Amazon Interview Question


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
1
of 1 vote

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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import java.util.TreeMap;
public class PrimeSum {

    TreeMap<Integer, Integer> 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);
    }

}

- aonecoding4 January 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

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)

- tarptaeya January 14, 2019 | Flag Reply


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