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 <= 1090 <= start <= 1000start != goal- All the integers in
numsare distinct.
BFS
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 nx >= 0 and nx <= 1000 and not vis[nx]:
q.append((nx, step + 1))
vis[nx] = True
return -1class 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;
}
};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;
}