# Hash Function

Hash Function

In data structure Hash, hash function is used to convert a string(or any other type) into an integer smaller than hash size and bigger or equal to zero. The objective of designing a hash function is to “hash” the key as unreasonable as possible. A good hash function can avoid collision as less as possible. A widely used hash function algorithm is using a magic number 33, consider any string as a 33 based big integer like follow:

hashcode(“abcd”) = (ascii(a) * 333 + ascii(b) * 332 + ascii(c) *33 + ascii(d)) % HASH_SIZE

= (97* 333 + 98 * 332 + 99 * 33 +100) % HASH_SIZE

= 3595978 % HASH_SIZE

here HASH_SIZE is the capacity of the hash table (you can assume a hash table is like an array with index 0 ~ HASH_SIZE-1).

Given a string as a key and the size of hash table, return the hash value of this key.f

Example

For key=”abcd” and size=100, return 78

Clarification

For this problem, you are not necessary to design your own hash algorithm or consider any collision issue, you just need to implement the algorithm as described.

Solution: Use a helper to avoid overflow.  To calculate a*33%HASH_SIZE, instead of multiply by 33, we add a to itself 33 times. if(result+num>mod) we only add num-mod it to the result, otherwise, just need to add num to result. In this case, we are either adding or removing HASH_SIZE from the final result so it won’t overflow.

```public int hashCode(char[] key, int HASH_SIZE) {
int result = 0;
for (int i = 0; i < key.length; i++) {
result = helper(result, 33, HASH_SIZE);
result += key[i];
result %= HASH_SIZE;
}
return result;
}

int helper(int num, int base, int mod) {
int result = 0;
for (int i = 0; i < base; i++) {
if (result + num > mod) {
result += num - mod;
} else {
result += num;
}
}
return result;
}```