2769. Find the Maximum Achievable Number


Problem Description

In this problem, we are provided with two integers, num and t. We have to understand the concept of an "achievable" integer x. An integer x is considered achievable if we can make it equal to num by using a certain operation at most t times. This operation involves changing the value of x by 1 (either increasing or decreasing it) and simultaneously changing num by 1 in the opposite direction (if x is decreased, num is increased, and vice versa). Our task is to find the maximum possible achievable integer x.

To solve this, we need to think about how the operation affects the relationship between x and num. Each operation changes the difference between x and num by 2 units. Because we have the freedom to do this t times, we must evaluate how far we can push x away from num initially so that after at most t operations they can still become equal.

Intuition

To derive the intuition behind the solution, it's crucial to look at the effect of the operation on the difference between x and num. Since each operation can be performed at most t times, we want to maximize x before we start performing operations.

If we increase x by 1 and decrease num by 1, the difference between them increases by 2. Conversely, if we decrease x by 1 and increase num by 1, the difference decreases by 2. Since we're looking for the maximum achievable x, we'll want to increase x as much as possible before hitting the limit of t operations. Since t is the maximum number of times we can perform this operation, the farthest we can stretch x from num initially is by t operations, each of which will increase the difference by 2. Therefore, the maximum achievable x can simply be calculated by taking num and adding t*2 to it.

Thus, even without running through the operations, we can directly find the maximum achievable x by this straightforward mathematical relationship.

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Solution Approach

The given problem is simple in nature and doesn't require complex algorithms or data structures. It's essentially a mathematical problem that we solve through an understanding of integers and their relationships after a series of operations.

Given that we can either increase or decrease x by 1 and simultaneously do the opposite to num in the same operation, we realize that each operation will either widen or narrow the gap between x and num by 2. To reach the highest achievable x, we need to maximize x from the onset.

As per the Reference Solution Approach, we see that this operation of increasing x by 1 and decreasing num by 1 (or vice versa) can be done up to t times. Therefore, to maximize x, we would theoretically perform all t operations in the direction that increases x (and decreases num).

The solution code implements this straightforward insight:

1class Solution:
2    def theMaximumAchievableX(self, num: int, t: int) -> int:
3        # By performing the operation t times, each time increasing x by 1,
4        # the maximum achievable x is calculated.
5        return num + t * 2

Here, we take the original num and add t * 2 to account for t operations, each increasing the initial value of x by 2. There's no iteration, conditional logic, or use of additional space for data structures, making the entire solution a single mathematical expression.

In conclusion, the solution relies on the understanding that each operation can happen t times, changing the value of x by a total of t * 2. We immediately return num + t * 2 as the result, representing the maximum achievable x. This simplicity is what makes the problem solvable in constant time and space complexity, i.e., O(1).

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Suppose we have num = 10 and t = 3. Our goal is to find the maximum possible achievable integer x.

Recall that the operation we can perform changes x by 1 and simultaneously changes num by 1 in the opposite direction. Let's see how we can use this operation to our advantage:

  1. We start with num = 10 and we choose an initial value for x. As we want to find the maximum achievable x, it's intuitive to start by increasing x.
  2. For the first operation, if we increase x by 1 (making x initially 11), we have to decrease num by 1 (making num now 9).
  3. After the first operation, the difference between x and num is 11 - 9 = 2. We can perform the operation two more times since t=3.
  4. For the second operation, we increase x by 1 to 12 and decrease num by 1 to 8. The difference is now 12 - 8 = 4.
  5. Finally, for the third operation, we increase x by 1 to 13 and decrease num by 1 to 7. The difference is now 13 - 7 = 6.

Since we have now performed the operation t times (which is 3 times in our case), we have reached our limit, and the resulting value of x is the maximum achievable. Every time we performed an operation, we increased the difference by 2, and since we've done that t times, the maximum increase in x compared to the original num is t * 2.

In our example, the maximum achievable x is therefore 10 + (3 * 2) = 16.

This simple example illustrates that we can calculate the maximum achievable x, without having to actually increment and decrement x and num step by step, by simply taking the original num and adding t * 2 to it, as shown in the solution code. Thus, our answer for this example is x = 16.

Solution Implementation

1class Solution:
2    def theMaximumAchievableX(self, num: int, t: int) -> int:
3        # This method calculates the maximum value achievable for 'x'
4        # by adding the given number 'num' and twice the value of 't'.
5      
6        # Calculate the result by adding 'num' with double the value of 't'
7        result = num + t * 2
8      
9        # Return the computed result
10        return result
11
1class Solution {
2  
3    // Method to calculate the maximum achievable value of X.
4    // Parameters:
5    //   int num: The base number from which we start.
6    //   int t: The number of times we perform the operation to increase 'num'.
7    // Returns:
8    //   The maximum achievable value of X after performing the operation 't' times.
9    public int theMaximumAchievableX(int num, int t) {
10        // The operation consists of doubling the increment each time (hence, t * 2)
11        // Add the total increment to the original 'num' to get the maximum achievable X.
12        return num + t * 2;
13    }
14}
15
1class Solution {
2public:
3    // Function to calculate the maximum achievable value of x
4    int theMaximumAchievableX(int num, int t) {
5        // The formula to compute the maximum achievable value
6        int maximizedX = num + t * 2;
7
8        // Return the calculated value
9        return maximizedX;
10    }
11};
12
1// This function calculates the maximum achievable value of x given a number 'num' and a multiplier 't'.
2// The value of 'x' is found by adding 'num' to twice the value of 't'.
3// @param {number} num - The base number to which the result of 't * 2' will be added.
4// @param {number} t - The multiplier that is used to find the additional value to add to 'num'.
5// @returns {number} - The maximum achieved value of 'x'.
6
7function theMaximumAchievableX(num: number, t: number): number {
8    // Calculate the result by adding num to twice the value of t and return it.
9    return num + t * 2;
10}
11

Time and Space Complexity

The time complexity of the given code is O(1). This is because the operation num + t * 2 is performed in constant time, regardless of the size of the input values for num and t.

Similarly, the space complexity of the code is also O(1). The function does not allocate any additional space that grows with the input size. It only uses a fixed amount of space to store the input parameters and the result of the computation.

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


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