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Count minimum right flips to set all values in an array

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  • Difficulty Level : Medium
  • Last Updated : 01 Nov, 2022
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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: A[] = [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: A[] = [1 0 0 0 0] 
Output: 1

Recommended Practice

Naive Approach:

Keep a count variable and Iterate from 0 to N-1 and check for every i if A[i] is 0 increment count by 1 and switch all values from i to N-1 as 0 to 1 or 1 to 0  

Below is the Implementation of above Idea:

C++




// C++ code for the Idea
// turn all bulbs on.
 
#include <bits/stdc++.h>
using namespace std;
 
int bulbs(int A[], int N)
{
 
    // To keep track of switch presses so far
    int count = 0;
 
    for (int i = 0; i < N; i++) {
        if (A[i] == 0) {
            A[i] = 1;
            for (int j = i + 1; j < N; j++) {
                if (A[j] == 1)
                    A[j] = 0;
                else
                    A[j] = 1;
            }
            count++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
 
    int states[] = { 0, 1, 0, 1 };
    int N = sizeof(states) / sizeof(states[0]);
 
    // Function Code
    cout << "The minimum number of switches needed are "
         << bulbs(states, N);
}

Java




// Java code for the Idea turn all bulbs on.
 
import java.io.*;
 
class GFG {
 
    static int bulbs(int[] A, int N)
    {
        // To keep track of switch presses so far
        int count = 0;
        for (int i = 0; i < N; i++) {
            if (A[i] == 0) {
                A[i] = 1;
                for (int j = i + 1; j < N; j++) {
                    if (A[j] == 1) {
                        A[j] = 0;
                    }
                    else {
                        A[j] = 1;
                    }
                }
                count++;
            }
        }
        return count;
    }
 
    public static void main(String[] args)
    {
        int[] states = { 0, 1, 0, 1 };
        int N = states.length;
 
        // Function Code
        System.out.print(
            "The minimum number of switches needed are "
            + bulbs(states, N));
    }
}
 
// This code is contributed by lokeshmvs21.

Python3




class GFG :
    @staticmethod
    def  bulbs( A,  N) :
       
        # To keep track of switch presses so far
        count = 0
        i = 0
        while (i < N) :
            if (A[i] == 0) :
                A[i] = 1
                j = i + 1
                while (j < N) :
                    if (A[j] == 1) :
                        A[j] = 0
                    else :
                        A[j] = 1
                    j += 1
                count += 1
            i += 1
        return count
    @staticmethod
    def main( args) :
        states = [0, 1, 0, 1]
        N = len(states)
         
        # Function Code
        print("The minimum number of switches needed are " + str(GFG.bulbs(states, N)), end ="")
     
if __name__=="__main__":
    GFG.main([])
     
    # This code is contributed by aaityaburujwale.

C#




// C# code for the Idea turn all bulbs on.
using System;
public class GFG {
 
  static int bulbs(int[] A, int N)
  {
 
    // To keep track of switch presses so far
    int count = 0;
    for (int i = 0; i < N; i++) {
      if (A[i] == 0) {
        A[i] = 1;
        for (int j = i + 1; j < N; j++) {
          if (A[j] == 1) {
            A[j] = 0;
          }
          else {
            A[j] = 1;
          }
        }
        count++;
      }
    }
    return count;
  }
 
  static public void Main()
  {
 
    // Code
    int[] states = { 0, 1, 0, 1 };
    int N = states.Length;
 
    // Function Code
    Console.Write(
      "The minimum number of switches needed are "
      + bulbs(states, N));
  }
}
 
// This code is contributed by lokeshmvs21.

Javascript




// Javascript code for the Idea
// turn all bulbs on.
function bulbs(A, N)
{
 
    // To keep track of switch presses so far
    let count = 0;
 
    for (let i = 0; i < N; i++) {
        if (A[i] == 0) {
            A[i] = 1;
            for (let j = i + 1; j < N; j++) {
                if (A[j] == 1)
                    A[j] = 0;
                else
                    A[j] = 1;
            }
            count++;
        }
    }
    return count;
}
 
// Driver code
let states = [ 0, 1, 0, 1 ];
let N = 4;
 
// Function Code
document.write("The minimum number of switches needed are "
               + bulbs(states, N));
                
// This code is contributed by garg28harsh.

Output

The minimum number of switches needed are 4

Time Complexity: O(N2).
Auxiliary Space: O(1).

Efficient Approach:

Traverse given array from left to right and keep pressing switch for off bulbs. Keep track of the number of switch presses so far. If the number of presses are even, 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. 

Below is the Implementation Of above Idea:

C++




// CPP program to find number switch presses to
// turn all bulbs on.
 
#include<bits/stdc++.h>
using namespace std;
 
int bulbs(int a[],int n)
{
    // To keep track of switch presses so far
    int count = 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)
        {
            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)
        {
            count++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int states[] = {0,1,0,1};
    int n = sizeof(states)/sizeof(states[0]);
   
      // Function Code
    cout << "The minimum number of switches needed are " << bulbs(states,n);
}

Java




// Java program to find number switch presses to
// turn all bulbs on.
import java.util.*;
 
public class GFG
{
    public int bulbs(ArrayList<Integer> a)
    {
        // To keep track of switch presses so far
        int count = 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)
            {
                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)
            {
                count++;
            }
        }
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        GFG gfg = new GFG();
 
        ArrayList<Integer> states = new ArrayList<Integer>();
 
        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));
    }
}

Python3




# 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
    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):
            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):
            count += 1
    return count
 
 
# 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#




// 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<int> a)
{
    // To keep track of switch presses so far
    int count = 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)
        {
            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)
        {
            count++;
        }
    }
    return count;
}
 
// Driver code
public static void Main(string[] args)
{
    GFG gfg = new GFG();
 
    List<int> states = new List<int>();
 
    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

Javascript




<script>
// javascript program to find number switch presses to
// turn all bulbs on.
function bulbs(a, n)
{
 
    // To keep track of switch presses so far
    let count = 0;
 
    for (let 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)
        {
            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)
        {
            count++;
        }
    }
    return count;
}
 
// Driver code
 
    let states = [0,1,0,1];
    let n = states.length;
    document.write("The minimum number of switches needed are " + bulbs(states,n));
     
    // This code is contributed by vaibhavrabadiya117.
</script>

Output

The minimum number of switches needed are 4

Time Complexity: O(n)
Auxiliary Space: O(1)

This article is contributed by Saloni Baweja. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 

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