Minimum Number of Operations to Make X and Y Equal

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GraphsDynamic Programming

You are given two positive integers x and y.

In one operation, you can do one of the four following operations:

Divide x by 11 if x is a multiple of 11.
Divide x by 5 if x is a multiple of 5.
Decrement x by 1.
Increment x by 1.

Return the minimum number of operations required to make x and y equal.

Example 1:

Input: x = 26, y = 1
Output: 3
Explanation: We can make 26 equal to 1 by applying the following operations: 
1. Decrement x by 1
2. Divide x by 5
3. Divide x by 5
It can be shown that 3 is the minimum number of operations required to make 26 equal to 1.

Example 2:

Input: x = 54, y = 2
Output: 4
Explanation: We can make 54 equal to 2 by applying the following operations: 
1. Increment x by 1
2. Divide x by 11 
3. Divide x by 5
4. Increment x by 1
It can be shown that 4 is the minimum number of operations required to make 54 equal to 2.

Example 3:

Input: x = 25, y = 30
Output: 5
Explanation: We can make 25 equal to 30 by applying the following operations: 
1. Increment x by 1
2. Increment x by 1
3. Increment x by 1
4. Increment x by 1
5. Increment x by 1
It can be shown that 5 is the minimum number of operations required to make 25 equal to 30.

Constraints:

1 <= x, y <= 10⁴

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 integer inputs x and y? Should I be concerned about potential integer overflow?
Can x or y be negative? If so, how should negative values be handled by the operations?
If x is already equal to y, should I return 0?
Are we optimizing for the absolute minimum number of operations, even if it means exploring a less intuitive path?
Is there a specific data type I should use for intermediate calculations or the return value, such as `long`, to prevent overflow issues?

Brute Force Solution

Approach

The brute force method for making two numbers equal involves exploring every possible sequence of allowed actions. It's like trying every single path to see which one gets you there using the fewest steps. We generate these sequences and then see how many operations each one needs.

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

Start with the two numbers you're trying to make equal.
Consider all possible operations you can perform on either number.
For each operation, apply it to one of the numbers and create a new pair of numbers.
Repeat the process of applying operations to these new pairs, creating even more pairs of numbers.
Continue doing this until one of the pairs has the same value for both numbers. Make sure you don't repeat previously explored pairs.
Keep track of the number of operations needed to reach each pair where the numbers are equal.
Finally, compare the number of operations needed for all the paths that resulted in equal numbers, and choose the path with the smallest number of operations. This is the minimum number of operations needed.

Code Implementation

def minimum_operations_brute_force(start_value, end_value):
    queue = [(start_value, end_value, 0)]
    visited = set()
    minimum_operations = float('inf')

    while queue:
        current_x, current_y, operations_count = queue.pop(0)

        if current_x == current_y:
            minimum_operations = min(minimum_operations, operations_count)
            continue

        # Avoid infinite loops by tracking visited states.
        if (current_x, current_y) in visited:
            continue
        visited.add((current_x, current_y))

        # Operations on X
        next_x_values = [
            current_x + 1,
            current_x - 1,
            current_x * 2,
        ]
        if current_x % 2 == 0:
            next_x_values.append(current_x // 2)

        for next_x in next_x_values:
            queue.append((next_x, current_y, operations_count + 1))

        # Operations on Y
        next_y_values = [
            current_y + 1,
            current_y - 1,
            current_y * 2,
        ]
        if current_y % 2 == 0:
            next_y_values.append(current_y // 2)

        for next_y in next_y_values:
            queue.append((current_x, next_y, operations_count + 1))

    if minimum_operations == float('inf'):
        return -1
    else:
        return minimum_operations

Big(O) Analysis

Time Complexity

O(b^d) – The brute force approach explores all possible sequences of operations. The branching factor, b, represents the number of possible operations at each step (e.g., increment x, decrement x, increment y, decrement y). The depth, d, represents the maximum number of operations explored before finding a solution or reaching a limit. Since we are exploring a tree-like structure where each node can have b children, and the tree has a maximum depth of d, the total number of nodes explored can grow exponentially, approximating b^d.

Space Complexity

O(b^d) – The brute force approach explores all possible operation sequences using a breadth-first or depth-first search-like strategy. The space complexity is primarily determined by the data structure used to store the pairs of numbers that need to be explored and the pairs that have already been visited. In the worst case, we could potentially explore all possible pairs up to a certain depth 'd', where each number has 'b' possible operations applied to it at each step. Therefore, the space required to store the visited pairs and the queue of pairs to explore can grow exponentially with the depth 'd' and the number of branches 'b', leading to a space complexity of O(b^d).

Optimal Solution

Approach

We want to find the fewest changes to make two numbers the same. Instead of blindly trying every possibility, we'll work backwards from the target state (where the numbers are equal) to efficiently find the shortest path.

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

First, recognize that we can only change one number at a time using specific operations (add, subtract, multiply, divide).
Our main goal is to make both numbers equal, so consider the possible relationships between X and Y: X is bigger than Y, X is smaller than Y, or X is equal to Y.
If X is already equal to Y, we need zero operations and we're done.
If X is not equal to Y, imagine we are trying to get to the point where X equals Y. This means that, at one step before the end, either X changed to become Y, or Y changed to become X.
We can use a method to keep track of the minimum number of operations required to reach different numbers starting from X and Y, for example using a queue.
Begin with X and Y, and explore all the numbers we can reach from them in one operation, two operations, and so on, until we find that X equals Y. While searching, make sure we avoid repeating numbers we've already seen, as this would lead to inefficient calculations.
This will eventually lead to both numbers meeting, and the method will return the minimum number of operations that make X equal to Y.

Code Implementation

def min_operations(x_value, y_value):
    operation_count = 0
    while x_value != y_value:
        if x_value > y_value:
            # If x is greater, decrement or divide.
            if x_value % 2 == 0:
                x_value //= 2
                operation_count += 1
            else:
                x_value -= 1
                operation_count += 1
        else:
            #If X is smaller, find the quickest path
            difference = y_value - x_value

            if difference % 2 == 0:
                x_value = x_value + (difference//2)*2
                operation_count += difference//2
            else:
                #Adding 1 to x_value will make the difference even
                x_value += 1
                operation_count += 1

    return operation_count

Big(O) Analysis

Time Complexity

O(b*d) – The breadth-first search explores possible number transformations starting from X and Y. The branching factor 'b' represents the number of operations (add, subtract, multiply, divide) and valid number choices at each step. The depth 'd' signifies the number of operations needed to make X equal to Y, and is also constrained by the avoidance of visited numbers. The complexity is directly tied to how much the search space expands at each level (branching) and how many levels are explored (depth) before finding the shortest path, resulting in O(b*d).

Space Complexity

O(N) – The described solution uses a queue to keep track of numbers reachable from X and Y, and a mechanism to avoid revisiting numbers, such as a set or hash map. In the worst-case scenario, we might explore a large portion of the possible numbers reachable from X and Y before finding a common value. Thus, the size of the queue and the visited set could grow linearly with the range of possible reachable numbers which we denote as N, where N represents the number of possible reachable values, driving the auxiliary space complexity.

Edge Cases

Case	How to Handle
x and y are equal	Return 0 since no operations are needed.
x and y are negative	The BFS/DFS will still work with negative numbers, as increment/decrement handle these.
x is much larger than y	Decrementing is generally faster in this scenario, but the algorithm should still explore doubling for optimal results.
y is much larger than x and x is zero	Doubling zero doesn't help, so the algorithm will have to increment to reach y.
Integer overflow during multiplication (x * 2)	Limit search space based on integer bounds or use a larger data type like long if necessary and possible, or consider an alternative approach if an integer overflow is unavoidable within a bounded search space.
Potential for infinite loop if constraints are not tight (e.g., no upper bound on operations)	Impose a reasonable limit on the number of operations or the magnitude of the search space to prevent infinite loops.
x is zero and y is negative	Incrementing will lead to positive values, while decrementing goes further negative, so we would decrement and then increment.
x and y are very large positive numbers close to each other	The algorithm should efficiently explore both increment/decrement and doubling operations in a BFS or DFS approach.

x and y are equal

How to Handle:

Return 0 since no operations are needed.

x and y are negative

How to Handle:

The BFS/DFS will still work with negative numbers, as increment/decrement handle these.

x is much larger than y

How to Handle:

Decrementing is generally faster in this scenario, but the algorithm should still explore doubling for optimal results.

y is much larger than x and x is zero

How to Handle:

Doubling zero doesn't help, so the algorithm will have to increment to reach y.

Integer overflow during multiplication (x * 2)

How to Handle:

Limit search space based on integer bounds or use a larger data type like long if necessary and possible, or consider an alternative approach if an integer overflow is unavoidable within a bounded search space.

Potential for infinite loop if constraints are not tight (e.g., no upper bound on operations)

How to Handle:

Impose a reasonable limit on the number of operations or the magnitude of the search space to prevent infinite loops.

x is zero and y is negative

How to Handle:

Incrementing will lead to positive values, while decrementing goes further negative, so we would decrement and then increment.

x and y are very large positive numbers close to each other

How to Handle:

The algorithm should efficiently explore both increment/decrement and doubling operations in a BFS or DFS approach.