LeetCode #957 — MEDIUM

Prison Cells After N Days

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

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 i^th cell is occupied and cells[i] == 0 if the i^th 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

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math · Bit Manipulation

Example 1

[0,1,0,1,1,0,0,1]
7

Example 2

[1,0,0,1,0,0,1,0]
1000000000

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #957: Prison Cells After N Days
class Solution {
  public int[] prisonAfterNDays(int[] cells, int n) {
    int[] firstDayCells = new int[cells.length];
    int[] nextDayCells = new int[cells.length];

    for (int day = 0; n-- > 0; cells = nextDayCells.clone(), ++day) {
      for (int i = 1; i + 1 < cells.length; ++i)
        nextDayCells[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
      if (day == 0)
        firstDayCells = nextDayCells.clone();
      else if (Arrays.equals(nextDayCells, firstDayCells))
        n %= day;
    }

    return cells;
  }
}

// Accepted solution for LeetCode #957: Prison Cells After N Days
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #957: Prison Cells After N Days
// class Solution {
//   public int[] prisonAfterNDays(int[] cells, int n) {
//     int[] firstDayCells = new int[cells.length];
//     int[] nextDayCells = new int[cells.length];
// 
//     for (int day = 0; n-- > 0; cells = nextDayCells.clone(), ++day) {
//       for (int i = 1; i + 1 < cells.length; ++i)
//         nextDayCells[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
//       if (day == 0)
//         firstDayCells = nextDayCells.clone();
//       else if (Arrays.equals(nextDayCells, firstDayCells))
//         n %= day;
//     }
// 
//     return cells;
//   }
// }

# Accepted solution for LeetCode #957: Prison Cells After N Days
class Solution:
  def prisonAfterNDays(self, cells: list[int], n: int) -> list[int]:
    nextDayCells = [0] * len(cells)
    day = 0

    while n > 0:
      n -= 1
      for i in range(1, len(cells) - 1):
        nextDayCells[i] = 1 if cells[i - 1] == cells[i + 1] else 0
      if day == 0:
        firstDayCells = nextDayCells.copy()
      elif nextDayCells == firstDayCells:
        n %= day
      cells = nextDayCells.copy()
      day += 1

    return cells

// Accepted solution for LeetCode #957: Prison Cells After N Days
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn prison_after_n_days(mut cells: Vec<i32>, mut n: i32) -> Vec<i32> {
        let mut hm: HashMap<Vec<i32>, i32> = HashMap::new();
        while n > 0 {
            hm.insert(cells.to_vec(), n);
            let mut next = vec![0; 8];
            for i in 1..7 {
                next[i] = 1 - (cells[i - 1] ^ cells[i + 1]);
            }
            cells = next;
            n -= 1;
            if let Some(m) = hm.get(&cells) {
                n %= m - n;
            }
        }
        cells
    }
}

#[test]
fn test() {
    let cells = vec![0, 1, 0, 1, 1, 0, 0, 1];
    let n = 7;
    let res = vec![0, 0, 1, 1, 0, 0, 0, 0];
    assert_eq!(Solution::prison_after_n_days(cells, n), res);
    let cells = vec![1, 0, 0, 1, 0, 0, 1, 0];
    let n = 1_000_000_000;
    let res = vec![0, 0, 1, 1, 1, 1, 1, 0];
    assert_eq!(Solution::prison_after_n_days(cells, n), res);
}

// Accepted solution for LeetCode #957: Prison Cells After N Days
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #957: Prison Cells After N Days
// class Solution {
//   public int[] prisonAfterNDays(int[] cells, int n) {
//     int[] firstDayCells = new int[cells.length];
//     int[] nextDayCells = new int[cells.length];
// 
//     for (int day = 0; n-- > 0; cells = nextDayCells.clone(), ++day) {
//       for (int i = 1; i + 1 < cells.length; ++i)
//         nextDayCells[i] = cells[i - 1] == cells[i + 1] ? 1 : 0;
//       if (day == 0)
//         firstDayCells = nextDayCells.clone();
//       else if (Arrays.equals(nextDayCells, firstDayCells))
//         n %= day;
//     }
// 
//     return cells;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.