How do you solve Minimum Number of Operations to Make X and Y Equal on LeetCode?

Minimum Number of Operations to Make X and Y Equal (LeetCode #2998) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: When y >= x, the quickest way to reach y is to increment x directly.

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

The optimal solution for Minimum Number of Operations to Make X and Y Equal runs in O(n) time and uses O(n) space. The time complexity is O(n) because we are exploring each state at most once, and the space complexity is O(n) due to the queue and visited set.

What is the brute force solution for Minimum Number of Operations to Make X and Y Equal?

The brute force approach for Minimum Number of Operations to Make X and Y Equal is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves exploring all possible operations to transform x into y. This means trying every operation and tracking the number of steps taken until we reach y. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Number of Operations to Make X and Y Equal?

Always consider edge cases, such as when y is greater than or equal to x. Explain your thought process clearly while coding; it helps interviewers understand your approach. Don't forget to optimize your solution after the brute-force approach; interviewers appreciate candidates who can improve their solutions.

What are common mistakes when solving Minimum Number of Operations to Make X and Y Equal?

Not considering the case where y >= x and returning y - x directly. Failing to track visited states can lead to infinite loops or excessive computations.

#2998

Minimum Number of Operations to Make X and Y Equal

Medium

Dynamic ProgrammingBreadth-First SearchMemoizationBreadth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a breadth-first search (BFS) but with a more targeted approach. We only explore valid operations and track the minimum steps needed to reach y from x efficiently.

⚙️

Algorithm

3 steps

1Step 1: If y >= x, return y - x directly since we can only increment x.
2Step 2: Initialize a queue for BFS starting from x and a set to track visited nodes.
3Step 3: For each number dequeued, apply valid operations and enqueue new states until y is reached.

solution.py27 lines

1from collections import deque
2
3def min_operations(x, y):
4    if y >= x:
5        return y - x
6    queue = deque([(x, 0)])  # (current value, steps)
7    visited = set([x])
8    while queue:
9        current, steps = queue.popleft()
10        if current == y:
11            return steps
12        # Possible operations
13        for next_val in [current - 1, current + 1]:
14            if next_val not in visited:
15                visited.add(next_val)
16                queue.append((next_val, steps + 1))
17        if current % 5 == 0:
18            next_val = current // 5
19            if next_val not in visited:
20                visited.add(next_val)
21                queue.append((next_val, steps + 1))
22        if current % 11 == 0:
23            next_val = current // 11
24            if next_val not in visited:
25                visited.add(next_val)
26                queue.append((next_val, steps + 1))
27    return -1

ℹ

Complexity note: The time complexity is O(n) because we are exploring each state at most once, and the space complexity is O(n) due to the queue and visited set.

1When y >= x, the quickest way to reach y is to increment x directly.
2The operations of dividing by 5 or 11 can significantly reduce x, making them powerful tools for large reductions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.