Step by Step solution about How to resolve the algorithm 100 doors step by step in the Kotlin programming language

The provided Kotlin code simulates a classic puzzle involving 100 closed doors. The goal of the puzzle is to determine which doors are open after a series of steps.

Here's how the code works:

1. Initialization: First, an array named doors is created, containing 100 Boolean values, each representing a door. Initially, all doors are set to false, indicating that they are closed.

2. Simulation Loop: The repeat loop iterates from 0 to 99 (the indices of the doors array). Each iteration represents a step in the puzzle:

   • Inside the loop, another for loop iterates over the doors array with a step of (i + 1). This inner loop simulates a person passing through the doors and changing their states.

   • The inner loop starts at index i and goes up to the end of the array, skipping (i + 1) doors at each step. This ensures that each door is visited and its state is flipped (from false to true or vice versa).

3. Result: After the simulation, the doors array contains the final states of the doors. A foldIndexed operation is used to create a list of open door numbers. The foldIndexed function takes an initial empty list and an accumulator (acc) initialized to this empty list. It iterates over the doors array and checks if a door is open (i.e., door is true). If it is, the index of the open door (i + 1) is added to the accumulator. This process continues for all doors, resulting in a list containing the numbers of the open doors.

4. Return Value: The function returns the list of open door numbers.

In summary, the code simulates the classic 100 doors puzzle and returns a list containing the numbers of the open doors after the simulation.

fun oneHundredDoors(): List<Int> {
    val doors = BooleanArray(100) { false }

    repeat(doors.size) { i ->
        for (j in i until doors.size step (i + 1)) {
            doors[j] = !doors[j]
        }
    }

    return doors
        .foldIndexed(emptyList()) { i, acc, door ->
            if (door) acc + (i + 1) else acc
        }
}