java codility training Genomic-range-query

Question

The task is:

A non-empty zero-indexed string S is given. String S consists of N characters from the set of upper-case English letters A, C, G, T.

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Answer 1

Python Solution with explanation

The idea is to hold an auxiliary array per nucleotide X, with position i (ignoring zero) is how many times X has occurred as of now. And so if we need the number of occurrences of X from position f to position t, we could take the following equation:

aux(t) - aux(f)

Time complexity is:

O(N+M)

def solution(S, P, Q):
    n = len(S)
    m = len(P)

    aux = [[0 for i in range(n+1)] for i in [0,1,2]]

    for i,c in enumerate(S):
        aux[0][i+1] = aux[0][i] + ( c == 'A' )
        aux[1][i+1] = aux[1][i] + ( c == 'C' )
        aux[2][i+1] = aux[2][i] + ( c == 'G' )

    result = []

    for i in range(m):
        fromIndex , toIndex = P[i] , Q[i] +1
        if   aux[0][toIndex] - aux[0][fromIndex] > 0:
            r = 1
        elif aux[1][toIndex] - aux[1][fromIndex] > 0:
            r = 2
        elif aux[2][toIndex] - aux[2][fromIndex] > 0:
            r = 3
        else:
            r = 4

        result.append(r)

    return result

Answer 2

   static public int[] solution(String S, int[] P, int[] Q) {
    // write your code in Java SE 8

    int A[] = new int[S.length() + 1], C[] = new int[S.length() + 1], G[] = new int[S.length() + 1];

    int last_a = 0, last_c = 0, last_g = 0;

    int results[] = new int[P.length];
    int p = 0, q = 0;
    for (int i = S.length() - 1; i >= 0; i -= 1) {
        switch (S.charAt(i)) {
            case 'A': {
                last_a += 1;
                break;
            }
            case 'C': {
                last_c += 1;
                break;
            }

            case 'G': {
                last_g += 1;
                break;
            }

        }
        A[i] = last_a;
        G[i] = last_g;
        C[i] = last_c;
    }


    for (int i = 0; i < P.length; i++) {
        p = P[i];
        q = Q[i];

        if (A[p] - A[q + 1] > 0) {
            results[i] = 1;
        } else if (C[p] - C[q + 1] > 0) {
            results[i] = 2;
        } else if (G[p] - G[q + 1] > 0) {
            results[i] = 3;
        } else {
            results[i] = 4;
        }

    }
    return results;
}

Answer 3

Here's a simple javascript solution which got 100%.

function solution(S, P, Q) {
    var A = [];
    var C = [];
    var G = [];
    var T = [];
    var result = [];
    var i = 0;

    S.split('').forEach(function(a) {
        if (a === 'A') {
            A.push(i);
        } else if (a === 'C') {
            C.push(i);
        } else if (a === 'G') {
            G.push(i);
        } else {
            T.push(i);
        }

        i++;
    });

    function hasNucl(typeArray, start, end) {
        return typeArray.some(function(a) {
            return a >= P[j] && a <= Q[j];
        });
    }

    for(var j=0; j<P.length; j++) {
        if (hasNucl(A, P[j], P[j])) {
            result.push(1)
        } else if (hasNucl(C, P[j], P[j])) {
            result.push(2);
        } else if (hasNucl(G, P[j], P[j])) {
            result.push(3);
        } else {
            result.push(4);
        }
    }

    return result;
}

Answer 4

simple php 100/100 solution

function solution($S, $P, $Q) {
    $result = array();
    for ($i = 0; $i < count($P); $i++) {
        $from = $P[$i];
        $to = $Q[$i];
        $length = $from >= $to ? $from - $to + 1 : $to - $from + 1;
        $new = substr($S, $from, $length);

        if (strpos($new, 'A') !== false) {
            $result[$i] = 1;
        } else {
            if (strpos($new, 'C') !== false) {
                $result[$i] = 2;
            } else {
                if (strpos($new, 'G') !== false) {
                    $result[$i] = 3;
                } else {
                   $result[$i] = 4;
                }
            }
        }
    }
    return $result;
}

Answer 5

Here's 100% Scala solution:

def solution(S: String, P: Array[Int], Q: Array[Int]): Array[Int] = {


    val resp = for(ind <- 0 to P.length-1) yield {

      val sub= S.substring(P(ind),Q(ind)+1)


      var factor = 4

      if(sub.contains("A")) {factor=1}
      else{
        if(sub.contains("C")) {factor=2}
        else{
          if(sub.contains("G")) {factor=3}
        }
      }
      factor

    }

    return resp.toArray

  }

And performance: https://codility.com/demo/results/trainingEUR4XP-425/

Answer 6

pshemek's solution constrains itself to the space complexity (O(N)) - even with the 2-d array and the answer array because a constant (4) is used for the 2-d array. That solution also fits in with the computational complexity - whereas mine is O (N^2) - though the actual computational complexity is much lower because it skips over entire ranges that include minimal values.

I gave it a try - but mine ends up using more space - but makes more intuitive sense to me (C#):

public static int[] solution(String S, int[] P, int[] Q)
{
    const int MinValue = 1;
    Dictionary<char, int> stringValueTable = new Dictionary<char,int>(){ {'A', 1}, {'C', 2}, {'G', 3}, {'T', 4} };

    char[] inputArray = S.ToCharArray();
    int[,] minRangeTable = new int[S.Length, S.Length]; // The meaning of this table is [x, y] where x is the start index and y is the end index and the value is the min range - if 0 then it is the min range (whatever that is)
    for (int startIndex = 0; startIndex < S.Length; ++startIndex)
    {
        int currentMinValue = 4;
        int minValueIndex = -1;
        for (int endIndex = startIndex; (endIndex < S.Length) && (minValueIndex == -1); ++endIndex)
        {
            int currentValue = stringValueTable[inputArray[endIndex]];
            if (currentValue < currentMinValue)
            {
                currentMinValue = currentValue;
                if (currentMinValue == MinValue) // We can stop iterating - because anything with this index in its range will always be minimal
                    minValueIndex = endIndex;
                else
                    minRangeTable[startIndex, endIndex] = currentValue;
            }
            else
                minRangeTable[startIndex, endIndex] = currentValue;
        }

        if (minValueIndex != -1) // Skip over this index - since it is minimal
            startIndex = minValueIndex; // We would have a "+ 1" here - but the "auto-increment" in the for statement will get us past this index
    }

    int[] result = new int[P.Length];
    for (int outputIndex = 0; outputIndex < result.Length; ++outputIndex)
    {
        result[outputIndex] = minRangeTable[P[outputIndex], Q[outputIndex]];
        if (result[outputIndex] == 0) // We could avoid this if we initialized our 2-d array with 1's
            result[outputIndex] = 1;
    }

    return result;
}

In pshemek's answer - the "trick" in the second loop is simply that once you've determined you've found a range with the minimal value - you don't need to continue iterating. Not sure if that helps.