How do you solve Find the Pivot Integer on LeetCode?

Find the Pivot Integer (LeetCode #2485) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using mathematical properties can significantly reduce the time complexity.

What is the brute force solution for Find the Pivot Integer?

The brute force approach for Find the Pivot Integer is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each integer from 1 to n to see if it can be the pivot by calculating the sums on both sides. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Pivot Integer?

Find the Pivot Integer uses the following concepts: Math, Prefix Sum. The recommended patterns to study are: Mathematical Summation, Prefix Sum.

What are the interview tips for Find the Pivot Integer?

Always start with a brute force solution to understand the problem better. Look for patterns or mathematical relationships that can simplify calculations. Be prepared to explain your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Find the Pivot Integer?

Not realizing that calculating sums repeatedly is inefficient. Overlooking the mathematical relationships that can simplify the problem.

#2485

Find the Pivot Integer

Easy

MathPrefix SumMathematical SummationPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of calculating sums repeatedly, we can derive the pivot condition using a mathematical formula, which allows us to find the pivot in linear time.

⚙️

Algorithm

4 steps

1Step 1: Calculate the total sum of integers from 1 to n using the formula n * (n + 1) / 2.
2Step 2: Iterate through each integer x from 1 to n.
3Step 3: For each x, check if 2 * sum(1 to x) == total_sum. If true, return x.
4Step 4: If no pivot is found after checking all integers, return -1.

solution.py10 lines

1# Full working Python code
2
3def find_pivot_integer(n):
4    total_sum = n * (n + 1) // 2
5    sum_left = 0
6    for x in range(1, n + 1):
7        sum_left += x
8        if 2 * sum_left == total_sum:
9            return x
10    return -1

ℹ

Complexity note: The time complexity is O(n) because we only loop through the integers once, and the space complexity is O(1) since we use a constant amount of space.

1Using mathematical properties can significantly reduce the time complexity.
2Understanding the relationship between the sums can lead to a more efficient solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.