Minimum Operations to Convert Number

Medium

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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.

Try coding it in LeetCode

Minimum Operations to Convert Number

Medium

Asked by:

4 views

Topics:

Graphs

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the `start`, `goal`, and the numbers within the `nums` array?
Can the `nums` array be empty?
If it's impossible to reach the `goal` from the `start` using the given operations, what value should I return?
Are there any constraints on the number of operations I can perform? Is there a maximum depth I should consider during the search?
Are the operations performed in `nums` applied in the order presented in the `nums` array, or can I choose any order?

Brute Force Solution

Approach

The brute force approach to this problem involves exploring every possible combination of operations to reach the target number. We will simulate each possible sequence of operations, starting from the initial number, until we find the shortest path to the target. This is like trying every single route on a map to find the fastest one.

Here's how the algorithm would work step-by-step:

Start with the initial number and zero operations.
Consider applying each possible operation (+a, -b, XOR c) to the current number.
For each operation, create a new number and increment the operation count by one.
Keep track of all the numbers you've created and the number of operations it took to reach them.
Repeat the process of applying operations to the newly created numbers, making sure not to repeat any numbers you've already visited (to avoid infinite loops).
If you reach the target number, record the number of operations it took.
Continue exploring all possible combinations of operations until you have explored all reasonable possibilities.
Once you've explored enough possibilities, pick the path that reaches the target number with the fewest operations.

Code Implementation

def minimum_operations_to_convert_number(number_start, number_target, operation_a, operation_b, operation_c):
    queue = [(number_start, 0)]
    visited_numbers = {number_start}

    while queue:
        current_number, operations_count = queue.pop(0)

        if current_number == number_target:
            return operations_count

        # Explore adding operation_a to the current number.
        next_number_add = current_number + operation_a

        if 0 <= next_number_add <= 1000 and next_number_add not in visited_numbers:
            queue.append((next_number_add, operations_count + 1))
            visited_numbers.add(next_number_add)

        # Explore subtracting operation_b from the current number.
        next_number_subtract = current_number - operation_b

        if 0 <= next_number_subtract <= 1000 and next_number_subtract not in visited_numbers:
            queue.append((next_number_subtract, operations_count + 1))
            visited_numbers.add(next_number_subtract)

        # Explore XORing operation_c with the current number.
        next_number_xor = current_number ^ operation_c

        # Use bitwise XOR operation.
        if 0 <= next_number_xor <= 1000 and next_number_xor not in visited_numbers:
            queue.append((next_number_xor, operations_count + 1))
            visited_numbers.add(next_number_xor)

    return -1

Big(O) Analysis

Time Complexity

O(3^N) – The brute force approach explores all possible combinations of operations (+a, -b, XOR c) at each step. In the worst case, without constraints to avoid revisiting numbers, we can perform any of the three operations on each newly generated number. If we consider N to be the number of operations needed to reach the target, the algorithm branches out into 3 possibilities for each step, leading to 3 * 3 * 3 ... (N times), which is 3^N. Because of the exponential branching, the time complexity is O(3^N).

Space Complexity

O(1) – The brute force approach explores possible combinations of operations. This involves keeping track of visited numbers to avoid infinite loops. The maximum number of unique numbers is limited by the range of possible values, so we might have a visited set that contains up to some constant number of numbers, say C. We also need to store the operation counts associated with each of these visited numbers. Therefore, the auxiliary space is limited by this constant number, making the space complexity O(1).

Optimal Solution

Approach

The most efficient way to solve this is to explore all possible paths from the starting number to the target number, but without re-exploring the same numbers unnecessarily. We track how many operations it takes to reach each number, so we don't waste time going in circles.

Here's how the algorithm would work step-by-step:

Start with your initial number and a count of zero operations.
Create a way to keep track of the minimum number of operations needed to reach any number we encounter. Initially, we only know how to reach the starting number (with zero operations).
Consider all possible operations you can perform on the current number (addition, subtraction, or bitwise XOR).
For each resulting number from these operations, check: have we reached this number before? If not, or if we've reached it before but with more operations, then update the minimum operations needed to reach this number.
Add these newly reached numbers to our list of numbers to explore.
Repeat the process of exploring operations and updating operation counts until you reach the target number or you have no more numbers to explore.
If you reach the target, the recorded operation count is the minimum number of operations needed. If you run out of numbers to explore before reaching the target, it's impossible to reach the target from the starting number.

Code Implementation

def minimum_operations_to_convert_number(start_number, target_number, numbers):
    queue = [(start_number, 0)]
    visited = {start_number: 0}

    while queue:
        current_number, operation_count = queue.pop(0)

        if current_number == target_number:
            return operation_count

        for number in numbers:
            # Explore all operations: add, subtract, xor
            next_numbers = [
                current_number + number,
                current_number - number,
                current_number ^ number
            ]

            for next_number in next_numbers:
                # We limit the search space to 0 <= number <= 1000
                if 0 <= next_number <= 1000:
                    # Check if we've reached this number before
                    if next_number not in visited or visited[next_number] > operation_count + 1:
                        visited[next_number] = operation_count + 1

                        queue.append((next_number, operation_count + 1))

    return -1

Big(O) Analysis

Time Complexity

O(2^32) – The algorithm explores a search space defined by all possible numbers reachable from the start using the given operations. In the worst-case scenario, every possible integer between 0 and 2^32-1 could be visited. For each number visited, three operations (add, subtract, XOR) are performed with elements from the nums array. The size of the nums array is constant so it does not affect Big O. Therefore, the complexity is determined by the size of the possible integer space, leading to O(2^32).

Space Complexity

O(1) – The algorithm uses a queue to store numbers to explore and a data structure (e.g., a set or a map) to track visited numbers and their corresponding operation counts. In the worst case, we might explore all possible numbers within a certain range, let's say up to 1000 (or based on constraints not explicitly mentioned in the problem description). The size of the queue and the visited set/map depend on this range. Therefore the space usage does not scale with an input size parameter and is constant.

Edge Cases

Case	How to Handle
start == goal	Return 0 immediately as no operations are required.
nums array is empty or null	Return -1, indicating that the goal cannot be reached from the start using an empty set of operations.
nums array contains very large numbers leading to potential integer overflow during addition or multiplication	Use modulo operator with a sufficiently large prime number to prevent overflow, ensuring numbers stay within a manageable range.
The goal is unreachable from the start using the given operations	Return -1 after exploring all possible paths using BFS or DFS if the goal is not found.
One of the nums is 0	Multiplication by 0 can significantly alter the search space, handle this case as a normal calculation within the BFS/DFS approach.
start, goal or elements of nums are negative	The problem statement doesn't exclude negative numbers; handle negative results appropriately in calculations (especially modulo) and ensure the search space includes negative numbers if potentially reachable.
nums contains duplicate elements	Duplicate numbers don't change the fundamental logic; the algorithm will consider each unique operation independently and the BFS/DFS will manage visited states correctly, thus duplicates are handled without special consideration.
Prevent revisiting states to avoid infinite loops within the BFS/DFS search	Use a visited set or array to keep track of visited numbers and avoid redundant calculations, improving efficiency.