How do you solve Minimum Operations to Convert Number on LeetCode?

Minimum Operations to Convert Number (LeetCode #2059) 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: Using BFS ensures we explore the shortest path to the goal efficiently.

What is the time complexity of Minimum Operations to Convert Number?

The optimal solution for Minimum Operations to Convert Number runs in O(n) time and uses O(n) space. The complexity is linear because we are processing each number in nums and each state only once due to the visited set.

What is the brute force solution for Minimum Operations to Convert Number?

The brute force approach for Minimum Operations to Convert Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible values of x by applying every operation on it repeatedly. This method is straightforward but inefficient as it can lead to many redundant calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Convert Number?

Minimum Operations to Convert Number uses the following concepts: Array, Breadth-First Search. The recommended patterns to study are: Breadth-First Search, Graph Traversal.

What are the interview tips for Minimum Operations to Convert Number?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as when the start is very close to the goal. Explain your thought process and approach as you code; it shows your understanding.

What are common mistakes when solving Minimum Operations to Convert Number?

Not using a visited set, leading to infinite loops. Forgetting to check if the current value equals the goal before applying operations.

#2059

Minimum Operations to Convert Number

Medium

ArrayBreadth-First SearchBreadth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a breadth-first search (BFS) approach allows us to explore the shortest path to the goal efficiently. We can track visited states to avoid redundant calculations and ensure we find the minimum operations needed.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue and add the starting value x (start) along with the operation count (0).
2Step 2: While the queue is not empty, dequeue an element and check if it equals the goal.
3Step 3: If it does, return the operation count. If not, apply all operations (+, -, ^) for each number in nums and enqueue the new values with incremented operation count.
4Step 4: Use a set to track visited values to prevent cycles and redundant processing.

solution.py15 lines

1from collections import deque
2
3def minOperations(nums, start, goal):
4    queue = deque([(start, 0)])
5    visited = set([start])
6    while queue:
7        x, ops = queue.popleft()
8        if x == goal:
9            return ops
10        for num in nums:
11            for next_x in (x + num, x - num, x ^ num):
12                if next_x not in visited:
13                    visited.add(next_x)
14                    queue.append((next_x, ops + 1))
15    return -1

ℹ

Complexity note: The complexity is linear because we are processing each number in nums and each state only once due to the visited set.

1Using BFS ensures we explore the shortest path to the goal efficiently.
2Tracking visited states prevents cycles and redundant processing.

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