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:

class Solution { public int[] solution(int N, int[] P, 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 Java 21

Time spent on task 2 minutes

Notes

not defined yet

Code: 07:55:54 UTC, java, verify, result: Failed

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 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89

// you can also use imports, for example:
// import java.util.*;

// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
   public int[] solution(int N, int[] P, int[] Q) {
        // write your code in Java SE 8
        if (N<4) return new int[P.length]; 
        
        boolean[] notPrime = new boolean[N+1]; 
        notPrime[0] = true; notPrime[1] = true; notPrime[2] = false; 
        int numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (notPrime[i]) continue;
            numPrimes++; 
            for (int k=i*2; k<=N/2; k+=i){
                notPrime[k] = true; 
            }
        }
        int[] primes = new int[numPrimes];
        numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (!notPrime[i]) primes[numPrimes++]=i; 
        }
        boolean[] isSemiPrime = new boolean[N+1]; 
        int numSemiPrimes=0; 
        for (int i=0; i<numPrimes; i++){
            for (int j=i; j<numPrimes && primes[j]<=N/primes[i]; j++){
               // if (N/primes[i] < primes[j]) continue; 
                int k = primes[i]*primes[j]; 
                isSemiPrime[k] = true;
                numSemiPrimes++; 
            }
        }
        int[] semiPrimes = new int[numSemiPrimes]; 
        numSemiPrimes = 0; 
        for (int i=4; i<=N; i++){
            if (isSemiPrime[i]) semiPrimes[numSemiPrimes++]=i; 
        }
        
        int[] ret = new int[P.length]; 
        for (int i=0; i<P.length; i++){
            int left = findNext(P[i], semiPrimes); 
            if (left<0) {ret[i]=0; continue; } 
            int right = findPrev(Q[i], semiPrimes); 
            if (right<0) {ret[i]=0; continue; } 
            int num = right-left+1; 
            ret[i] = (num<0)?0:num; 
        }
        return ret; 
    }
    
    int findNext(int val, int[] A){
        //find the smallest x such that A[x]>=val, elements in A are distinct
        int start = 0, end = A.length-1;
        if (A[end]<val) return -1; 
        while (start<end){
            if (A[start]>=val) return start; 
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return start; 
    }
    int findPrev(int val, int[] A){
        //find the largest x such that A[x]<=val, elements in A are distinct
        int start = 0, end = A.length-1; 
        if (A[start]>val) return -1;  
        while (start<end){
            if (A[end]<=val) return end;
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid-1; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return end; 
    }
}

Code: 07:56:34 UTC, java, verify, result: Passed

// you can also use imports, for example:
// import java.util.*;

// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
   public int[] solution(int N, int[] P, int[] Q) {
        // write your code in Java SE 8
        if (N<4) return new int[P.length]; 
        
        boolean[] notPrime = new boolean[N+1]; 
        notPrime[0] = true; notPrime[1] = true; notPrime[2] = false; 
        int numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (notPrime[i]) continue;
            numPrimes++; 
            for (int k=i*2; k<=N/2; k+=i){
                notPrime[k] = true; 
            }
        }
        int[] primes = new int[numPrimes];
        numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (!notPrime[i]) primes[numPrimes++]=i; 
        }
        boolean[] isSemiPrime = new boolean[N+1]; 
        int numSemiPrimes=0; 
        for (int i=0; i<numPrimes; i++){
            for (int j=i; j<numPrimes && primes[j]<=N/primes[i]; j++){
               // if (N/primes[i] < primes[j]) continue; 
                int k = primes[i]*primes[j]; 
                isSemiPrime[k] = true;
                numSemiPrimes++; 
            }
        }
        int[] semiPrimes = new int[numSemiPrimes]; 
        numSemiPrimes = 0; 
        for (int i=4; i<=N; i++){
            if (isSemiPrime[i]) semiPrimes[numSemiPrimes++]=i; 
        }
        
        int[] ret = new int[P.length]; 
        for (int i=0; i<P.length; i++){
            int left = findNext(P[i], semiPrimes); 
            if (left<0) {ret[i]=0; continue; } 
            int right = findPrev(Q[i], semiPrimes); 
            if (right<0) {ret[i]=0; continue; } 
            int num = right-left+1; 
            ret[i] = (num<0)?0:num; 
        }
        return ret; 
    }
    
    int findNext(int val, int[] A){
        //find the smallest x such that A[x]>=val, elements in A are distinct
        int start = 0, end = A.length-1;
        if (A[end]<val) return -1; 
        while (start<end){
            if (A[start]>=val) return start; 
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return start; 
    }
    int findPrev(int val, int[] A){
        //find the largest x such that A[x]<=val, elements in A are distinct
        int start = 0, end = A.length-1; 
        if (A[start]>val) return -1;  
        while (start<=end){
            if (A[end]<=val) return end;
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid-1; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return end; 
    }
}

Code: 07:56:41 UTC, java, verify, result: Passed

// you can also use imports, for example:
// import java.util.*;

// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
   public int[] solution(int N, int[] P, int[] Q) {
        // write your code in Java SE 8
        if (N<4) return new int[P.length]; 
        
        boolean[] notPrime = new boolean[N+1]; 
        notPrime[0] = true; notPrime[1] = true; notPrime[2] = false; 
        int numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (notPrime[i]) continue;
            numPrimes++; 
            for (int k=i*2; k<=N/2; k+=i){
                notPrime[k] = true; 
            }
        }
        int[] primes = new int[numPrimes];
        numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (!notPrime[i]) primes[numPrimes++]=i; 
        }
        boolean[] isSemiPrime = new boolean[N+1]; 
        int numSemiPrimes=0; 
        for (int i=0; i<numPrimes; i++){
            for (int j=i; j<numPrimes && primes[j]<=N/primes[i]; j++){
               // if (N/primes[i] < primes[j]) continue; 
                int k = primes[i]*primes[j]; 
                isSemiPrime[k] = true;
                numSemiPrimes++; 
            }
        }
        int[] semiPrimes = new int[numSemiPrimes]; 
        numSemiPrimes = 0; 
        for (int i=4; i<=N; i++){
            if (isSemiPrime[i]) semiPrimes[numSemiPrimes++]=i; 
        }
        
        int[] ret = new int[P.length]; 
        for (int i=0; i<P.length; i++){
            int left = findNext(P[i], semiPrimes); 
            if (left<0) {ret[i]=0; continue; } 
            int right = findPrev(Q[i], semiPrimes); 
            if (right<0) {ret[i]=0; continue; } 
            int num = right-left+1; 
            ret[i] = (num<0)?0:num; 
        }
        return ret; 
    }
    
    int findNext(int val, int[] A){
        //find the smallest x such that A[x]>=val, elements in A are distinct
        int start = 0, end = A.length-1;
        if (A[end]<val) return -1; 
        while (start<end){
            if (A[start]>=val) return start; 
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return start; 
    }
    int findPrev(int val, int[] A){
        //find the largest x such that A[x]<=val, elements in A are distinct
        int start = 0, end = A.length-1; 
        if (A[start]>val) return -1;  
        while (start<=end){
            if (A[end]<=val) return end;
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid-1; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return end; 
    }
}

Code: 07:56:46 UTC, java, final, score: 100

// you can also use imports, for example:
// import java.util.*;

// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
   public int[] solution(int N, int[] P, int[] Q) {
        // write your code in Java SE 8
        if (N<4) return new int[P.length]; 
        
        boolean[] notPrime = new boolean[N+1]; 
        notPrime[0] = true; notPrime[1] = true; notPrime[2] = false; 
        int numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (notPrime[i]) continue;
            numPrimes++; 
            for (int k=i*2; k<=N/2; k+=i){
                notPrime[k] = true; 
            }
        }
        int[] primes = new int[numPrimes];
        numPrimes = 0; 
        for (int i=2; i<=N; i++){
            if (!notPrime[i]) primes[numPrimes++]=i; 
        }
        boolean[] isSemiPrime = new boolean[N+1]; 
        int numSemiPrimes=0; 
        for (int i=0; i<numPrimes; i++){
            for (int j=i; j<numPrimes && primes[j]<=N/primes[i]; j++){
               // if (N/primes[i] < primes[j]) continue; 
                int k = primes[i]*primes[j]; 
                isSemiPrime[k] = true;
                numSemiPrimes++; 
            }
        }
        int[] semiPrimes = new int[numSemiPrimes]; 
        numSemiPrimes = 0; 
        for (int i=4; i<=N; i++){
            if (isSemiPrime[i]) semiPrimes[numSemiPrimes++]=i; 
        }
        
        int[] ret = new int[P.length]; 
        for (int i=0; i<P.length; i++){
            int left = findNext(P[i], semiPrimes); 
            if (left<0) {ret[i]=0; continue; } 
            int right = findPrev(Q[i], semiPrimes); 
            if (right<0) {ret[i]=0; continue; } 
            int num = right-left+1; 
            ret[i] = (num<0)?0:num; 
        }
        return ret; 
    }
    
    int findNext(int val, int[] A){
        //find the smallest x such that A[x]>=val, elements in A are distinct
        int start = 0, end = A.length-1;
        if (A[end]<val) return -1; 
        while (start<end){
            if (A[start]>=val) return start; 
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return start; 
    }
    int findPrev(int val, int[] A){
        //find the largest x such that A[x]<=val, elements in A are distinct
        int start = 0, end = A.length-1; 
        if (A[start]>val) return -1;  
        while (start<=end){
            if (A[end]<=val) return end;
            int mid = (start+end)/2; 
            if (A[mid]>val){
                end = mid-1; 
            }
            else if (A[mid]<val){
                start = mid+1; 
            }
            else return mid; 
        }
        return end; 
    }
}

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

✔