Is Minimum Number of Operations to Make X and Y Equal easy or hard?

Minimum Number of Operations to Make X and Y Equal is rated Medium difficulty on LeetCode. The main challenge is recognizing the problem as a shortest-path search on integers and efficiently exploring valid operations without revisiting states.

Minimum Number of Operations to Make X and Y Equal Python or Java solution

Python and Java implementations typically use a queue for BFS and a set to track visited numbers. Each step generates neighbors using +1, -1, divide by 5, and divide by 11 when divisible. The algorithm stops once Y is reached, returning the number of BFS levels as the answer.

What is the best approach for Minimum Number of Operations to Make X and Y Equal?

Breadth-First Search (BFS) is the most reliable approach because the problem can be modeled as a shortest-path search where each integer value is a node and each operation is an edge. BFS explores states level by level and guarantees the first time you reach Y uses the minimum number of operations. Time complexity is O(N) with O(N) space for visited states.

What data structure is used in Minimum Number of Operations to Make X and Y Equal?

The primary data structure is a queue used for Breadth-First Search traversal. A hash set or boolean visited array tracks numbers that were already processed to prevent cycles. These structures ensure each state is explored once while maintaining optimal BFS traversal order.

What is the time complexity of Minimum Number of Operations to Make X and Y Equal?

The BFS solution runs in O(N) time where N is the range of numbers explored between X and Y. Each state is visited at most once and generates up to four transitions. A greedy division-based strategy can reduce the complexity to about O(log max(x, y)) in practice because division operations shrink the value quickly.

Is Minimum Number of Operations to Make X and Y Equal asked at Google or Amazon interviews?

Problems involving transforming numbers using minimum operations frequently appear in interviews at companies like Google, Amazon, and Meta. They test whether candidates recognize the hidden graph structure and apply BFS or dynamic programming techniques to compute the shortest transformation sequence.

Description Hints 5 Solution Video 16:09

2998. Minimum Number of Operations to Make X and Y Equal

Q: How to solve Minimum Number of Operations to Make X and Y Equal efficiently?

Use BFS to treat each number as a graph node and apply transitions for +1, -1, divide by 5, and divide by 11 when valid. Push the starting value into a queue and explore neighbors while tracking visited numbers. The first time you reach Y gives the minimum operations because BFS processes nodes by increasing distance.

Medium48.4% acceptance2 companies

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

Asked at 2 companies

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Topics

Hints open one at a time

Hint 1

The only way to make x larger is to increase it by 1 so if y >= x the answer is y - x.

Hint 2

For y < x, x - y is always a candidate answer since we can repeatedly decrease x by one to reach y.

Hint 3

We can also increase x and then use the division operations. For example, if x = 10 and y = 1, we can increment x by 1 then divide it by 11.

Hint 4

Find an upper bound U on the maximum value of x we will reach an optimal solution. Since all values of x will be in the range [1, U], we can use BFS to find the answer.

Hint 5

One possible upper bound on x is U = x + (x - y) . To reach any number strictly greater than U from x, we will need more than x - y operations which is not optimal since we can always reach y in x - y operations.

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