Test results - Codility

Tasks Details

medium

1. CountSemiprimes

Count the semiprime numbers in the given range [a..b]

100%

A prime is a positive integer X that has exactly two distinct divisors: 1 and X. The first few prime integers are 2, 3, 5, 7, 11 and 13.

A semiprime is a natural number that is the product of two (not necessarily distinct) prime numbers. The first few semiprimes are 4, 6, 9, 10, 14, 15, 21, 22, 25, 26.

You are given two non-empty arrays P and Q, each consisting of M integers. These arrays represent queries about the number of semiprimes within specified ranges.

Query K requires you to find the number of semiprimes within the range (P[K], Q[K]), where 1 ≤ P[K] ≤ Q[K] ≤ N.

For example, consider an integer N = 26 and arrays P, Q such that:

P[0] = 1 Q[0] = 26 P[1] = 4 Q[1] = 10 P[2] = 16 Q[2] = 20

The number of semiprimes within each of these ranges is as follows:

(1, 26) is 10,

(4, 10) is 4,

(16, 20) is 0.

Write a function:

vector<int> solution(int N, vector<int> &P, vector<int> &Q);

that, given an integer N and two non-empty arrays P and Q consisting of M integers, returns an array consisting of M elements specifying the consecutive answers to all the queries.

For example, given an integer N = 26 and arrays P, Q such that:

P[0] = 1 Q[0] = 26 P[1] = 4 Q[1] = 10 P[2] = 16 Q[2] = 20

the function should return the values [10, 4, 0], as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..50,000];

M is an integer within the range [1..30,000];

each element of arrays P and Q is an integer within the range [1..N];

P[i] ≤ Q[i].

Solution

Programming language used C++

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 18:21:35 UTC, cpp, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

#include <cassert>
#include <math.h>
#include <vector>

vector<int> solution(int N, vector<int> &P, vector<int> &Q) {

  vector<int> result(P.size());

  // array of smallest factor of numbers
  int factors[N + 1];
  for (int fact = 0; fact <= N; fact++) {
    factors[fact] = 0;
  }
  // only need to go up to sq root as smallest factor will be <= sq root of number
  int largest = ceil(sqrt(N));
  for (int fact = 2; fact <= largest; fact++) {
    // primes only
    if (factors[fact] == 0) {
      // can start from sq of factor as smallest factors of values < that will already be found
      int mult = fact * fact;
      while (mult <= N) {
        if (factors[mult] == 0) {
          // found smallest factor
          factors[mult] = fact;
        }
        mult += fact;
      }
    }
  } // for
  // partSums = partial sum of number of semiprimes
  int partSums[N + 1];
  partSums[0] = partSums[1] = 0;
  int countSemi = 0; // count of number of semiprimes found so far
  for (int fact = 2; fact <= N; fact++) {
    int nextFact = factors[fact];
    if (nextFact != 0) {
      // not prime
      if (factors[nextFact] == 0) {
        // smallest factor is prime, what about other
        nextFact = fact / nextFact;
        if (factors[nextFact] == 0) {
          countSemi++;
        }
      }
    }
    partSums[fact] = countSemi;
  }
  // now do result
  for (ulong expr = 0; expr < P.size(); expr++) {
    result[expr] = partSums[Q[expr]] - partSums[P[expr] - 1];
  }

  return result;
}

Analysis

▶

example
example test

✔

Code: 18:21:41 UTC, cpp, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

#include <cassert>
#include <math.h>
#include <vector>

vector<int> solution(int N, vector<int> &P, vector<int> &Q) {

  vector<int> result(P.size());

  // array of smallest factor of numbers
  int factors[N + 1];
  for (int fact = 0; fact <= N; fact++) {
    factors[fact] = 0;
  }
  // only need to go up to sq root as smallest factor will be <= sq root of number
  int largest = ceil(sqrt(N));
  for (int fact = 2; fact <= largest; fact++) {
    // primes only
    if (factors[fact] == 0) {
      // can start from sq of factor as smallest factors of values < that will already be found
      int mult = fact * fact;
      while (mult <= N) {
        if (factors[mult] == 0) {
          // found smallest factor
          factors[mult] = fact;
        }
        mult += fact;
      }
    }
  } // for
  // partSums = partial sum of number of semiprimes
  int partSums[N + 1];
  partSums[0] = partSums[1] = 0;
  int countSemi = 0; // count of number of semiprimes found so far
  for (int fact = 2; fact <= N; fact++) {
    int nextFact = factors[fact];
    if (nextFact != 0) {
      // not prime
      if (factors[nextFact] == 0) {
        // smallest factor is prime, what about other
        nextFact = fact / nextFact;
        if (factors[nextFact] == 0) {
          countSemi++;
        }
      }
    }
    partSums[fact] = countSemi;
  }
  // now do result
  for (ulong expr = 0; expr < P.size(); expr++) {
    result[expr] = partSums[Q[expr]] - partSums[P[expr] - 1];
  }

  return result;
}

Analysis

▶

example
example test

✔

Code: 18:21:44 UTC, cpp, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

#include <cassert>
#include <math.h>
#include <vector>

vector<int> solution(int N, vector<int> &P, vector<int> &Q) {

  vector<int> result(P.size());

  // array of smallest factor of numbers
  int factors[N + 1];
  for (int fact = 0; fact <= N; fact++) {
    factors[fact] = 0;
  }
  // only need to go up to sq root as smallest factor will be <= sq root of number
  int largest = ceil(sqrt(N));
  for (int fact = 2; fact <= largest; fact++) {
    // primes only
    if (factors[fact] == 0) {
      // can start from sq of factor as smallest factors of values < that will already be found
      int mult = fact * fact;
      while (mult <= N) {
        if (factors[mult] == 0) {
          // found smallest factor
          factors[mult] = fact;
        }
        mult += fact;
      }
    }
  } // for
  // partSums = partial sum of number of semiprimes
  int partSums[N + 1];
  partSums[0] = partSums[1] = 0;
  int countSemi = 0; // count of number of semiprimes found so far
  for (int fact = 2; fact <= N; fact++) {
    int nextFact = factors[fact];
    if (nextFact != 0) {
      // not prime
      if (factors[nextFact] == 0) {
        // smallest factor is prime, what about other
        nextFact = fact / nextFact;
        if (factors[nextFact] == 0) {
          countSemi++;
        }
      }
    }
    partSums[fact] = countSemi;
  }
  // now do result
  for (ulong expr = 0; expr < P.size(); expr++) {
    result[expr] = partSums[Q[expr]] - partSums[P[expr] - 1];
  }

  return result;
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N * log(log(N)) + M)

▶

example
example test

✔

▶

extreme_one
small N = 1

✔

▶

extreme_four
small N = 4

✔

▶

small_functional
small functional

✔

▶

small_random
small random, length = ~40

✔

▶

medium_random
small random, length = ~300

✔

▶

large_small_slices
large with very small slices, length = ~30,000

✔

▶

large_random1
large random, length = ~30,000

✔

▶

large_random2
large random, length = ~30,000

✔

▶

extreme_large
all max ranges

✔