Problem Description
Given two integers, num and t, determine the maximum possible value of a number x such that after applying the following operation at most t times, x can be made equal to num: • In one operation, increase or decrease x by 1 while simultaneously doing the opposite (decrease or increase) to num. Return the maximum possible initial value of x that is achievable.
Key Insights
- The operation changes both x and num simultaneously, maintaining a predictable difference change.
- Each operation can change the difference between x and num by exactly 2.
- To equalize x and num, if the initial difference d = x - num is positive, one can repeatedly decrease x by 1 and increase num by 1 to reduce the difference by 2.
- Consequently, to make the difference 0, the minimum required moves is d/2, provided d is even.
- The maximum x we can choose such that d = 2t is x = num + 2t.
Space and Time Complexity
Time Complexity: O(1) – Only a constant number of operations are performed.
Space Complexity: O(1) – Only a fixed amount of extra space is used.
Solution
The key observation is that each allowed operation adjusts the difference between x and num by 2. To achieve equality (final difference 0), if the initial difference d = x - num is positive, you would need to apply the operation "decrease x by 1 and increase num by 1" exactly d/2 times. Since you can perform at most t operations, the maximum valid difference is 2t. Hence, the maximum achievable x is computed as num + 2t.
The solution uses a direct mathematical formula, avoiding loops or extra data structures.