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Prison Cells After N Days in C++
Suppose there are 8 prison cells in a row, and in each cell there is a prisoner or that is empty. In each day, whether the cell is occupied or vacant changes according to the following rules −
If one cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
Otherwise, it becomes empty.
We will describe the current state of the prison in the following way: cells[i] will be 1 if the i-th cell is occupied, else cells[i] will be 0.
So we have the initial state of the prison, then return the state of the prison after N days.
So if the input is like: [0,1,0,1,1,0,0,1], and N = 7, then the output will be [0,0,1,1,0,0,0,0]. So this is because of the following. For seven days −
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]
To solve this, we will follow these steps −
Create a map m, and create one set called visited.
if N is 0, then return cells
insert cells into visited set
for i in range 1 to 14
create an array called temp of size 8
for j in range 1 to 6
if cells[j – 1] XOR cells[j + 1] = 0, then temp[j] := 1
cells := temp
m[i] := temp
insert temp into visited
if N is divisible by 14, then return m[14], otherwise m[N mod 14]
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<int> prisonAfterNDays(vector<int>& cells, int N) {
map <int, vector <int> > m;
if(N == 0) return cells;
set <vector <int> > visited;
visited.insert(cells);
for(int i = 1; i<=14 ; i++ ){
vector <int> temp(8);
for(int j = 1; j < 7; j++){
if(cells[j - 1] ^ cells[j + 1] == 0){
temp[j] = 1;
}
}
cells = temp;
m[i] = temp;
visited.insert(temp);
}
return m[N % 14 == 0? 14 : N % 14];
}
};
main(){
vector<int> v1 = {0,1,0,1,1,0,0,1};
Solution ob;
print_vector(ob.prisonAfterNDays(v1, 7));
}Input
[0,1,0,1,1,0,0,1] 7
Output
[0,0,1,1,0,0,0,0]
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