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2059. Minimum Operations to Convert Number

Difficulty Topics

Description

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

minimum-operations-to-convert-number.py
class Solution:
    def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
        queue = collections.deque([(start, 0)])
        seen = set()

        while queue:
            x, steps = queue.popleft()

            for num in nums:
                operations = [x + num, x - num, x ^ num]
                for o in operations:
                    if o == goal: return steps + 1
                    if 0 <= o <= 1000 and o not in seen:
                        queue.append((o, steps + 1))
                        seen.add(o)

        return -1