## Amazon Interview Question

Software Engineers**Country:**United States

**Interview Type:**Phone Interview

```
#!/usr/bin/python
"""
This is Sieve Of Eratosthenes Algorithm
"""
def main():
n = 30
print "Sum of prime numbers up to {0} is : {1}".format(n, prime_numbers_SieveOfEratosthenes(n))
def prime_numbers_SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while(p*p <= n):
if prime[p] == True:
for i in range(p*2, n+1, p):
prime[i] = False
p += 1
result = []
sum = 0
for p in range(2, n):
if prime[p]:
sum += p
return sum
if __name__ == '__main__':
main()
```

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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)

- aonecoding July 28, 2017