Minimum Operations to Convert Number

You are given a 0-indexed integer array nums containing distinct numbers, an integer start, and an integer goal. There is an integer x that is initially set to start, and you want to perform operations on x such that it is converted to goal. You can perform the following operation repeatedly on the number x:

If 0 <= x <= 1000, then for any index i in the array (0 <= i < nums.length), you can set x to any of the following:

• x + nums[i]
• x - nums[i]
• x ^ nums[i] (bitwise-XOR)

Note that you can use each nums[i] any number of times in any order. Operations that set x to be out of the range 0 <= x <= 1000 are valid, but no more operations can be done afterward.

Return the minimum number of operations needed to convert x = start into goal, and -1 if it is not possible.

Example 1:

Input: nums = [2,4,12], start = 2, goal = 12
Output: 2
Explanation: We can go from 2 → 14 → 12 with the following 2 operations.

• 2 + 12 = 14
• 14 - 2 = 12

Example 2:

Input: nums = [3,5,7], start = 0, goal = -4
Output: 2
Explanation: We can go from 0 → 3 → -4 with the following 2 operations.

• 0 + 3 = 3
• 3 - 7 = -4 Note that the last operation sets x out of the range 0 <= x <= 1000, which is valid.

Example 3:

Input: nums = [2,8,16], start = 0, goal = 1
Output: -1
Explanation: There is no way to convert 0 into 1.

Constraints:

• 1 <= nums.length <= 1000
• -109 <= nums[i], goal <= 109
• 0 <= start <= 1000
• start != goal
• All the integers in nums are distinct.

SOLUTION:

class Solution:
    def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
        paths = [(start, 0)]
        visited = {start}
        i = 0
        while i < len(paths):
            curr, steps = paths[i]
            if curr == goal:
                return steps
            if 0 <= curr <= 1000:
                for num in nums:
                    if (curr + num) not in visited:
                        visited.add((curr + num))
                        paths.append((curr + num, steps + 1))
                    if (curr - num) not in visited:
                        visited.add((curr - num))
                        paths.append((curr - num, steps + 1))
                    if (curr ^ num) not in visited:
                        visited.add((curr ^ num))
                        paths.append((curr ^ num, steps + 1))
            i += 1
        return -1