LeetCode in Kotlin

2059. Minimum Operations to Convert Number

Medium

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 {
    fun minimumOperations(nums: IntArray, start: Int, goal: Int): Int {
        val seen = BooleanArray(1001)
        var q = listOf(goal)
        var cnt = 0
        while (q.isNotEmpty()) {
            ++cnt
            val q1: MutableList<Int> = ArrayList()
            for (x in q) {
                for (n in nums) {
                    for (xn in intArrayOf(x + n, x - n, x xor n)) {
                        if (xn in 0..1000 && !seen[xn]) {
                            if (xn == start) {
                                return cnt
                            }
                            seen[xn] = true
                            q1.add(xn)
                        }
                    }
                }
                q = q1
            }
        }
        return -1
    }
}

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