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You are given a positive integer n. A pivot integer is an integer x such that the sum of all integers from 1 to x is equal to the sum of all integers from x to n. Return the pivot integer x. If no such integer exists, return -1. If there are multiple pivot integers, return the largest one.

Clarifying Questions:

  1. Are there any constraints on the value of n?
    • Typically, what’s the range of n? (E.g., small values below 100, or very large values up to 10^9)
  2. Can the input n be 1?
  3. Is n guaranteed to be a positive integer?

Assumptions:

  1. n is a positive integer.
  2. The function should return the largest pivot integer if multiple such integers exist.
  3. If no pivot integer exists, return -1.

Strategy:

  1. Define the sum of the first x integers: ( \text{Sum}(1 \text{ to } x) = \frac{x \times (x + 1)}{2} )
  2. Define the sum of integers from x to n: [ \text{Sum}(x \text{ to } n) = \text{Sum}(1 \text{ to } n) - \text{Sum}(1 \text{ to } (x-1)) ] [ = \frac{n \times (n + 1)}{2} - \frac{(x-1) \times x}{2} ]
  3. Set those two sums equal to each other and solve for x: [ \frac{x \times (x + 1)}{2} = \frac{n \times (n + 1)}{2} - \frac{(x-1) \times x}{2} ]
  4. Rearrange and solve the equation to find valid x.

Code Implementation:

def find_pivot_integer(n: int) -> int:
    left_sum = 0
    total_sum = n * (n + 1) // 2
    
    for x in range(1, n + 1):
        left_sum += x
        right_sum = total_sum - left_sum + x
        
        if left_sum == right_sum:
            return x
        
    return -1

# Test the function with an example
print(find_pivot_integer(8))  # Example test case

Explanation:

  1. Calculate the total sum of numbers from 1 to n.
  2. Iterate over each integer x from 1 to n:
    • Incrementally calculate the left sum.
    • Calculate the right sum as the remaining total sum after subtracting the left sum.
    • If the left sum equals the right sum, return x as the pivot integer.
  3. If no pivot is found, return -1.

Time Complexity:

The time complexity of this solution is (O(n)) because it involves iterating through the integers from 1 to n. The space complexity is (O(1)) because the space required does not depend on the input size and only a few variables are used.

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