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Find the Maximum Achievable Number

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

Given two integers, num and t. A number is achievable if it can become equal to num after applying the following operation:

  • Increase or decrease the number by 1, and simultaneously increase or decrease num by 1.

Return the maximum achievable number after applying the operation at most t times.

Examples

Example 1


Input: num = 4, t = 1

Output: 6

Explanation:

Apply the following operation once to make the maximum achievable number equal
to `num`:

  * Decrease the maximum achievable number by 1, and increase `num` by 1.

Example 2


Input: num = 3, t = 2

Output: 7

Explanation:

Apply the following operation twice to make the maximum achievable number
equal to `num`:

  * Decrease the maximum achievable number by 1, and increase `num` by 1.

Constraints

  • 1 <= num, t <= 50

Solution

Method 1 – Math Formula

Intuition

Each operation allows us to increase or decrease both the number and num by 1. To maximize the achievable number, we should always increase both as much as possible, so the answer is simply num + 2 * t.

Approach

  1. The maximum achievable number is obtained by increasing both the number and num by 1 in each operation, for a total of t operations.
  2. Thus, the answer is num + 2 * t.

Code

C++
class Solution {
public:
    int theMaximumAchievableX(int num, int t) {
        return num + 2 * t;
    }
};
Go
func theMaximumAchievableX(num int, t int) int {
    return num + 2*t
}
Java
class Solution {
    public int theMaximumAchievableX(int num, int t) {
        return num + 2 * t;
    }
}
Kotlin
class Solution {
    fun theMaximumAchievableX(num: Int, t: Int): Int {
        return num + 2 * t
    }
}
Python
class Solution:
    def theMaximumAchievableX(self, num: int, t: int) -> int:
        return num + 2 * t
Rust
impl Solution {
    pub fn the_maximum_achievable_x(num: i32, t: i32) -> i32 {
        num + 2 * t
    }
}
TypeScript
class Solution {
    theMaximumAchievableX(num: number, t: number): number {
        return num + 2 * t;
    }
}

Complexity

  • ⏰ Time complexity: O(1), just a simple calculation.
  • 🧺 Space complexity: O(1), no extra space used.

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