Problem Description

You are given two integers: num and t.

A number x is considered achievable if it can become equal to num after applying the following operation at most t times:

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

Your task is to find the maximum possible value of x that is achievable.

The key insight is understanding how the operation works: when you change x and num simultaneously, you're affecting the gap between them. If you decrease x by 1 and increase num by 1, the distance between them decreases by 2. Conversely, if you increase x by 1 and decrease num by 1, the distance increases by 2.

To maximize x, you want to start with x as large as possible while still being able to make it equal to num within t operations. Since each operation can reduce the gap by 2, and you have t operations available, the maximum initial gap can be 2 * t. Therefore, the maximum achievable x is num + 2 * t.

Intuition

Let's think about what happens during each operation. When we apply the operation, we can either:

• Decrease x by 1 and increase num by 1, or
• Increase x by 1 and decrease num by 1

The crucial observation is that in both cases, the distance between x and num changes by exactly 2. In the first case, the gap decreases by 2, and in the second case, it increases by 2.

Since we want to find the maximum possible x, we should start with x as large as possible. But how large can it be? Well, we need to ensure that after at most t operations, x equals num.

If we start with a large x (larger than num), we need to bring it down to meet num. The most efficient way is to decrease x while increasing num in each operation, which closes the gap by 2 per operation.

With t operations available, we can close a gap of at most 2 * t. Therefore, the maximum starting value of x would be num + 2 * t.

To verify: Starting with x = num + 2 * t, after t operations of decreasing x by 1 and increasing num by 1:

• x becomes: (num + 2 * t) - t = num + t
• num becomes: num + t
• They meet at num + t, confirming our answer.

This explains why the solution is simply num + 2 * t.

Learn more about Math patterns.

Solution Approach

The solution is remarkably straightforward once we understand the mathematical relationship between the operations and the achievable values.

From our analysis, we know that:

• Each operation can change the distance between x and num by exactly 2
• To maximize x, we want to start with the largest possible value
• With t operations, we can bridge a maximum gap of 2 * t

Therefore, the maximum achievable x is simply num + 2 * t.

The implementation is a direct translation of this mathematical formula:

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

This is a constant time O(1) solution with O(1) space complexity. No loops, no complex data structures, just a simple mathematical calculation.

The elegance of this solution lies in recognizing that we don't need to simulate the operations or try different starting values. The mathematical insight that each operation affects the gap by 2 immediately gives us the answer: start x at num + 2 * t, and use all t operations to decrease x while increasing num, meeting in the middle at num + t.

Example Walkthrough

Let's walk through a concrete example with num = 4 and t = 3.

We want to find the maximum possible value of x that can become equal to num after at most 3 operations.

Step 1: Calculate the maximum achievable x Using our formula: x = num + 2 * t = 4 + 2 * 3 = 10

Step 2: Verify this works Let's start with x = 10 and num = 4, and show how they can meet:

• Initial state: x = 10, num = 4 (gap = 6)
• Operation 1: Decrease x by 1 (→ 9), increase num by 1 (→ 5), gap = 4
• Operation 2: Decrease x by 1 (→ 8), increase num by 1 (→ 6), gap = 2
• Operation 3: Decrease x by 1 (→ 7), increase num by 1 (→ 7), gap = 0

After exactly 3 operations, both values meet at 7. Notice that:

• x went from 10 to 7 (decreased by 3)
• num went from 4 to 7 (increased by 3)
• The initial gap of 6 was closed by 2 per operation (6 → 4 → 2 → 0)

Step 3: Confirm no larger x is possible Could we start with x = 11 instead? Let's check:

• Initial gap would be 11 - 4 = 7
• With 3 operations, we can close a gap of at most 2 * 3 = 6
• Since 7 > 6, we cannot make x = 11 equal to num = 4 within 3 operations

This confirms that num + 2 * t = 10 is indeed the maximum achievable value of x.

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the function performs a fixed number of operations regardless of the input size. It simply executes one multiplication (t * 2) and one addition (num + t * 2), which are constant-time arithmetic operations.

The space complexity is O(1) because the function only uses a constant amount of extra memory. No additional data structures are created, and the space used does not depend on the input values. The function directly returns the calculated result without storing any intermediate values in variables.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Operation Direction

Pitfall: Many people initially think that to maximize x, they should always increase both x and num simultaneously, leading to the incorrect formula x = num + t.

Why it happens: The problem statement says we can "increase or decrease x by 1, and simultaneously increase or decrease num by 1", which might be misread as both values moving in the same direction.

Correct Understanding: The operations are:

• Increase x by 1 AND decrease num by 1 (or vice versa)
• The values move in opposite directions, not the same direction

Solution: Remember that each operation changes the gap between x and num by 2, not by 0. If both moved in the same direction, the gap would remain constant.

2. Overthinking with Complex Algorithms

Pitfall: Attempting to use binary search, dynamic programming, or greedy algorithms to find the maximum x.

Why it happens: The problem appears complex at first glance, and the mention of "maximum" often triggers thoughts of optimization algorithms.

Example of overthinking:

# Unnecessary complex approach
def theMaximumAchievableX(self, num: int, t: int) -> int:
    left, right = num - t * 2, num + t * 2
    while left < right:
        mid = (left + right + 1) // 2
        if self.canReach(mid, num, t):
            left = mid
        else:
            right = mid - 1
    return left

Solution: Recognize that this is a pure mathematical problem. Once you understand that each operation affects the distance by 2, the answer becomes a simple formula: num + 2 * t.

3. Integer Overflow Concerns (Language-Specific)

Pitfall: In languages with fixed integer sizes (like C++ or Java), computing num + 2 * t might cause overflow if the values are large.

Solution: In Python, this isn't an issue due to arbitrary precision integers. In other languages, consider using appropriate data types:

// C++ solution with overflow consideration
long long theMaximumAchievableX(int num, int t) {
    return static_cast<long long>(num) + 2LL * t;
}

4. Off-by-One Errors in Understanding

Pitfall: Confusing whether we need exactly t operations or at most t operations, potentially leading to formulas like num + 2 * (t - 1) or num + 2 * (t + 1).

Why it happens: Some problems require exact operation counts, while others allow "up to" a certain number.

Solution: The problem clearly states "at most t times", meaning we can use anywhere from 0 to t operations. To maximize x, we use all t operations, giving us num + 2 * t.