2059. 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.

class Solution {
public:
    int minimumOperations(vector<int>& A, int start, int goal) {
        queue<int> q;
        q.push(start);
        unordered_set<int> vis;
        vis.insert(start);
        int res = 1;
        while(!q.empty() and res <= 1000) {
            int sz = q.size();
            while(sz--) {
                auto n = q.front(); q.pop();
                for(auto& a : A) {
                    if(a + n == goal or n - a == goal or (a ^ n) == goal) {
                        return res;
                    }
                    if(0 <= a + n and a + n <= 1000 and !vis.count(a+n)) {
                        vis.insert(a+n);
                        q.push(a+n);
                    }
                    if(0 <= n - a and n - a <= 1000 and !vis.count(n - a)) {
                        vis.insert(n-a);
                        q.push(n-a);
                    }
                    if(0 <= (a ^ n) and (a ^ n) <= 1000 and !vis.count(a ^ n)) {
                        vis.insert(a^n);
                        q.push(a^n);
                    }
                }
            }
            res++;
        }
        return -1;
    }
};