LeetCode 2769: Find the Maximum Achievable Number (Math Derivation)

Today we solve LeetCode 2769 - Find the Maximum Achievable Number.

Source: https://leetcode.com/problems/find-the-maximum-achievable-number/

English

Problem Summary

You are given integers num and t. In one move, you can increase or decrease x by 1, and do the opposite operation on num. After exactly t moves, return the maximum possible x.

Key Insight

In one operation, if x increases by 1 then num decreases by 1 (opposite directions).
To maximize x, we always choose that direction for all t moves. Each move effectively gives a net gain of 2 toward achievable x, so the final answer is num + 2t.

Derivation

Start with target candidate x = num. After one optimal move, x can be pushed one step higher while preserving the allowed relation. Repeating this for t moves adds 2t total headroom from the original num baseline.
Therefore, x_max = num + 2t.

Complexity Analysis

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

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

class Solution {
    public int theMaximumAchievableX(int num, int t) {
        return num + 2 * t;
    }
}

func theMaximumAchievableX(num int, t int) int {
    return num + 2*t
}

class Solution {
public:
    int theMaximumAchievableX(int num, int t) {
        return num + 2 * t;
    }
};

class Solution:
    def theMaximumAchievableX(self, num: int, t: int) -> int:
        return num + 2 * t

var theMaximumAchievableX = function(num, t) {
  return num + 2 * t;
};

中文

题目概述

给定整数 num 和 t。每次操作里，你可以让 x 加 1 或减 1， 同时让 num 做相反操作。恰好做 t 次后，求 x 的最大值。

核心思路

要让 x 最大，每一步都应该让 x 增加 1，而 num 减少 1。
这样做 t 次后，x 一共增加 t，同时由题意关系可化简为最终答案 num + 2t。

推导说明

每一步把 x 往上推 1，总共推 t 次，并与 num 的反向移动配对。
代数化简后最大可达值就是 num + 2t，无需模拟过程。

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 theMaximumAchievableX(int num, int t) {
        return num + 2 * t;
    }
}

func theMaximumAchievableX(num int, t int) int {
    return num + 2*t
}

class Solution {
public:
    int theMaximumAchievableX(int num, int t) {
        return num + 2 * t;
    }
};

class Solution:
    def theMaximumAchievableX(self, num: int, t: int) -> int:
        return num + 2 * t

var theMaximumAchievableX = function(num, t) {
  return num + 2 * t;
};