LeetCode 2059 - Minimum Operations to Convert Number

LeetCode Problem 2059

Difficulty: 🟡 Medium
Topics: Array, Breadth-First Search

Solution

Problem Understanding

This problem asks us to transform a starting integer into a target integer using the minimum number of operations. At every step, we may choose any number from the nums array and apply one of three operations to the current value:

• Addition
• Subtraction
• Bitwise XOR

The current value is stored in a variable x, which initially equals start.

There is one very important restriction. We are only allowed to continue performing operations while the current value remains in the range [0, 1000]. However, an operation is still considered valid even if it moves the value outside this range. The difference is that once the value goes outside [0, 1000], the process immediately stops and no further operations may be performed.

The goal is to determine the minimum number of operations required to reach goal. If it is impossible, we return -1.

The input consists of:

• nums, an array of distinct integers
• start, the initial value
• goal, the desired target value

The output is a single integer representing the smallest number of operations needed.

The constraints reveal an important detail. Even though goal and values in nums may be extremely large, the only values from which we can continue exploring are integers between 0 and 1000. That means the searchable state space is actually very small, only 1001 possible reusable states.

This strongly suggests modeling the problem as a graph traversal problem.

Several edge cases are important:

• The target may lie outside [0, 1000]
• Intermediate states may go outside the valid range
• XOR operations can dramatically change values
• Some states may repeat endlessly if we do not track visited values
• The shortest sequence of operations is required, not just any sequence

Because the problem asks for the minimum number of operations in an unweighted graph of states, Breadth-First Search is the natural solution.

Approaches

Brute Force Approach

A brute-force solution would recursively try every possible operation at every step.

From a value x, we could generate:

• x + nums[i]
• x - nums[i]
• x ^ nums[i]

for every index i.

This explores all possible operation sequences until either:

• We reach goal
• We exceed some arbitrary recursion depth

This approach is correct in theory because it eventually explores every possible path. However, it is computationally infeasible because the number of possible sequences grows exponentially.

Another major problem is repeated states. For example, we may repeatedly move between two numbers:

5 -> 7 -> 5 -> 7

Without memoization or visited tracking, the recursion may never terminate efficiently.

Key Insight

The important observation is that only values in the range [0, 1000] can generate further states.

This means the entire searchable graph contains at most 1001 reusable nodes.

Each valid number represents a graph node. From every node, we can transition to new numbers using the three allowed operations.

Since every operation has equal cost, one step, Breadth-First Search guarantees that the first time we reach the goal is the minimum number of operations required.

BFS is ideal because:

• It explores states level by level
• The first visit to a state uses the shortest path
• The searchable state space is very small

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Explores all operation sequences without efficient pruning
Optimal BFS	O(1000 × n)	O(1000)	Explores each valid state at most once

Here, n is len(nums).

Algorithm Walkthrough

1. Create a queue for Breadth-First Search.

BFS processes states level by level, which naturally corresponds to the number of operations performed. 2. Create a visited set.

Since states can repeat, we must avoid revisiting the same value multiple times. Otherwise, cycles could cause infinite processing. 3. Insert the starting value into the queue with distance 0.

The queue stores pairs:

(current_value, operations_used)

4. While the queue is not empty, remove the front state.

This gives us the next state reachable with the smallest number of operations. 5. For every number in nums, generate the three possible next states:

• current + num
• current - num
• current ^ num

6. If any generated value equals goal, immediately return:

operations + 1

BFS guarantees this is the shortest possible path. 7. Otherwise, if the generated value lies within [0, 1000] and has not been visited yet:

• Mark it visited
• Add it to the queue

Only these values may continue generating future states. 8. If BFS finishes without finding goal, return -1.

Why it works

Breadth-First Search explores all states reachable in k operations before exploring states requiring k + 1 operations. Therefore, the first time we encounter goal, we have found the minimum number of operations.

The visited set guarantees that every reusable state is processed at most once, preventing cycles and redundant work.

Python Solution

from collections import deque
from typing import List

class Solution:
    def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
        queue = deque([(start, 0)])
        visited = {start}

        while queue:
            current, steps = queue.popleft()

            for num in nums:
                next_values = [
                    current + num,
                    current - num,
                    current ^ num
                ]

                for next_value in next_values:
                    if next_value == goal:
                        return steps + 1

                    if 0 <= next_value <= 1000 and next_value not in visited:
                        visited.add(next_value)
                        queue.append((next_value, steps + 1))

        return -1

The implementation begins by initializing a BFS queue with the starting value and zero operations used.

A visited set is used to prevent processing the same state multiple times. Since only values within [0, 1000] may continue generating new states, only those values are stored and explored.

Inside the BFS loop, we remove the current state from the queue and generate all possible next values using addition, subtraction, and XOR operations.

If any generated value equals goal, we immediately return the current distance plus one. This works because BFS guarantees the shortest path.

If a generated value remains within the valid exploration range and has not been visited before, it is added to the queue for future processing.

If the queue becomes empty without reaching the target, no valid sequence exists, so we return -1.

Go Solution

package main

func minimumOperations(nums []int, start int, goal int) int {
	type State struct {
		value int
		steps int
	}

	queue := []State{{start, 0}}
	visited := make(map[int]bool)
	visited[start] = true

	for len(queue) > 0 {
		current := queue[0]
		queue = queue[1:]

		for _, num := range nums {
			nextValues := []int{
				current.value + num,
				current.value - num,
				current.value ^ num,
			}

			for _, nextValue := range nextValues {
				if nextValue == goal {
					return current.steps + 1
				}

				if nextValue >= 0 && nextValue <= 1000 && !visited[nextValue] {
					visited[nextValue] = true
					queue = append(queue, State{
						value: nextValue,
						steps: current.steps + 1,
					})
				}
			}
		}
	}

	return -1
}

The Go implementation follows the same BFS strategy as the Python version.

Since Go does not provide a built-in deque, the queue is implemented using a slice. We remove elements from the front by slicing:

queue = queue[1:]

The visited structure is implemented using a map[int]bool.

Go integers are sufficient for this problem because the constraints safely fit within standard integer ranges.

Worked Examples

Example 1

nums = [2,4,12]
start = 2
goal = 12

Initial queue:

Queue	Visited
[(2,0)]	{2}

Process (2,0):

Possible next values:

Operation	Result
2 + 2	4
2 - 2	0
2 ^ 2	0
2 + 4	6
2 - 4	-2
2 ^ 4	6
2 + 12	14
2 - 12	-10
2 ^ 12	14

Valid unexplored states added:

0, 4, 6, 14

Queue becomes:

Queue
[(4,1), (0,1), (6,1), (14,1)]

Process (14,1):

Operation	Result
14 - 2	12

Goal found.

Answer:

2

Example 2

nums = [3,5,7]
start = 0
goal = -4

Initial state:

Queue	Visited
[(0,0)]	{0}

Process (0,0):

Generated states:

3, -3, 5, -5, 7, -7

Only positive in-range values are queued:

3, 5, 7

Process (3,1):

Operation	Result
3 - 7	-4

Goal reached.

Answer:

2

Example 3

nums = [2,8,16]
start = 0
goal = 1

BFS explores all reachable states within [0,1000].

No generated operation ever produces 1.

Eventually the queue becomes empty.

Answer:

-1

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1000 × n)	Each valid state is processed once, and each processing tries all numbers
Space	O(1000)	Queue and visited set store at most 1001 states

The algorithm is efficient because the searchable state space is bounded. Even though arithmetic operations may temporarily produce huge numbers, only values inside [0,1000] are expanded further.

Since each of the 1001 possible reusable states is processed at most once, the algorithm runs very quickly.

Test Cases

from typing import List

class Solution:
    def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
        from collections import deque

        queue = deque([(start, 0)])
        visited = {start}

        while queue:
            current, steps = queue.popleft()

            for num in nums:
                next_values = [
                    current + num,
                    current - num,
                    current ^ num
                ]

                for next_value in next_values:
                    if next_value == goal:
                        return steps + 1

                    if 0 <= next_value <= 1000 and next_value not in visited:
                        visited.add(next_value)
                        queue.append((next_value, steps + 1))

        return -1

sol = Solution()

assert sol.minimumOperations([2,4,12], 2, 12) == 2  # provided example 1

assert sol.minimumOperations([3,5,7], 0, -4) == 2  # provided example 2

assert sol.minimumOperations([2,8,16], 0, 1) == -1  # provided example 3

assert sol.minimumOperations([1], 0, 1) == 1  # single operation addition

assert sol.minimumOperations([1], 1, 0) == 1  # single operation subtraction

assert sol.minimumOperations([2], 0, 2) == 1  # direct reach

assert sol.minimumOperations([5], 0, -5) == 1  # target outside valid range

assert sol.minimumOperations([1,3], 0, 7) == 3  # multiple BFS layers

assert sol.minimumOperations([4], 0, 3) == -1  # unreachable target

assert sol.minimumOperations([2,4,6], 999, 1001) == 1  # valid final out-of-range move

assert sol.minimumOperations([7], 0, 0) == -1  # start != goal guaranteed, unreachable

Test	Why
[2,4,12], 2 -> 12	Validates standard BFS traversal
[3,5,7], 0 -> -4	Confirms out-of-range target handling
[2,8,16], 0 -> 1	Confirms impossible case detection
[1], 0 -> 1	Tests direct addition
[1], 1 -> 0	Tests direct subtraction
[2], 0 -> 2	Tests immediate success
[5], 0 -> -5	Tests single-step out-of-range target
[1,3], 0 -> 7	Tests deeper BFS traversal
[4], 0 -> 3	Tests unreachable odd target
[2,4,6], 999 -> 1001	Tests finishing with out-of-range value
[7], 0 -> 0	Confirms no accidental zero-step handling

Edge Cases

One important edge case occurs when the target lies outside the valid exploration range [0,1000]. A naive implementation might incorrectly reject such states immediately. However, the problem explicitly allows the final operation to move outside the range. The BFS correctly checks whether a generated value equals goal before enforcing the range restriction for future exploration.

Another tricky case involves cycles. For example, operations may repeatedly alternate between the same few values. Without a visited set, the algorithm could process the same state indefinitely. The implementation prevents this by marking every reusable state as visited the first time it is encountered.

A third important edge case occurs when no solution exists. Some targets are fundamentally unreachable because the allowed operations cannot generate them. The BFS naturally handles this situation by exploring all reachable states. If the queue eventually becomes empty, the algorithm correctly returns -1.

A fourth subtle case involves XOR operations. XOR can produce unintuitive transitions that differ greatly from arithmetic operations. A greedy approach based only on numeric distance would fail badly here. BFS avoids this problem entirely because it systematically explores every reachable state in order of operation count.