CareerCup

This problem can be solved using prime numbers

Let's say A is an array which contains first 10 prime numbers
A = { 2,3,5,7,11,13,17,19,23,29}

unsigned long int multnum =1, multsearchnum =1 ;
foreach digit of number
{
multnum *= A[digit];
}

foreach digit of (searchNumber)
{
multsearchnum *= A[digit];
}

if { multnum % multsearchnum == 0} {
print ( "all digits of search string are present in the number");
}



I didn't find any failing case till now. Please let me know if you find any failing case.

We need to store 4 things
1. color
2. presence
3. type of the coin
4. location

Locations range is (0,0) = 0x00 to (7,7) = 0x77 which need 6bits. That means 32 * 6bits = 24bytes (for storing the locations)

1 bit for color => 32 bits = 4 bytes
1 bit for presence => 32 bits = 4 bytes
3 bits for type of coin (000 - pawn, 001 - bishop, 010 - knight , 011 -rook, 100- queen, 101 - king) => 3*32 = 12 bytes

Total = 44 bytes

@ Srinivas: Yes, I do agree. max sum is -2, min sum is -10. But, does this algo give right answer ?

@ karthikaditya


Your code doesn't seem to work if the array contains all -ve numbers

Ex : {-4 , -4 , -2}

Max Sum is -10... But, your algo give 0 as max sum

@arya

How does it work if the array contains all the negative numbers

for ex : { -4, -5, -2}

Max sum should be -11