Phone Interview Amazon, Seattle

I. Get the sum of all prime numbers up to N. primeSum(N).

Follow-up: If primeSum(N) is frequently called, how to optimize it.

II. OODesign Parking Lot

I. At first find found all primes <= N (sieve of Eratosthenes). Getting the sum will be easy then.

Follow-up:

Cache the sums for any given N to save time. {N:SUM}

Optimization: Don't have to store sums for every N.

When N = 7, N = 8, N = 9, N = 10, the prime sum remains 17.

For N between 11 to 12, the prime sum is 28.

For N between 13 to 16, the sum is 41.

Use a BST structure as the cache. For N = 16, cache:

{2:3, 4:6, 6:11, 10:17, 12:28, 16:41}

For a given N, call cache.ceilingKey(N) to find the bucket for N.

N/log(n) * log(N)

Complexity

Time:

sieve of Eratosthenes takes O(NloglogN) time.

Insert an element into BST takes O(logN), there are N/logN primes in total to be added.

So building the cache takes logN * N / LogN = O(N) time

requesting primeSum(N) takes O(logN)

Space:

sieve of Eratosthenes takes O(N) extra space which will later be release after the cache is created.

Cache: O(N/logN)

```
public class PrimeSum {
TreeMap
``` sums;
public PrimeSum(int n) { //input the upper limit for all Ns
sums = new TreeMap<>();
// init an array to track prime numbers
boolean[] primes = new boolean[n + 1];
for (int i = 2; i < n; i++)
primes[i] = true;
for (int i = 2; i <= Math.sqrt(n); i++) {
if (primes[i]) {
for (int j = i + i; j < n; j += i)
primes[j] = false;
}
}
// insert sums into cache
int sum = 0;
for(int i = 2; i <= n; i++) {
if(primes[i]) {
sums.put(i - 1, sum);
sum += i;
}
}
if(primes[n]) {
sums.put(n, sum);
}
}
public int primeSum(int N) {
Integer ceiling = sums.ceilingKey(N);
//if(ceiling == null) {
//Exception("input value overflows");
//}
return sums.get(ceiling);
}
}

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