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Hi Partha, dont make low of your self. Whenever you dont get it take a small example. I will try to help with one.
consider 2,x; The main thing you want to do is make a singleton( or the minimum posssible number of) choice per a candidate.
a)If differnce between x and 2 is 5:
-adding 0 makes it divisble by 5. (you only made one choice on x).
-adding 1 or 2 makes it you to have more number of choices.
b)If differnce between x and 2 is 4:
-adding 1 makes it divisble by 5. (you only made one choice on x).
-adding 0 or 2 makes it you to have more number of choices.
c)If differnce between x and 2 is 3:
-adding 2 makes it divisble by 5. (you only made one choice on x).
-adding 0 or 1 makes it you to have more number of choices.
d)If differnce between x and 2 is 2:
-adding 0 makes it divisble by 2. (you only made one choice on x).
-adding 1 or 2 makes it you to have more number of choices.
e)If differnce between x and 2 is 1:
-adding 0 makes it divisble by 1. (you only made one choice on x).
-adding 1 or 2 makes it you to have more number of choices.
f)If differnce between x and 2 is 6:
-adding 1 makes it divisble by 5. (you only made two choices on x, first divide by 5 then reminder by divided 2). //same number of choices in this case even by adding 0.
-adding 2 makes it you to have more number of choices.
So for every candidate you add some base (0 or 1 0r 2) and consider the only the base that gives you minimum choices.
But why I need add some base?
A basic divisibility exaple. 9 is not perfectly divisible 5 but adding 1 to 9 makes it 10 and now perfectly divisble by 5.
But why base only till 0,1,2 and not more than 2?
if the diffence between x and y is:
a)4 adding 1 is div by 5
b)3 adding 2 is div by 5
c)2 it is div by 2
d)7 is div by 5 followed by 2
e)8 adding 2 is div by 5 followed by 2
f)9 adding 1 is div by 5 followed by 2
Did you observe the base is cycling around 0,1,2.so you never need not go beyond 2
This is a perfect explaination of adding base and finding the best case the logic except base addition would get you clear some test cases by the reducing approach but you would definitely need to add base so that you get the best apporach out of the three possible solutions. Good approach rohit! Thank You
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Hi Partha, dont make low of your self. Whenever you dont get it take a small example. I will try to help with one.
consider 2,x; The main thing you want to do is make a singleton( or the minimum posssible number of) choice per a candidate. a)If differnce between x and 2 is 5: -adding 0 makes it divisble by 5. (you only made one choice on x). -adding 1 or 2 makes it you to have more number of choices. b)If differnce between x and 2 is 4:
c)If differnce between x and 2 is 3:
-adding 2 makes it divisble by 5. (you only made one choice on x).
-adding 0 or 1 makes it you to have more number of choices.
d)If differnce between x and 2 is 2:
-adding 0 makes it divisble by 2. (you only made one choice on x). -adding 1 or 2 makes it you to have more number of choices. e)If differnce between x and 2 is 1:
-adding 0 makes it divisble by 1. (you only made one choice on x).
-adding 1 or 2 makes it you to have more number of choices.
f)If differnce between x and 2 is 6:
-adding 1 makes it divisble by 5. (you only made two choices on x, first divide by 5 then reminder by divided 2). //same number of choices in this case even by adding 0.
-adding 2 makes it you to have more number of choices.
So for every candidate you add some base (0 or 1 0r 2) and consider the only the base that gives you minimum choices.
But why I need add some base?
A basic divisibility exaple. 9 is not perfectly divisible 5 but adding 1 to 9 makes it 10 and now perfectly divisble by 5.
But why base only till 0,1,2 and not more than 2?
if the diffence between x and y is:
a)4 adding 1 is div by 5
b)3 adding 2 is div by 5
c)2 it is div by 2
d)7 is div by 5 followed by 2
e)8 adding 2 is div by 5 followed by 2
f)9 adding 1 is div by 5 followed by 2
Did you observe the base is cycling around 0,1,2.so you never need not go beyond 2
Thanks for your encouragement and support Rahul. I am slowly getting there.
This is a perfect explaination of adding base and finding the best case the logic except base addition would get you clear some test cases by the reducing approach but you would definitely need to add base so that you get the best apporach out of the three possible solutions. Good approach rohit! Thank You