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March 26, 2018 12:08
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Codlity
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def solution(A, B, K): | |
qa, ra = divmod(A, K) | |
qb, rb = divmod(B, K) | |
count = qb - qa | |
if (qa >= 0 and ra == 0): | |
count += 1 | |
return count |
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def solution(S, P, Q): | |
# write your code in Python 3.6 | |
s_len = len(S) | |
g = [[0], [0], [0], [0]] | |
for i, s in enumerate(S): | |
for vector in g: | |
vector.append(vector[-1]) | |
if s == 'A': | |
g[0][i + 1] = g[0][i] + 1 | |
elif s == 'C': | |
g[1][i + 1] = g[1][i] + 1 | |
elif s == 'G': | |
g[2][i + 1] = g[2][i] + 1 | |
elif s == 'T': | |
g[3][i + 1] = g[3][i] + 1 | |
result = [] | |
for index in range(len(P)): | |
from_i = P[index] | |
to_i = Q[index] + 1 | |
if g[0][to_i] - g[0][from_i] > 0: | |
result.append(1) | |
elif g[1][to_i] - g[1][from_i] > 0: | |
result.append(2) | |
elif g[2][to_i] - g[2][from_i] > 0: | |
result.append(3) | |
else: | |
result.append(4) | |
return result |
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def solution(A): | |
# write your code in Python 3.6 | |
factor = 0 | |
sum = 0 | |
for a in A: | |
if a == 0: | |
factor += 1 | |
else: | |
sum += factor | |
if sum > 1000000000: | |
return -1 | |
return sum |
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