# Solving Leetcode Interviews in Seconds with AI: Minimum Operations to Convert Number


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2059" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.  

	# Explanation
	Here's the solution:

*   **High-level approach:**
    *   Use Breadth-First Search (BFS) to explore possible values of `x` reachable from `start`.
    *   Maintain a queue of `(value, steps)` representing the current value of `x` and the number of operations taken to reach that value.
    *   Keep track of visited values within the range \[0, 1000] to avoid cycles. Once out of bound, stop and don't add to queue.

*   **Complexity:**
    *   Runtime Complexity: O(N \* R), where N is the length of `nums` and R is the range of values visited (up to 1001 within bounds, potentially more outside).
    *   Storage Complexity: O(R), mainly for the `visited` set/array.

	
	# Code
	```python
	from collections import deque

def minimumOperations(nums, start, goal):
    """
    Finds the minimum number of operations to convert start to goal using nums.

    Args:
        nums: A list of distinct integers.
        start: The starting integer.
        goal: The target integer.

    Returns:
        The minimum number of operations, or -1 if not possible.
    """

    queue = deque([(start, 0)])
    visited = {start}
    
    while queue:
        x, steps = queue.popleft()

        if x == goal:
            return steps

        for num in nums:
            next_values = [x + num, x - num, x ^ num]
            for next_x in next_values:
                if 0 <= next_x <= 1000 and next_x not in visited:
                    visited.add(next_x)
                    queue.append((next_x, steps + 1))
                elif not (0 <= next_x <= 1000): # Allow out of bound, but don't add to queue
                    if next_x == goal:
                        return steps + 1


    return -1
	```