution beyond brute force.
Overcomplicating the problem with unnecessary calculations.
Missing edge cases where no pivot integer exists.

Follow-up variants

Handling cases where no pivot integer exists.
Optimizing for larger values of n (up to 1000).
Extending the problem to larger ranges of n or different sum conditions.

FAQ

What is the pattern behind finding the pivot integer?

The pattern involves using prefix sums and the formula for the sum of integers to check if an integer divides the sum of the sequence into two equal parts.

How can I optimize my solution for larger inputs of n?

To optimize, avoid brute force. Use the formula for the sum of an arithmetic series to directly compute the pivot integer, reducing the time complexity.

What happens when there is no pivot integer?

If no such integer exists, return -1. The sum of the integers cannot be split into two equal parts.

Can this problem be solved using brute force?

Yes, brute force can check every integer from 1 to n, but it is inefficient for larger values of n. A mathematical approach is preferred.

What is the time complexity of solving the Find the Pivot Integer problem?

The time complexity is O(1) since you can calculate the pivot integer using a direct formula without iterating through all integers.

terminal

Solution

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 $x$.

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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

Solution 2: Mathematics

We can transform the above equation to get:

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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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