Hi Geeks, One of The My Friend had This Question in His Technical
Round of Facebook, I m going to share with you.lest see how geek
approach this...Plz don't make this post spam..by discussing whats ur
friend name, wich colge, etc etc..just share your approach, think &
solve the question, even  Google search wont give you correct &
efficient approach ,answer for this question..so think your self..

O(n^2) Solution is Obvious ..but .it wont work for 10 million  as a
limit so not a good solution

we have to solve it using best approach & algo..as we have

so here is the question...From Facebook...

/*
A non-negative integer is called heavy if the average value of its
digits in decimal representation exceeds 7. Assume that 0 has average
value of its digits equal to 0.

For example the number 8698 is heavy, because the average value of its
digits equal to (8+6+9+8)/4 = 7.75. The number 53141 has the average
value of its digits equal to (5+3+1+4+1)/5 = 2.6, so it is not heavy.

Write a function

int heavy_decimal_count(int a,int b);

that given two non-negative integers A and B returns the number of
heavy integers in the interval [A..B] (both ends included). Assume
that 0 <=A <= B <= 200,000,000 Range Given ..It Really Matters Your
Program should not give time out & memory error

For example, given A=8,675 and B=8,689 the function should return 5,
because there are 5 heavy integers in range [8,675..8,689]:

8675   avg=6.5
8676   avg=6.75
8677   avg=7
8678   avg=7.25    HEAVY
8679   avg=7.5     HEAVY
8680   avg=5.5
8681   avg=5.75
8682   avg=6
8683   avg=6.25
8684   avg=6.5
8685   avg=6.75
8686   avg=7
8687   avg=7.25    HEAVY
8688   avg=7.5     HEAVY
8689   avg=7.75    HEAVY

you have to keep in mind for given range  e.g given  B<=2 Billion Its
Man Thing  so what happen when
A=1 Billion & B=2 Billion

*/

Go Ahead

Thanks & Regards
Shashank Mani
Cell 9740852296

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