Codility CountDiv/

def solution(A, B, K):
    # write your code in Python 3.6
    if A % K ==0:
        return (B - A) // K + 1
    else:
        return (B - (A - A % K)) // K
def solution(A, B, K):
    # write your code in Python 3.6
    border = 0
    if A % K ==0 or B % K == 0:
        border = 1
    if A < K and B < K:
        return border
    elif A < K and B > K:
        return int((B - K)/K) + 1 + border
    else:
        return int((B-A)/K) + border

def solution(A):
    # write your code in Python 3.6
    count = 0
    zero_count = 0
    one_count = 0
    for i in range(len(A)):
        if A[i] == 0:
            zero_count = zero_count + 1
        elif A[i] == 1:
            count = count + zero_count
	    if count > 1000000000:
	        return -1	
    return count

def solution(A):
    # write your code in Python 3.6[]
    min = (A[0]+A[1])/2.0
    start = 0
    for i in range(len(A)-1):
        if (A[i]+A[i+1])/2.0 <min:
            min = (A[i]+A[i+1])/2.0
            start = i
        if i+2 <=len(A)-1 and (A[i]+A[i+1]+A[i+2])/3.0 < min:
            min = (A[i]+A[i+1]+A[i+2])/3.0
            start = i
        
    return start

solution 1:
def solution(S, P, Q):
    # write your code in Python 3.6
    dicx = {'A':1,'C':2,'G':3,'T':4}
    Smin = [0 for x in range(len(P))]
    min = 4

    for x in range(len(P)):
        left = P[x]
        right = Q[x]
        if left == right:
            Smin[x] = dicx[S[left]]
        else:
            for y in range(left,right+1):
                if dicx[S[y]] < min:
                    min = dicx[S[y]]
            Smin[x] = min
        min = 4
    return Smin

def compare(A,i,j,k):
    if A[i]+ A[j] > A[k] and A[k]+ A[j] > A[k] and A[i]+ A[k] > A[j]:
        return True

def solution(A):
    # write your code in Python 3.6
    A.sort()
    for i in range(len(A)-2):
        j = i + 1
        k = i + 2
        if compare(A,i,j,k):
            return 1
    return 0