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Bulb Switcher II in C++
Suppose there is a room with n lights which are switched on initially and 4 buttons present on the wall. After performing exactly m unknown operations towards buttons, we need to return how many different kinds of status of the n lights could be. So consider n lights are labeled as number [1, 2, 3 ..., n], function of these 4 buttons are as follows −
- Flip all the lights.
- Flip lights with even numbers.
- Flip lights with odd numbers.
- Flip lights with (3k + 1) numbers, k = 0, 1, 2, ...
Now if n = 3 and m = 1, then there will be 4 operations, these are, [off, on, off], [on, off, on], [off, off, off], [off, on, on]
To solve this, we will follow these steps −
- if n is 0 or m is 0, then return 1
- if n is 1, then return 2
- if n is 2, then return 3, when m is 1, otherwise return 4
- if m is 1, then return 4
- if m is 2, then return 7, otherwise return 8.
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int flipLights(int n, int m) {
if (m == 0 || n == 0) return 1;
if (n == 1) return 2;
if (n == 2) return m == 1? 3:4;
if (m == 1) return 4;
return m == 2? 7:8;
}
};
main(){
Solution ob;
cout << (ob.flipLights(3, 1));
}Input
3 1
Output
4
Updated on: 04-May-2020
189 Views
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