Leetcode 957. Prison Cells After N Days

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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 = 7Output: [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 = 1000000000Output: [0,0,1,1,1,1,1,0]`

Constraints:

• `cells.length == 8`
• `cells[i]` is either `0` or `1`.
• `1 <= n <= 109`

[Gary- brute force- Java]

1. can’t pass all the test case — TLE
2. need to find pattern

[Second] — using hashmap to store all the possible string pattern

1. Trick part: n %= map.get(gen) — n

CS new grad, 6 years experience related to supply chain management. Located in Bay area

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CS new grad, 6 years experience related to supply chain management. Located in Bay area