2059

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 = [1,3], start = 6, goal = 4
Output: 2
Explanation:
We can go from 6 → 7 → 4 with the following 2 operations.
- 6 ^ 1 = 7
- 7 ^ 3 = 4

Example 2:

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 3:

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 4:

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

Example 5:

Input: nums = [1], start = 0, goal = 3
Output: 3
Explanation: 
We can go from 0 → 1 → 2 → 3 with the following 3 operations. 
- 0 + 1 = 1 
- 1 + 1 = 2
- 2 + 1 = 3

 

Constraints:

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

Similar Questions:

• Minimum Operations to Reduce X to Zero (Medium)

Solution 1. BFS

// OJ: https://leetcode.com/problems/minimum-operations-to-convert-number/
// Author: github.com/lzl124631x
// Time: O(1000 * N)
// Space: O(1000)
class Solution {
public:
    int minimumOperations(vector<int>& A, int start, int goal) {
        queue<int> q{{start}};
        bool seen[1001] = {};
        seen[start] = true;
        int step = 0;
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                int n = q.front();
                q.pop();
                for (int d : A) {
                    for (int next : {n + d, n - d, n ^ d}) {
                        if (next == goal) return step + 1;
                        if (next < 0 || next > 1000 || seen[next]) continue;
                        seen[next] = true;
                        q.push(next);
                    }
                }
            }
            ++step;
        }
        return -1;
    }
};