http://codegolf.stackexchange.com/questions/2952/count-number-of-hefty-decimals-between-2-numbers
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best approach for such problems is try to solve for very small numbers
then do DP for solve large numbers. as follows
single digit = 7,8,9 answer = 3 [i am taking 7 inclussive]
double digit= 7*2 = 14. now smallest number can be.
59=1
68, 69 = 2
77 78, 79,=3
86, 87, 88, 89=4
95,96,97,98,99 =5
3
Ok. Here's a possible O(n) solution.
Assuming last digit of a is 0.
for(n=a;n=b;n+=10)
{
Calculate the sum of digits, leaving the last digit.
Find the minimum value of last digit for it to be a heavy number.
Increment count by 10-that number.
}
So here, complexity will be: O(n/10*(d-1))
where, d
@all , Nikhi Jindal ,.as i already told that O(n^2) Solution is
Obvious ..but .it wont work for 1Biillion as a time
limit exceeds , memory Error occur, so its not a good solution ..I
think There is DP Solution Exist For Thats We Need to Figure it Out to
resolve this problem
@anand what r u
OK, here's my rather rambling theory on how to approach it:
First break the range down into a series of smaller ranges into form:
xxyzz
where:
xx is 0 to n digits which are fixed throughout the range.
y is 0 or 1 digit in the range 0..n or m..9 (with a special case *)
zz is 0 to n digits in
Hey James
Would that work equally well for 59 up to 1000?
Sent from my Windows Phone From: James Curran
Sent: Tuesday, April 05, 2011 7:50 AM
To: Algorithm Geeks
Subject: [algogeeks] Re: Facebook Interview QuestionHeavy Number
OK, here's my rather rambling theory on how to approach
Hey , i m a new follower to this group.. i was actually looking for a nice
group that would contain a nice challenge to sharpen the brain. and i
finally ended with this group... i m happy to follo this group along with
geeks like you.
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Sent from my Windows Phone From: James Curran
Sent: Tuesday, April 05, 2011 7:50 AM
To: Algorithm Geeks
Subject: [algogeeks] Re: Facebook Interview QuestionHeavy Number
OK, here's my rather rambling theory on how to approach it:
First break the range down into a series
