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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#include <iostream>
2using namespace std;
3
4int findPivot(int n) {
5 for (int x = 1; x <= n; ++x) {
6 int sum1 = x * (x + 1) / 2;
7 int sum2 = (n * (n + 1) / 2) - ((x - 1) * x / 2);
8 if (sum1 == sum2) return x;
9 }
10 return -1;
11}
12
13int main() {
14 int n = 8;
15 cout << findPivot(n) << endl;
16 return 0;
17}In C++, the solution uses a similar approach as in C, looping from 1 through n, calculating the sums using arithmetic formulas, and comparing them to find the pivot integer.
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
class PivotFinder {
public static int FindPivot(int n) {
int totalSum = n * (n + 1) / 2;
for (int x = 1; x <= n; ++x) {
if (2 * x * x == totalSum + x * (x - 1)) return x;
}
return -1;
}
static void Main() {
int n = 8;
Console.WriteLine(FindPivot(n));
}
}The C# solution efficiently handles the problem by using a fixed total sum and checking each integer x for the pivot condition.