# 2059. Minimum Operations to Convert Number

## 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.

## Solutions

BFS.

• class Solution {
public int minimumOperations(int[] nums, int start, int goal) {
IntBinaryOperator op1 = (x, y) -> x + y;
IntBinaryOperator op2 = (x, y) -> x - y;
IntBinaryOperator op3 = (x, y) -> x ^ y;
IntBinaryOperator[] ops = {op1, op2, op3};
boolean[] vis = new boolean[1001];
Queue<int[]> queue = new ArrayDeque<>();
queue.offer(new int[] {start, 0});
while (!queue.isEmpty()) {
int[] p = queue.poll();
int x = p[0], step = p[1];
for (int num : nums) {
for (IntBinaryOperator op : ops) {
int nx = op.applyAsInt(x, num);
if (nx == goal) {
return step + 1;
}
if (nx >= 0 && nx <= 1000 && !vis[nx]) {
queue.offer(new int[] {nx, step + 1});
vis[nx] = true;
}
}
}
}
return -1;
}
}

• class Solution {
public:
int minimumOperations(vector<int>& nums, int start, int goal) {
using pii = pair<int, int>;
vector<function<int(int, int)>> ops{
[](int x, int y) { return x + y; },
[](int x, int y) { return x - y; },
[](int x, int y) { return x ^ y; },
};
vector<bool> vis(1001, false);
queue<pii> q;
q.push({start, 0});
while (!q.empty()) {
auto [x, step] = q.front();
q.pop();
for (int num : nums) {
for (auto op : ops) {
int nx = op(x, num);
if (nx == goal) {
return step + 1;
}
if (nx >= 0 && nx <= 1000 && !vis[nx]) {
q.push({nx, step + 1});
vis[nx] = true;
}
}
}
}
return -1;
}
};

• class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
op1 = lambda x, y: x + y
op2 = lambda x, y: x - y
op3 = lambda x, y: x ^ y
ops = [op1, op2, op3]
vis = [False] * 1001
q = deque([(start, 0)])
while q:
x, step = q.popleft()
for num in nums:
for op in ops:
nx = op(x, num)
if nx == goal:
return step + 1
if 0 <= nx <= 1000 and not vis[nx]:
q.append((nx, step + 1))
vis[nx] = True
return -1


• func minimumOperations(nums []int, start int, goal int) int {
type pair struct {
x    int
step int
}

ops := []func(int, int) int{
func(x, y int) int { return x + y },
func(x, y int) int { return x - y },
func(x, y int) int { return x ^ y },
}
vis := make([]bool, 1001)
q := []pair{ {start, 0} }

for len(q) > 0 {
x, step := q[0].x, q[0].step
q = q[1:]
for _, num := range nums {
for _, op := range ops {
nx := op(x, num)
if nx == goal {
return step + 1
}
if nx >= 0 && nx <= 1000 && !vis[nx] {
q = append(q, pair{nx, step + 1})
vis[nx] = true
}
}
}
}
return -1
}

• function minimumOperations(nums: number[], start: number, goal: number): number {
const n = nums.length;
const op1 = function (x: number, y: number): number {
return x + y;
};
const op2 = function (x: number, y: number): number {
return x - y;
};
const op3 = function (x: number, y: number): number {
return x ^ y;
};
const ops = [op1, op2, op3];
let vis = new Array(1001).fill(false);
let quenue: Array<Array<number>> = [[start, 0]];
vis[start] = true;
while (quenue.length) {
let [x, step] = quenue.shift();
for (let i = 0; i < n; i++) {
for (let j = 0; j < ops.length; j++) {
const nx = ops[j](x, nums[i]);
if (nx == goal) {
return step + 1;
}
if (nx >= 0 && nx <= 1000 && !vis[nx]) {
vis[nx] = true;
quenue.push([nx, step + 1]);
}
}
}
}
return -1;
}