## Amazon Interview Question

**Country:**United States

**Interview Type:**Phone Interview

With O(Max_N) preprocessing time and O(Max_N) storage with subsequent queries in O(1) time -

```
class Primes:
__primeUptoN = [0, 0]
__primeSum = [0]
def __init__(self):
self.__generatePrimesIfNotGeneratedBefore()
def primeSum(self, N):
return self.__primeSum[self.__primeUptoN[N]]
@classmethod
def __generatePrimesIfNotGeneratedBefore(Primes):
if len(Primes.__primeSum) > 1:
return
MAX_N = 10 ** 6
non_primes = set()
for i in range(2, MAX_N + 1):
if i not in non_primes:
Primes.__primeSum.append(Primes.__primeSum[-1] + i)
j = i + i
while j <= MAX_N:
non_primes.add(j)
j += i
Primes.__primeUptoN.append(len(Primes.__primeSum) - 1)
```

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)

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- aonecoding4 January 07, 2019