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
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++ 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 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. |
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# 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 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. |
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:
// 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 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));
}
} |
# 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# 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 |
<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> |
The minimum number of switches needed are 4
Time Complexity: O(n)
Auxiliary Space: O(1)