Find the Pivot Integer

Find the Pivot Integer - Problem

Given a positive integer n, find the pivot integer x such that:

The sum of all elements between 1 and x inclusively equals the sum of all elements between x and n inclusively.

Return the pivot integer x. If no such integer exists, return -1.

Note: It is guaranteed that there will be at most one pivot index for the given input.

Input & Output

Example 1 — Basic Case

$ Input: n = 8

› Output: 6

💡 Note: The pivot integer is 6. Sum from 1 to 6: 1+2+3+4+5+6 = 21. Sum from 6 to 8: 6+7+8 = 21. Both sums equal 21.

Example 2 — Single Element

$ Input: n = 1

› Output: 1

💡 Note: The pivot integer is 1. Sum from 1 to 1: 1. Sum from 1 to 1: 1. Both sums equal 1.

Example 3 — No Pivot Exists

$ Input: n = 4

› Output: -1

💡 Note: No pivot exists. For any x: sum(1 to x) ≠ sum(x to 4). Total sum is 10, and √10 ≈ 3.16 is not an integer.

Constraints

1 ≤ n ≤ 1000

Visualization

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Asked in

M Microsoft 15 a Amazon 12 G Google 8

The key insight is recognizing that the pivot condition sum(1 to x) = sum(x to n) simplifies to x² = total_sum. Best approach is the mathematical formula with direct square root calculation. Time: O(1), Space: O(1)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force - Check Every Possible Pivot

⏱️ Time: O(n²) Space: O(1)

For each potential pivot x from 1 to n, calculate the sum from 1 to x and the sum from x to n. If they are equal, return x. If no pivot is found, return -1.

Mathematical Formula (Optimal)

⏱️ Time: O(1) Space: O(1)

From the pivot condition: sum(1 to x) = sum(x to n), we get x*(x+1)/2 = total_sum - x*(x-1)/2. Solving this algebraically gives us x² = total_sum. So we can directly check if sqrt(total_sum) is an integer.

Prefix Sum Optimization

⏱️ Time: O(n) Space: O(1)

Calculate the total sum first, then iterate through potential pivots. For each pivot x, the left sum is x*(x+1)/2 and right sum is total_sum - left_sum + x. This avoids recalculating sums from scratch each time.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int pivotInteger(int n) {
    // For pivot x: sum(1..x) = sum(x..n)
    // sum(1..x) = x*(x+1)/2
    // sum(x..n) = sum(1..n) - sum(1..x-1) = n*(n+1)/2 - (x-1)*x/2
    // Setting equal: x*(x+1)/2 = n*(n+1)/2 - (x-1)*x/2
    // Simplifying: x*(x+1)/2 = n*(n+1)/2 - x*(x-1)/2
    // x*(x+1) = n*(n+1) - x*(x-1)
    // x^2 + x = n*(n+1) - x^2 + x
    // 2*x^2 = n*(n+1)
    // x^2 = n*(n+1)/2
    // x = sqrt(n*(n+1)/2)
    
    long long total = (long long)n * (n + 1) / 2;
    
    // Check if total is a perfect square
    long long x = (long long)sqrt(total);
    
    // Check x-1, x, x+1 to handle floating point precision issues
    for (long long candidate = x - 1; candidate <= x + 1; candidate++) {
        if (candidate <= 0 || candidate > n) continue;
        if (candidate * candidate == total) {
            return (int)candidate;
        }
    }
    
    return -1;
}

int main() {
    int n;
    if (scanf("%d", &n) != 1) {
        return 1;
    }
    
    printf("%d\n", pivotInteger(n));
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

💡 Explanation

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