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This approach involves iterating through each number from 1 to n and checking if it satisfies the given condition of a pivot integer. For each candidate pivot integer x, calculate the sum of numbers from 1 to x using the formula for the sum of an arithmetic series. Similarly, calculate the sum from x to n. Compare these sums to determine if x is a pivot integer.
Time Complexity: O(n).
Space Complexity: O(1).
1def find_pivot(n):
2 for x in range(1, n + 1):
3 sum1 = x * (x + 1) // 2
4 sum2 = (n * (n + 1) // 2) - ((x - 1) * x // 2)
5 if sum1 == sum2:
6 return x
7 return -1
8
9n = 8
10result = find_pivot(n)
11print(result)The function find_pivot iteratively checks each possible pivot integer from 1 through n, calculating the necessary sums and checking for equality.
Instead of iterating through each possible x, we can use mathematical formulation. As the sum from 1 to n is fixed, let's denote the total sum as S. Then, for pivot x, the condition becomes:
(x * (x + 1)) / 2 = (S - ((x - 1) * x) / 2)
From this, derive a formula to find x directly if it exists. Solving algebraically can help us identify the pivot integer without iteration.
Time Complexity: O(n).
Space Complexity: O(1).
1
This optimized solution uses algebra to directly verify if a number x is a pivot without recalculating sums within the loop.