Introduction

In this blog post, we will explore how to solve the LeetCode problem "957" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There are 8 prison cells in a row and each cell is either occupied or vacant. Each day, whether the cell is occupied or vacant changes according to the following rules: If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied. Otherwise, it becomes vacant. Note that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors. You are given an integer array cells where cells[i] == 1 if the ith cell is occupied and cells[i] == 0 if the ith cell is vacant, and you are given an integer n. Return the state of the prison after n days (i.e., n such changes described above). Example 1: Input: cells = [0,1,0,1,1,0,0,1], n = 7 Output: [0,0,1,1,0,0,0,0] Explanation: The following table summarizes the state of the prison on each day: Day 0: [0, 1, 0, 1, 1, 0, 0, 1] Day 1: [0, 1, 1, 0, 0, 0, 0, 0] Day 2: [0, 0, 0, 0, 1, 1, 1, 0] Day 3: [0, 1, 1, 0, 0, 1, 0, 0] Day 4: [0, 0, 0, 0, 0, 1, 0, 0] Day 5: [0, 1, 1, 1, 0, 1, 0, 0] Day 6: [0, 0, 1, 0, 1, 1, 0, 0] Day 7: [0, 0, 1, 1, 0, 0, 0, 0] Example 2: Input: cells = [1,0,0,1,0,0,1,0], n = 1000000000 Output: [0,0,1,1,1,1,1,0] Constraints: cells.length == 8 cells[i] is either 0 or 1. 1 <= n <= 109

Explanation

Here's the breakdown of the solution:

• Detect Cycles: The state of the cells repeats after a certain number of days. We need to detect this cycle to efficiently compute the state after a very large number of days.
• Simulate and Store States: We simulate the cell state transitions and store the states we encounter. If a state repeats, we've found a cycle.

• Apply Modulo: Once a cycle is found, we can use the modulo operator to determine which state in the cycle corresponds to the target day n.

• Runtime Complexity: O(Number of unique states before cycle) which is at most O(2⁸) as the number of possible cell states is limited; thus, O(1) practically.

• Storage Complexity: O(Number of unique states before cycle), which is at most O(2⁸), thus O(1) practically.

Code

    def prisonAfterNDays(cells, n):
    seen = {}

    for i in range(n):
        cells_tuple = tuple(cells)  # Convert to tuple for hashability
        if cells_tuple in seen:
            loop_length = i - seen[cells_tuple]
            remaining_days = (n - i) % loop_length

            for _ in range(remaining_days):
                next_cells = [0] * 8
                for j in range(8):
                    if j == 0:
                        next_cells[j] = 0 if cells[1] == 0 else 1
                    elif j == 7:
                        next_cells[j] = 0 if cells[6] == 0 else 1
                    else:
                        next_cells[j] = 1 if cells[j-1] == cells[j+1] else 0
                cells = next_cells
            return cells

        else:
            seen[cells_tuple] = i
            next_cells = [0] * 8
            for j in range(8):
                if j == 0:
                    next_cells[j] = 0 if cells[1] == 0 else 1
                elif j == 7:
                    next_cells[j] = 0 if cells[6] == 0 else 1
                else:
                    next_cells[j] = 1 if cells[j-1] == cells[j+1] else 0
            cells = next_cells

    return cells