Welcome! Are you ready to embark on a captivating journey into the world of slice manipulations? Today, we're going to explore a fascinating scenario involving a wonderful small town, its houses, and a fun balloon game. Without further ado, let's dive right in!

Picture a quaint, small town where every house is numbered sequentially from 1 to n. One day, a festive town event is held, and balloons are tied to each house. The festivities do not end there. At the conclusion of the event, a fun game is played: at each step of the game, each house sends half of its balloons to the neighboring house simultaneously (the neighbor to the right side, and for the last house, the neighbor is the first house). The game goes on until, at some step, there are no changes in the number of balloons compared to the previous step.

The task is to create a function func solution(balloons []int) int in Go, where balloons is a slice representing the number of balloons at each house. The function should simulate this game and return the number of steps in the game.

For example, if balloons := []int{4, 1, 2}, the output should be solution([]int{4, 1, 2}) = 3. After the first step, the slice becomes []int{3, 3, 1}. This is because the first house sends 2 balloons and gets 1, the second house sends nothing but gets 2, and the third house sends 1 but receives nothing. Note that when the number of balloons x is odd, the house sends (x - 1) / 2 balloons. After the second step, the slice becomes []int{2, 3, 2} and never changes after that. So, after the third step, the process finishes.

Firstly, it's essential to note that we're dealing with a cyclical event. In other words, when iterating over our balloons slice, we need to perceive the slice as circular, meaning the last element should pass balloons to the first element. This concept of cyclicity becomes crucial when we consider the last house passing balloons to the first.

Confident in our understanding of the problem, we proceed to set up a loop to iterate the rounds of balloon sharing. This loop should continue as long as the slice changes, and we will use a steps counter to keep track of how many iterations have occurred.

Go
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1func solution(balloons []int) int {
2    steps := 0
3    for {
4        steps++
5        newBalloons := make([]int, len(balloons))
6        copy(newBalloons, balloons)
7        // TODO: Share the balloons
8        if slicesEqual(newBalloons, balloons) {
9            break // break if there are no changes in the balloons slice
10        }
11        balloons = newBalloons
12    }
13    return steps
14}

In this code snippet, we initialize a steps variable to count iterations. We then create a new slice newBalloons to store the updated state after each step via copy(newBalloons, balloons). The loop continues until there are no changes in the balloons slice.

To determine when the game finishes, we need a utility function slicesEqual that checks if two slices are identical. This helps us identify when the simulation reaches a steady state.

Go
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1func slicesEqual(a, b []int) bool {
2    if len(a) != len(b) {
3        return false
4    }
5    for i := range a {
6        if a[i] != b[i] {
7            return false
8        }
9    }
10    return true
11}

This utility function first compares the length of both slices, and if they are equal, it iterates over the elements to check for equality. If all elements match, it returns true, otherwise false.

Our next step delves into the core game mechanics: sharing the balloons. Throughout each cycle, each house must share half of its balloons with the next house.

We must also ensure that the last house shares balloons with the first house at the end of each cycle — for this, we can use modular arithmetic.

Go
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1func solution(balloons []int) int {
2    steps := 0
3    n := len(balloons)
4    for {
5        steps++
6        newBalloons := make([]int, n)
7        copy(newBalloons, balloons)
8        for i := 0; i < n; i++ {
9            share := balloons[i] / 2 // Balloons to share
10            newBalloons[i] -= share  // Decrease balloons of current house
11            newBalloons[(i+1)%n] += share // Increase balloons of next house
12        }
13        if slicesEqual(newBalloons, balloons) {
14            break
15        }
16        balloons = newBalloons
17    }
18    return steps
19}
20
21func main() {
22    balloons := []int{4, 1, 2}
23    fmt.Println(solution(balloons))
24}

In this updated solution, a loop iterates over each house to calculate the number of balloons each house shares with its neighbor. The use of (i+1)%n ensures cyclic sharing, connecting the last house back to the first. This cycle continues until slicesEqual returns true, indicating no further changes.

Bravo! We've navigated through the maze of slice manipulation and successfully simulated an intriguing game event.

Congratulations on mastering this crucial programming scenario! You've successfully navigated a task involving the simulation of real-world events using slice manipulation.

What's next? Now is the time to put into practice everything we've learned today. Try designing different versions of this balloon-sharing game. As always, happy coding!