Question:
Minimum number of clicks required to navigate through the list of channels passed as a string.
Goldman sachs hackerrank questions
Example the channels range from ("1,100")
and we have to navigate from the begining till the end for the folowing list of channels provided as a string ("24,35,64,35") with few operations avaliable.
Left shift : to move channel from 14 to 13 will take one click
Right shift : to move channel from 15 to 17 will take two click
and other way is to type the number.That will take the number of click equal to the number of digit in the channel number.
Eg if the input is ("1,100","24,35,64,35")
then the output will be 7 clicks.
Solution is provided in java :
package testing;
public class channel {
public static void main(String[] args)
{
int count=min_count("1,100","24,35,64,35");
System.out.println(count);
}
public static int min_count(String range,String channel)
{
int flag=1;
String array[]=range.split(",");
String array2[]=channel.split(",");
int first_channel=Integer.parseInt(array[0]);
int last_channel=Integer.parseInt(array[1]);
int current_channel=first_channel;
int count_clicks=0;
int click2=10000;
int click3=10000;
for(int x=0;x<array2.length;x++)
{
int change_channel=Integer.parseInt(array2[x]);
if(change_channel>last_channel || change_channel<first_channel)
{
flag=0;
}
int click1=array2[x].length();
if(x>=2)
{
if(array2[x-2].equals(array2[x]))
{
click2=1;
}
}
if(current_channel<change_channel)
{
int temp_count=Math.abs(current_channel-change_channel);
int temp_count2=Math.abs(first_channel-current_channel);
int temp_count3=Math.abs(change_channel-last_channel);
int temp_count4=temp_count2+temp_count3;
click3=Math.min(temp_count, temp_count4);
}
if(current_channel>change_channel)
{
int temp_count=Math.abs(current_channel-change_channel);
int temp_count2=Math.abs(first_channel-change_channel);
int temp_count3=Math.abs(current_channel-last_channel);
int temp_count4=temp_count2+temp_count3;
click3=Math.min(temp_count, temp_count4);
}
int minimum_count=Math.min(click1, Math.min(click2,click3));
count_clicks=count_clicks+minimum_count;
current_channel=change_channel;
}
if(flag==1)
{
return count_clicks;
}
else
{
return 0;
}
}
}
Minimum number of clicks required to navigate through the list of channels passed as a string.
Goldman sachs hackerrank questions
Example the channels range from ("1,100")
and we have to navigate from the begining till the end for the folowing list of channels provided as a string ("24,35,64,35") with few operations avaliable.
Left shift : to move channel from 14 to 13 will take one click
Right shift : to move channel from 15 to 17 will take two click
and other way is to type the number.That will take the number of click equal to the number of digit in the channel number.
Eg if the input is ("1,100","24,35,64,35")
then the output will be 7 clicks.
Solution is provided in java :
package testing;
public class channel {
public static void main(String[] args)
{
int count=min_count("1,100","24,35,64,35");
System.out.println(count);
}
public static int min_count(String range,String channel)
{
int flag=1;
String array[]=range.split(",");
String array2[]=channel.split(",");
int first_channel=Integer.parseInt(array[0]);
int last_channel=Integer.parseInt(array[1]);
int current_channel=first_channel;
int count_clicks=0;
int click2=10000;
int click3=10000;
for(int x=0;x<array2.length;x++)
{
int change_channel=Integer.parseInt(array2[x]);
if(change_channel>last_channel || change_channel<first_channel)
{
flag=0;
}
int click1=array2[x].length();
if(x>=2)
{
if(array2[x-2].equals(array2[x]))
{
click2=1;
}
}
if(current_channel<change_channel)
{
int temp_count=Math.abs(current_channel-change_channel);
int temp_count2=Math.abs(first_channel-current_channel);
int temp_count3=Math.abs(change_channel-last_channel);
int temp_count4=temp_count2+temp_count3;
click3=Math.min(temp_count, temp_count4);
}
if(current_channel>change_channel)
{
int temp_count=Math.abs(current_channel-change_channel);
int temp_count2=Math.abs(first_channel-change_channel);
int temp_count3=Math.abs(current_channel-last_channel);
int temp_count4=temp_count2+temp_count3;
click3=Math.min(temp_count, temp_count4);
}
int minimum_count=Math.min(click1, Math.min(click2,click3));
count_clicks=count_clicks+minimum_count;
current_channel=change_channel;
}
if(flag==1)
{
return count_clicks;
}
else
{
return 0;
}
}
}
