Find the Maximum Achievable Number

Difficulty: Easy | Acceptance: 91.30% | Paid: No Topics: Math

You are given two integers, num and t.

An integer x is called achievable if you can perform the following operation exactly t times:

Increment num by 1 and decrement x by 1.

Return the maximum achievable x.

• Examples
• Constraints
• Mathematical Formula
• Iterative Simulation
• Bitwise Operation

Examples

Example 1:

Input: num = 4, t = 1
Output: 6
Explanation: At the end of the operation, num becomes 4 + 1 = 5 and x becomes 6 - 1 = 5. Since num == x, 6 is achievable.

Example 2:

Input: num = 3, t = 2
Output: 7
Explanation: At the end of the operation, num becomes 3 + 2 = 5 and x becomes 7 - 2 = 5. Since num == x, 7 is achievable.

Constraints

0 <= num, t <= 100

Mathematical Formula

Intuition We need to find an integer x such that after incrementing num by t and decrementing x by t, the values are equal. This leads to the equation num + t = x - t. Solving for x gives x = num + 2t.

Steps

• Calculate num + 2 * t.
• Return the result.

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

Complexity

• Time: O(1)
• Space: O(1)
• Notes: This is the optimal solution with constant time complexity.

Iterative Simulation

Intuition We can simulate the process described in the problem. In each of the t operations, num increases by 1. To ensure num and x become equal after x is decremented by 1, x must effectively be 2 units larger than num for every operation step.

Steps

• Initialize a loop that runs t times.
• In each iteration, add 2 to num (representing the net difference needed).
• Return the final value of num.

class Solution:
    def theMaximumAchievableX(self, num: int, t: int) -> int:
        for _ in range(t):
            num += 2
        return num

Complexity

• Time: O(t)
• Space: O(1)
• Notes: While functionally correct for small constraints, the mathematical approach is preferred for O(1) performance.

Bitwise Operation

Intuition Multiplying an integer by 2 is equivalent to shifting its binary representation left by 1 bit. We can use this bitwise operation to calculate 2 * t.

Steps

• Calculate num + (t << 1).
• Return the result.

class Solution:
    def theMaximumAchievableX(self, num: int, t: int) -> int:
        return num + (t << 1)

Complexity

• Time: O(1)
• Space: O(1)
• Notes: This is an alternative O(1) approach that leverages low-level bit manipulation.