prefix sums--codility

lesson 5: prefix sums

lesson 5: prefix sums

PassingCars

Count the number of passing cars on the road.

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.

Array A contains only 0s and/or 1s:

0 represents a car traveling east,

1 represents a car traveling west.

The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.

For example, consider array A such that:

  A[0] = 0

  A[1] = 1

  A[2] = 0

  A[3] = 1

  A[4] = 1

We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).

Assume that:

N is an integer within the range [1..100,000];
each element of array A is an integer that can have one of the following values: 0, 1.

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1),

思路：

可以计算suffix sum的方式
然后，从前面开始遍历list，遇到a ＝ 0，result即加上当前的suffix sum的值
此题元素是0，1，故可以不用保留每一步的计算，题目有要求限制O(1)的space，也是给出提示，用一个变量retsum值来记录，每一步的prefix sum值，每移动一步，元素是1的话，将retsum 减1，即是下一个prefix sum 值。
Detected time complexity: O(N)
[100%]

def solution(A):

    # write your code in Python 2.7

    result = 0

    retsum = sum(A)

    for a in A:

        if a == 0:

            result += retsum

            if result > 1000000000:

                return -1

        else:

            retsum -= 1

    return result

2. CountDiv

Compute number of integers divisible by k in range [a..b].

given three integers A, B and K, returns the number of integers within the range [A..B] that are divisible by K, i.e.:

{ i : A ≤ i ≤ B, i mod K = 0 }

For example, for A = 6, B = 11 and K = 2, your function should return 3, because there are three numbers divisible by 2 within the range [6..11], namely 6, 8 and 10.

Assume that:

A and B are integers within the range [0..2,000,000,000];
K is an integer within the range [1..2,000,000,000];

A ≤ B.

Complexity:

expected worst-case time complexity is O(1);
expected worst-case space complexity is O(1).

CountDiv solution 1

Test score 100%

def solution(A, B, K):

    # write your code in Python 2.7

    ra = -1 if A == 0 else (A - 1)/K

    rb = B/K

    return rb -ra

solution 2

Test score 100%

def solution(A, B, K):

    # write your code in Python 2.7

    c = 1 if A%K == 0 else 0

    return B/K -A/K + c

def solution(A, B, K):

    # write your code in Python 2.7

    return (B/K - A/K) if (A%K != 0 ) else (B/K - A/K + 1)

3. GenomicRangeQuery

Find the minimal nucleotide from a range of sequence DNA.

A DNA sequence can be represented as a string consisting of the letters A, C, G and T, which correspond to the types of successive nucleotides in the sequence. Each nucleotide has an impact factor, which is an integer. Nucleotides of types A, C, G and T have impact factors of 1, 2, 3 and 4, respectively. You are going to answer several queries of the form: What is the minimal impact factor of nucleotides contained in a particular part of the given DNA sequence?

The DNA sequence is given as a non-empty string S = S[0]S[1]...S[N-1] consisting of N characters. There are M queries, which are given in non-empty arrays P and Q, each consisting of M integers. The K-th query (0 ≤ K < M) requires you to find the minimal impact factor of nucleotides contained in the DNA sequence between positions P[K] and Q[K](inclusive).

For example, consider string S = CAGCCTA and arrays P, Q such that:

P[0] = 2    Q[0] = 4

P[1] = 5    Q[1] = 5

P[2] = 0    Q[2] = 6

The answers to these M = 3 queries are as follows:

The part of the DNA between positions 2 and 4 contains nucleotides G and C (twice), whose impact factors are 3 and 2 respectively, so the answer is 2.
The part between positions 5 and 5 contains a single nucleotide T, whose impact factor is 4, so the answer is 4.
The part between positions 0 and 6 (the whole string) contains all nucleotides, in particular nucleotide A whose impact factor is 1, so the answer is 1.

the function should return the values [2, 4, 1], as explained above.

Assume that:

N is an integer within the range [1..100,000];
M is an integer within the range [1..50,000];
each element of arrays P, Q is an integer within the range [0..N − 1];
P[K] ≤ Q[K], where 0 ≤ K < M;
string S consists only of upper-case English letters A, C, G, T.

Complexity:

expected worst-case time complexity is O(N+M);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

解法一：

Test score 62%
Detected time complexity:

O(N * M)
最初的想法是：根据给出的范围，用set保存，看是否有各元素
时间复杂度，明显达不到O(N+M)

def getMinFactor(S,l,i,j):

    if j == (l-1):

        tmp = set(S[i:])  #

    else:

        tmp = set(S[i:(j+1)])

    if 'A' in tmp:

        return 1

    elif 'C' in tmp:

        return 2

    elif 'G' in tmp:

        return 3

    else:

        return 4

def solution(S, P, Q):

    # write your code in Python 2.7

    length = len(S)

    result = []

    for x,y in zip(P,Q):

        #print x,y

        result.append(getMinFactor(S,length,x,y))

    return result

解法二：

Test score 100%
used each list to save that states whether has element or not
在用prefix sum 做差的方式，依次检查是否存在A,C,G,T字符
注意数值关系

def calcPrefixSum(S):

    l = len(S)+1

    pa,pc,pg = [0]*l,[0]*l,[0]*l

    for idx,elem in enumerate(S):

        a,c,g = 0,0,0

        if elem == 'A':

            a = 1

        elif elem == 'C':

            c = 1

        elif elem == 'G':

            g = 1

        pa[idx+1] = pa[idx] + a

        pc[idx+1] = pc[idx] + c

        pg[idx+1] = pg[idx] + g

    return pa,pc,pg

def solution(S, P, Q):

    # write your code in Python 2.7

    pA,pC,pG = calcPrefixSum(S)

    result = []

    for i,j in zip(P,Q):

        if pA[j+1] - pA[i] > 0:

            ret = 1

        elif pC[j+1] - pC[i] > 0:

            ret = 2

        elif pG[j+1] - pG[i] > 0:

            ret = 3

        else:

            ret = 4

        result.append(ret)

    return result

根据prefix sum list:

pA = [0, 0, 1, 1, 1, 1, 1, 2]

pC = [0, 1, 1, 1, 2, 3, 3, 3]

pG = [0, 0, 0, 1, 1, 1, 1, 1]

故，是下标［j＋1］－［i］

4. MinAvgTwoSlice

Find the minimal average of any slice containing at least two elements.

A non-empty zero-indexed array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P < Q < N, is called a slice of array A (notice that the slice contains at least two elements). The average of a slice (P, Q) is the sum of A[P] + A[P + 1] + ... + A[Q] divided by the length of the slice. To be precise, the average equals (A[P] + A[P + 1] + ... + A[Q]) / (Q − P + 1).

For example, array A such that:

A[0] = 4

A[1] = 2

A[2] = 2

A[3] = 5

A[4] = 1

A[5] = 5

A[6] = 8

contains the following example slices:

slice (1, 2), whose average is (2 + 2) / 2 = 2;
slice (3, 4), whose average is (5 + 1) / 2 = 3;
slice (1, 4), whose average is (2 + 2 + 5 + 1) / 4 = 2.5.

The goal is to find the starting position of a slice whose average is minimal.

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N),

sloution

Test score 100%

note: transfer to 2/3 ,

只要查看相邻两个和三个的数的平均值即可
proof

def solution(A):

    # write your code in Python 2.7

    length = len(A)

    minStartPos = 0

    minSum = (A[0] + A[1])/2.0

    for i in xrange(length - 2):

        tmp = (A[i] + A[i+1])/2.0

        if tmp < minSum:

            minSum = tmp

            minStartPos = i

        tmp = (tmp*2 + A[i+2])/3.0

        if tmp < minSum:

            minSum = tmp

            minStartPos = i

    if (A[-1] + A[-2])/2.0 < minSum:

        minStartPos = length - 2

    return minStartPos