Leetcode #2059: Minimum Operations to Convert Number
In this guide, we solve Leetcode #2059 Minimum Operations to Convert Number in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Array
Intuition
We need level-by-level exploration or shortest steps, which is ideal for BFS.
A queue naturally models the frontier of the search.
Approach
Push initial nodes into a queue and expand in layers.
Track visited nodes to prevent cycles.
Steps:
- Initialize queue with start nodes.
- Process level by level.
- Track visited nodes.
Example
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
Python Solution
class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
op1 = lambda x, y: x + y
op2 = lambda x, y: x - y
op3 = lambda x, y: x ^ y
ops = [op1, op2, op3]
vis = [False] * 1001
q = deque([(start, 0)])
while q:
x, step = q.popleft()
for num in nums:
for op in ops:
nx = op(x, num)
if nx == goal:
return step + 1
if 0 <= nx <= 1000 and not vis[nx]:
q.append((nx, step + 1))
vis[nx] = True
return -1
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.