# Count minimum right flips to set all values in an array

N light bulbs are connected by a wire. Each bulb has a switch associated with it, however due to faulty wiring, a switch also changes the state of all the bulbs to the right of current bulb. Given an initial state of all bulbs, find the minimum number of switches you have to press to turn on all the bulbs. You can press the same switch multiple times.
Note: 0 represents the bulb is off and 1 represents the bulb is on.
Examples:

Input :  [0 1 0 1]
Output : 4
Explanation :
press switch 0 : [1 0 1 0]
press switch 1 : [1 1 0 1]
press switch 2 : [1 1 1 0]
press switch 3 : [1 1 1 1]

Input : [1 0 0 0 0]
Output : 1

We traverse given array from left to right and keep pressing switch for off bulbs. We keep track of the number of switch presses so far. If the number of presses are odd, that means the current switch is in its original state else it is in the other state. Depending on what state the bulb is in, we can increment the count of the number of presses.

 // CPP program to find number switch presses to // turn all bulbs on. #include using namespace std;   int bulbs(int a[],int n) {     // To keep track of switch presses so far     int count = 0;       int res = 0;     for (int i = 0; i < n; i++)     {         /* if the bulb's original state is on and count         is even, it is currently on...*/         /* no need to press this switch */         if (a[i] == 1 && count % 2 == 0)             continue;           /* If the bulb's original state is off and count         is odd, it is currently on...*/         /* no need to press this switch */         else if(a[i] == 0 && count % 2 != 0)             continue;           /* if the bulb's original state is on and count            is odd, it is currently off...*/         /* Press this switch to on the bulb and increase         the count */         else if (a[i] == 1 && count % 2 != 0)         {             res++;             count++;         }                   /* if the bulb's original state is off and         count is even, it is currently off...*/         /* press this switch to on the bulb and         increase the count */         else if (a[i] == 0 && count % 2 == 0)         {             res++;             count++;         }     }     return res; }   // Driver code int main() {     int states[] = {0,1,0,1};     int n = sizeof(states)/sizeof(states[0]);     cout << "The minimum number of switches needed are " << bulbs(states,n); }   // This code is contributed by // Sanjit_Prasad

 // Java program to find number switch presses to // turn all bulbs on. import java.util.*;   public class GFG {     public int bulbs(ArrayList a)     {         // To keep track of switch presses so far         int count = 0;           int res = 0;         for (int i = 0; i < a.size(); i++)         {             /* if the bulb's original state is on and count                is even, it is currently on...*/             /* no need to press this switch */             if (a.get(i) == 1 && count%2 == 0)                 continue;               /* If the bulb's original state is off and count                is odd, it is currently on...*/             /* no need to press this switch */             else if(a.get(i) == 0 && count%2 != 0)                 continue;               /* if the bulb's original state is on and count                is odd, it is currently off...*/             /* Press this switch to on the bulb and increase                the count */             else if (a.get(i) == 1 && count%2 != 0)             {                 res++;                 count++;             }               /* if the bulb's original state is off and                count is even, it is currently off...*/             /* press this switch to on the bulb and                increase the count */             else if (a.get(i) == 0 && count%2 == 0)             {                 res++;                 count++;             }         }         return res;     }       // Driver code     public static void main(String[] args)     {         GFG gfg = new GFG();           ArrayList states = new ArrayList();           states.add(0);         states.add(1);         states.add(0);         states.add(1);           System.out.println("The minimum number of switches" +                            " needed are " + gfg.bulbs(states));     } }

 # Python program to find number switch presses to # turn all bulbs on.     def bulbs(a, n):     # To keep track of switch presses so far     count = 0       res = 0     for i in range(n):         # if the bulb's original state is on and count         # is even, it is currently on...         # no need to press this switch         if (a[i] == 1 and count % 2 == 0):             continue           # If the bulb's original state is off and count         # is odd, it is currently on...         # no need to press this switch         elif(a[i] == 0 and count % 2 != 0):             continue           # if the bulb's original state is on and count         # is odd, it is currently off...         # Press this switch to on the bulb and increase         # the count         elif (a[i] == 1 and count % 2 != 0):             res += 1             count += 1           # if the bulb's original state is off and         # count is even, it is currently off...         # press this switch to on the bulb and         # increase the count         elif (a[i] == 0 and count % 2 == 0):             res += 1             count += 1     return res     # Driver code states = [0, 1, 0, 1] n = len(states) print("The minimum number of switches needed are", bulbs(states, n))   # This code is contributed by ankush_953

 // C# program to find number switch presses // to turn all bulbs on. using System; using System.Collections.Generic;   class GFG { public virtual int bulbs(List a) {     // To keep track of switch presses so far     int count = 0;       int res = 0;     for (int i = 0; i < a.Count; i++)     {         /* if the bulb's original state is on         and count is even, it is currently on...*/         /* no need to press this switch */         if (a[i] == 1 && count % 2 == 0)         {             continue;         }           /* If the bulb's original state is off         and count is odd, it is currently on...*/         /* no need to press this switch */         else if (a[i] == 0 && count % 2 != 0)         {             continue;         }           /* if the bulb's original state is on         and count is odd, it is currently off...*/         /* Press this switch to on the bulb and         increase the count */         else if (a[i] == 1 && count % 2 != 0)         {             res++;             count++;         }           /* if the bulb's original state is off and         count is even, it is currently off...*/         /* press this switch to on the bulb and         increase the count */         else if (a[i] == 0 && count % 2 == 0)         {             res++;             count++;         }     }     return res; }   // Driver code public static void Main(string[] args) {     GFG gfg = new GFG();       List states = new List();       states.Add(0);     states.Add(1);     states.Add(0);     states.Add(1);       Console.WriteLine("The minimum number of switches" +                     " needed are " + gfg.bulbs(states)); } }   // This code is contributed by Shrikant13



Output:

The minimum number of switches needed are 4.

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https://www.interviewbit.com/problems/bulbs/