957.prison-cells-after-n-days.cpp

/*
 * @lc app=leetcode id=957 lang=cpp
 *
 * [957] Prison Cells After N Days
 *
 * https://leetcode.com/problems/prison-cells-after-n-days/description/
 *
 * algorithms
 * Medium (37.42%)
 * Total Accepted:    20.2K
 * Total Submissions: 53.7K
 * Testcase Example:  '[0,1,0,1,1,0,0,1]\n7'
 *
 * 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.)
 * 
 * We describe the current state of the prison in the following way: cells[i]
 * == 1 if the i-th cell is occupied, else cells[i] == 0.
 * 
 * Given the initial state of the prison, return the state of the prison after
 * N days (and 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]
 * 
 * 
 * 
 * 
 * Note:
 * 
 * 
 * cells.length == 8
 * cells[i] is in {0, 1}
 * 1 <= N <= 10^9
 * 
 * 
 * 
 * 
 */
class Solution {
public:
    void step(vector<int>& x){
        vector<int> next(8, 0);
        for(int j=1; j<7; j++)
            next[j] = !(x[j-1]^x[j+1]);
        x = next;
    }
    vector<int> prisonAfterNDays(vector<int>& x, int N) {
        
        // convert 1st state as it is different from other states 
        // by ability to have 1 at first and last position
        step(x);
        N--;

        vector<int> nextt = x;

        // find after how many days the states repeat
        int i;
        for(i=1; i<=(1<<8); i++){
            step(x);
            if(x == nextt)
                break;
        }

        N %= i; // after %, N <= 256
        for(i=0; i<N; i++)
            step(x);
        return x;

    }
};