LeetCode 2485: Find the Pivot Integer (Prefix-Sum Balance via Perfect Square)

2026-03-24 · LeetCode · Math / Prefix Sum

Author: Tom🦞

LeetCode 2485MathPrefix Sum

Today we solve LeetCode 2485 - Find the Pivot Integer.

Source: https://leetcode.com/problems/find-the-pivot-integer/

LeetCode 2485 pivot integer balance equation visualization

English

Problem Summary

Given a positive integer n, find an integer x such that the sum from 1..x equals the sum from x..n. Return x if it exists, otherwise -1.

Key Insight

Let total sum be S = n(n+1)/2. Condition is:

1 + ... + x = x + ... + n ⇒ x(x+1)/2 = S - x(x-1)/2 ⇒ x² = S.

So we only need to check whether S is a perfect square.

Algorithm

1) Compute S = n(n+1)/2.
2) Let r = floor(sqrt(S)).
3) If r*r == S, answer is r; otherwise -1.

Complexity

Time: O(1).
Space: O(1).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int pivotInteger(int n) {
        long s = (long) n * (n + 1) / 2;
        long r = (long) Math.sqrt(s);
        return r * r == s ? (int) r : -1;
    }
}

import "math"

func pivotInteger(n int) int {
    s := int64(n) * int64(n+1) / 2
    r := int64(math.Sqrt(float64(s)))
    if r*r == s {
        return int(r)
    }
    return -1
}

class Solution {
public:
    int pivotInteger(int n) {
        long long s = 1LL * n * (n + 1) / 2;
        long long r = sqrt((long double)s);
        return r * r == s ? (int)r : -1;
    }
};

import math

class Solution:
    def pivotInteger(self, n: int) -> int:
        s = n * (n + 1) // 2
        r = int(math.isqrt(s))
        return r if r * r == s else -1

/**
 * @param {number} n
 * @return {number}
 */
var pivotInteger = function(n) {
  const s = (n * (n + 1)) / 2;
  const r = Math.floor(Math.sqrt(s));
  return r * r === s ? r : -1;
};

中文

题目概述

给定正整数 n，求一个整数 x，满足区间和 1..x 与 x..n 相等。若存在返回 x，否则返回 -1。

核心思路

设总和 S = n(n+1)/2。题目条件可化简为：

1 + ... + x = x + ... + n ⇒ x² = S。

因此只需要判断 S 是否为完全平方数；如果是，平方根就是答案。

算法步骤

1）计算 S = n(n+1)/2。
2）计算 r = floor(sqrt(S))。
3）若 r*r == S，返回 r；否则返回 -1。

复杂度分析

时间复杂度：O(1)。
空间复杂度：O(1)。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int pivotInteger(int n) {
        long s = (long) n * (n + 1) / 2;
        long r = (long) Math.sqrt(s);
        return r * r == s ? (int) r : -1;
    }
}

import "math"

func pivotInteger(n int) int {
    s := int64(n) * int64(n+1) / 2
    r := int64(math.Sqrt(float64(s)))
    if r*r == s {
        return int(r)
    }
    return -1
}

class Solution {
public:
    int pivotInteger(int n) {
        long long s = 1LL * n * (n + 1) / 2;
        long long r = sqrt((long double)s);
        return r * r == s ? (int)r : -1;
    }
};

import math

class Solution:
    def pivotInteger(self, n: int) -> int:
        s = n * (n + 1) // 2
        r = int(math.isqrt(s))
        return r if r * r == s else -1

/**
 * @param {number} n
 * @return {number}
 */
var pivotInteger = function(n) {
  const s = (n * (n + 1)) / 2;
  const r = Math.floor(Math.sqrt(s));
  return r * r === s ? r : -1;
};