Find the Maximum Achievable Number

• The main idea is to search for the maximum achievable number x by starting from a high value and iterating downwards.
• For each candidate x, we check if it's possible to make x and num equal in at most t steps.
• This check is performed by a recursive function, isAchievable(currentX, currentNum, stepsLeft), which simulates the process.
• isAchievable function logic:
  • Base Case 1: If currentX == currentNum, they are equal, so we return true.
  • Base Case 2: If stepsLeft == 0 and they are not equal, we've run out of operations, so we return false.
  • Recursive Step: The function explores all four possible operations by making a recursive call for each:
    1. isAchievable(currentX + 1, currentNum + 1, stepsLeft - 1)
    2. isAchievable(currentX + 1, currentNum - 1, stepsLeft - 1)
    3. isAchievable(currentX - 1, currentNum + 1, stepsLeft - 1)
    4. isAchievable(currentX - 1, currentNum - 1, stepsLeft - 1)
  • If any of these recursive calls return true, it means a path to equality exists, and the function returns true. Otherwise, it returns false.
• The first value of x (iterating downwards) for which isAchievable returns true is the maximum possible value.

We want to find the maximum achievable x. A simple strategy is to start checking from a large number and go downwards. The first number we find that is "achievable" will be our answer.

A safe upper bound to start from can be derived from the constraints. Since num <= 50 and t <= 50, num + 2*t is at most 50 + 100 = 150. We can start our search from a number like 200.

The core of this approach is the isAchievable(current_x, current_num, steps_left) function. This function checks if current_x and current_num can be made equal using steps_left operations.

1class Solution {2    public int theMaximumAchievableX(int num, int t) {3        // Start checking from a reasonably high number downwards.4        // num + 2*t is the actual answer, so starting there is efficient.5        // For a true brute-force, we might start from a hardcoded limit like 200.6        for (int x = num + 2 * t; x >= 1; x--) {7            if (isAchievable(x, num, t)) {8                return x;9            }10        }11        return -1; // Should not be reached12    }13 14    private boolean isAchievable(long currentX, long currentNum, int stepsLeft) {15        if (currentX == currentNum) {16            return true;17        }18        if (stepsLeft == 0) {19            return false;20        }21 22        // Explore all 4 possible operations23        if (isAchievable(currentX + 1, currentNum + 1, stepsLeft - 1)) return true;24        if (isAchievable(currentX + 1, currentNum - 1, stepsLeft - 1)) return true;25        if (isAchievable(currentX - 1, currentNum + 1, stepsLeft - 1)) return true;26        if (isAchievable(currentX - 1, currentNum - 1, stepsLeft - 1)) return true;27 28        return false;29    }30}