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] = 6The 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.
Write a function:
def solution(S, P, Q)
that, given a non-empty string S consisting of N characters and two non-empty arrays P and Q consisting of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.
Result array should be returned as an array of integers.
For example, given the 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] = 6the function should return the values [2, 4, 1], as explained above.
Write an efficient algorithm for the following assumptions:
- 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 and 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.
def solution(S, P, Q):
"""
S = CAGCCTA
P[0] = 2 Q[0] = 4
P[1] = 5 Q[1] = 5
P[2] = 0 Q[2] = 6
Given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting
of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.
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.
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.
:return: return the values [2, 4, 1]
"""
s_impact_factor = [[0 for i in range(len(S) + 1)] for j in range(3)]
print(s_impact_factor)
for i, item in enumerate(S):
a = 0
c = 0
g = 0
if 'A' == item:
a = 1
if 'B' == item:
c = 1
if 'C' == item:
g = 1
s_impact_factor[0][i + 1] = s_impact_factor[0][i] + a
s_impact_factor[1][i + 1] = s_impact_factor[1][i] + c
s_impact_factor[2][i + 1] = s_impact_factor[2][i] + g
print(s_impact_factor)
query_answers = [0] * len(P)
for i, item in enumerate(P):
from_index = P[i]
to_index = Q[i] + 1
if s_impact_factor[0][to_index] - s_impact_factor[0][from_index] > 0:
query_answers[i] = 1
elif s_impact_factor[1][to_index] - s_impact_factor[1][from_index] > 0:
query_answers[i] = 2
elif s_impact_factor[2][to_index] - s_impact_factor[2][from_index] > 0:
query_answers[i] = 3
else:
query_answers[i] = 4
return query_answers
[[0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0]] [[0, 0, 1, 1, 1, 1, 1, 2], [0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 1, 1, 2, 3, 3, 3]]
def solution(S, P, Q):
"""
Idea is consider solution as single dimensional array and use concept of prefix some ie.
stores the value in array for p,c and g based on frequency
array stores the frequency of p,c and g for all positions
Example -
# [0, 0, 1, 1, 1, 1, 1, 2] - prefix some of A - represents the max occurrence of A as 2 in array
# [0, 1, 1, 1, 2, 3, 3, 3] - prefix some of C - represents the max occurrence of A as 3 in array
# [0, 0, 0, 1, 1, 1, 1, 1] - prefix some of G - represents the max occurrence of A as 1 in array
# To find the query answers we can just use prefix some and find the distance between position
S = CAGCCTA
P[0] = 2 Q[0] = 4
P[1] = 5 Q[1] = 5
P[2] = 0 Q[2] = 6
Given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting
of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.
The part of the DNA between positions 2 and 4 contains nucleotide 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 nucleotide, in particular nucleotide A whose impact factor is 1, so the answer is 1.
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.
Ref - https://github.com/ghanan94/codility-lesson-solutions/blob/master/Lesson%2005%20-%20Prefix%20Sums/PrefixSums.pdf
:return: return the values [2, 4, 1]
"""
# two d array - column size is 3 for a,c,g - not taking size 4 since that will be part of else ie. don`t need to calculate
# row size is the length of DNA sequence
prefix_sum_two_d_array = [[0 for i in range(len(S) + 1)] for j in range(3)]
# find the prefix some of all nucleotide in given sequence
for i, nucleotide in enumerate(S):
# store prefix some of each
# nucleotide == 'A -> 1 if true 0 if false
# [0, 0, 1, 1, 1, 1, 1, 2] - prefix some of A - represents the max occurrence of A as 2 in array
prefix_sum_two_d_array[0][i + 1] = prefix_sum_two_d_array[0][i] + (nucleotide == 'A')
# store prefix some of c
# [0, 1, 1, 1, 2, 3, 3, 3] - prefix some of C - represents the max occurrence of A as 3 in array
prefix_sum_two_d_array[1][i + 1] = prefix_sum_two_d_array[1][i] + (nucleotide == 'C')
# store prefix some of g
# [0, 0, 0, 1, 1, 1, 1, 1] - prefix some of G - represents the max occurrence of A as 1 in array
prefix_sum_two_d_array[2][i + 1] = prefix_sum_two_d_array[2][i] + (nucleotide == 'G')
#print(prefix_sum_two_d_array)
# now to find the query answers we can just use prefix some and find the distance between position
query_answers = []
for position in range(len(P)):
# for each query of p
# find the start index from p
start_index = P[position]
# find the end index from Q
end_index = Q[position] + 1
# find the value from prefix some array - just subtract end index and start index to find the value
if prefix_sum_two_d_array[0][end_index] - prefix_sum_two_d_array[0][start_index]:
query_answers.append(1)
elif prefix_sum_two_d_array[1][end_index] - prefix_sum_two_d_array[1][start_index]:
query_answers.append(2)
elif prefix_sum_two_d_array[2][end_index] - prefix_sum_two_d_array[2][start_index]:
query_answers.append(3)
else:
query_answers.append(4)
return query_answers
def solution(S, P, Q):
"""
Idea is consider solution as single dimensional array and use concept of prefix some ie.
stores the value in array for p,c and g based on frequency
array stores the frequency of p,c and g for all positions
Example -
# [0, 0, 1, 1, 1, 1, 1, 2] - prefix some of A - represents the max occurrence of A as 2 in array
# [0, 1, 1, 1, 2, 3, 3, 3] - prefix some of C - represents the max occurrence of A as 3 in array
# [0, 0, 0, 1, 1, 1, 1, 1] - prefix some of G - represents the max occurrence of A as 1 in array
# To find the query answers we can just use prefix some and find the distance between position
S = CAGCCTA
P[0] = 2 Q[0] = 4
P[1] = 5 Q[1] = 5
P[2] = 0 Q[2] = 6
Given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting
of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.
The part of the DNA between positions 2 and 4 contains nucleotide 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 nucleotide, in particular nucleotide A whose impact factor is 1, so the answer is 1.
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.
Ref - https://github.com/ghanan94/codility-lesson-solutions/blob/master/Lesson%2005%20-%20Prefix%20Sums/PrefixSums.pdf
:return: return the values [2, 4, 1]
"""
# two d array - column size is 3 for a,c,g - not taking size 4 since that will be part of else ie. don`t need to calculate
# row size is the length of DNA sequence
prefix_sum_two_d_array = [[0 for i in range(len(S) + 1)] for j in range(3)]
# find the prefix some of all nucleotide in given sequence
for i, nucleotide in enumerate(S):
# store prefix some of each
# nucleotide == 'A -> 1 if true 0 if false
# [0, 0, 1, 1, 1, 1, 1, 2] - prefix some of A - represents the max occurrence of A as 2 in array
prefix_sum_two_d_array[0][i + 1] = prefix_sum_two_d_array[0][i] + (nucleotide == 'A')
# store prefix some of c
# [0, 1, 1, 1, 2, 3, 3, 3] - prefix some of C - represents the max occurrence of A as 3 in array
prefix_sum_two_d_array[1][i + 1] = prefix_sum_two_d_array[1][i] + (nucleotide == 'C')
# store prefix some of g
# [0, 0, 0, 1, 1, 1, 1, 1] - prefix some of G - represents the max occurrence of A as 1 in array
prefix_sum_two_d_array[2][i + 1] = prefix_sum_two_d_array[2][i] + (nucleotide == 'G')
#print(prefix_sum_two_d_array)
# now to find the query answers we can just use prefix some and find the distance between position
query_answers = []
for position in range(len(P)):
# for each query of p
# find the start index from p
start_index = P[position]
# find the end index from Q
end_index = Q[position] + 1
# find the value from prefix some array - just subtract end index and start index to find the value
if prefix_sum_two_d_array[0][end_index] - prefix_sum_two_d_array[0][start_index]:
query_answers.append(1)
elif prefix_sum_two_d_array[1][end_index] - prefix_sum_two_d_array[1][start_index]:
query_answers.append(2)
elif prefix_sum_two_d_array[2][end_index] - prefix_sum_two_d_array[2][start_index]:
query_answers.append(3)
else:
query_answers.append(4)
return query_answers
def solution(S, P, Q):
"""
Idea is consider solution as single dimensional array and use concept of prefix some ie.
stores the value in array for p,c and g based on frequency
array stores the frequency of p,c and g for all positions
Example -
# [0, 0, 1, 1, 1, 1, 1, 2] - prefix some of A - represents the max occurrence of A as 2 in array
# [0, 1, 1, 1, 2, 3, 3, 3] - prefix some of C - represents the max occurrence of A as 3 in array
# [0, 0, 0, 1, 1, 1, 1, 1] - prefix some of G - represents the max occurrence of A as 1 in array
# To find the query answers we can just use prefix some and find the distance between position
S = CAGCCTA
P[0] = 2 Q[0] = 4
P[1] = 5 Q[1] = 5
P[2] = 0 Q[2] = 6
Given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting
of M integers, returns an array consisting of M integers specifying the consecutive answers to all queries.
The part of the DNA between positions 2 and 4 contains nucleotide 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 nucleotide, in particular nucleotide A whose impact factor is 1, so the answer is 1.
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.
Ref - https://github.com/ghanan94/codility-lesson-solutions/blob/master/Lesson%2005%20-%20Prefix%20Sums/PrefixSums.pdf
:return: return the values [2, 4, 1]
"""
# two d array - column size is 3 for a,c,g - not taking size 4 since that will be part of else ie. don`t need to calculate
# row size is the length of DNA sequence
prefix_sum_two_d_array = [[0 for i in range(len(S) + 1)] for j in range(3)]
# find the prefix some of all nucleotide in given sequence
for i, nucleotide in enumerate(S):
# store prefix some of each
# nucleotide == 'A -> 1 if true 0 if false
# [0, 0, 1, 1, 1, 1, 1, 2] - prefix some of A - represents the max occurrence of A as 2 in array
prefix_sum_two_d_array[0][i + 1] = prefix_sum_two_d_array[0][i] + (nucleotide == 'A')
# store prefix some of c
# [0, 1, 1, 1, 2, 3, 3, 3] - prefix some of C - represents the max occurrence of A as 3 in array
prefix_sum_two_d_array[1][i + 1] = prefix_sum_two_d_array[1][i] + (nucleotide == 'C')
# store prefix some of g
# [0, 0, 0, 1, 1, 1, 1, 1] - prefix some of G - represents the max occurrence of A as 1 in array
prefix_sum_two_d_array[2][i + 1] = prefix_sum_two_d_array[2][i] + (nucleotide == 'G')
#print(prefix_sum_two_d_array)
# now to find the query answers we can just use prefix some and find the distance between position
query_answers = []
for position in range(len(P)):
# for each query of p
# find the start index from p
start_index = P[position]
# find the end index from Q
end_index = Q[position] + 1
# find the value from prefix some array - just subtract end index and start index to find the value
if prefix_sum_two_d_array[0][end_index] - prefix_sum_two_d_array[0][start_index]:
query_answers.append(1)
elif prefix_sum_two_d_array[1][end_index] - prefix_sum_two_d_array[1][start_index]:
query_answers.append(2)
elif prefix_sum_two_d_array[2][end_index] - prefix_sum_two_d_array[2][start_index]:
query_answers.append(3)
else:
query_answers.append(4)
return query_answers
The solution obtained perfect score.