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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).
1
The function find_pivot iterates through numbers from 1 to n. For each number, it calculates the sum of numbers from 1 to x (sum1) and from x to n (sum2). It returns x if the two sums are equal, otherwise returns -1 at the end.
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).
1using System;
2
3class PivotFinder {
4 public static int FindPivot(int n) {
5 int totalSum = n * (n + 1) / 2;
6 for (int x = 1; x <= n; ++x) {
7 if (2 * x * x == totalSum + x * (x - 1)) return x;
8 }
9 return -1;
10 }
11
12 static void Main() {
13 int n = 8;
14 Console.WriteLine(FindPivot(n));
15 }
16}The C# solution efficiently handles the problem by using a fixed total sum and checking each integer x for the pivot condition.
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