I'm trying to create a list (b) that is list (a) rotating a's members k times to the left. I came up with this on Python 3:
n = 5
k = 4
a = [1,2,3,4,5]
b = []
for i in a:
if (i + k) <= (n - 1):
b.append(a[i+k])
elif (i+k-n) < (n-1):
b.append(a[i+k-n])
print(b)
But for some reason, it doesn't work since when I tell print(b) it returns a list that is exactly like list a
What am I missing here?
A simple solution:
k = k % len(a) #we don't care about shifting x*len(a) times since it does not have any effect
b = a[k:] + a[:k]
Thanks to #RemcoGerlich for helping me see my mistake! Here's how I quickly fixed my code:
n = 5
k = 4
a = [1,2,3,4,5]
b = []
i = 0
while (i < n):
if (i + k) <= (n - 1):
b.append(a[i+k])
elif (i+k-n) < (n-1):
b.append(a[i+k-n])
i = i+1
print(b)
Related
I am trying to solve a linear programming problem with the following constraints:
for some given values of N and T. So suppose N={1,2} and T={1,2}
It is easy to write out for small |N|, but is impossible to write as N becomes large.
Im confused as to how to actually code these sums by indexing on my objective variables.
I used the following to create my objective variables:
for t in range(1,T+1):
for i in range(1,n+1):
for j in range(1,n+1):
# Skip the x_i,i,t entries
if j == i:
continue
element = "x"+str(i)+','+str(j)+','+str(t)
x_ijt_holding.append(element)
x_ijt =[]
for i in x_ijt_holding:
x_ijt.append(pulp.LpVariable(i, cat = "Binary"))
Initially I thought I could just define each constraint with LpVariable, but I realized the solver doesn't like that.
For example I did:
# Constraint 2 enter entries
x_ijt_2_holding = []
for j in range(1,n+1):
for i in range(1,j):
equations_2 = []
equations_3 = []
for t in range(1, T+1):
equations_2.append("x"+str(i)+','+str(j)+','+str(t))
equations_2.append("x"+str(j)+','+str(i)+','+str(t))
x_ijt_2_holding.append(equations_2)
# Constraint 2 as LpVariable:
for i in x_ijt_2_holding:
temp = []
for sublist in i:
temp.append(pulp.LpVariable(sublist, cat = "Binary"))
x_ijt_con2.append(temp)
So how would I then code the constraints into the problem?
Constraints defines relationships on previously defined variable written as equations. When you are building your constraints you need to reference the variables you defined in your first part. Saving your variables in dicts makes it much easier to reference your variables later.
Take a look at this
"""
Optimizing
sum(x[i][j][t] + x[j][i][t]) ==1 where j in N \ {i} for each i in N and for each t in T
sum(x[i][j][t] + x[j][i][t]) ==2 where t in T for each i,j in N, i < j
programmer Michael Gibbs
"""
import pulp
N = [1,2]
T = [1,2]
model = pulp.LpProblem("basis")
# variables
# x[i][j][t]
x = {
i:{
j:{
t:pulp.LpVariable(
'x_' + str(i) + '_' + str(j) + '_' + str(t),
cat=pulp.LpBinary
)
for t in T
}
for j in N if j != i
}
for i in N
}
# constraints
#sum(x[i][j][t] + x[j][i][t]) ==1 where j in N \ {i} for each i in N and for each t in T
for i in N:
for t in T:
c = pulp.lpSum([x[i][j][t] + x[j][i][t] for j in N if j != i]) == 1
model += c
#sum(x[i][j][t] + x[j][i][t]) ==2 where t in T for each i,j in N, i < j
for i in N:
for j in N:
if i < j:
c = pulp.lpSum([x[i][j][t] + x[j][i][t] for t in T]) == 2
model += c
# no objective
model.solve()
print("i","j","t","value")
print('-------------------')
[print(i,j,t, pulp.value(x[i][j][t])) for i in N for j in N if i != j for t in T]
I wrote the following code for a HackerRank problem involving repeated FOR loops with very similar syntax:
x, y = map(int, input().split())
inp = []
for _ in range(x):
inp.append(list(map(int, input().split())))
#I get the rest of the input as a nested list.
while len(inp) < 7:
inp.append([1, 0])
#Adding dummy values in the list
list = []
for a in inp[0][1:inp[0][0]+1]:
for b in inp[1][1:inp[1][0]+1]:
for c in inp[2][1:inp[2][0] + 1]:
for d in inp[3][1:inp[3][0] + 1]:
for e in inp[4][1:inp[4][0] + 1]:
for f in inp[5][1:inp[5][0] + 1]:
for g in inp[6][1:inp[6][0] + 1]:
list.append((a ** 2 + b ** 2 + c ** 2 + d ** 2 + e ** 2 + f ** 2 + g ** 2) % y)
#Given function
print(max(list))
I wonder if there's any way to do this in one go.
PS: I'm an absolute beginner in programming.
Use itertools.product can simplify the loops a little bits:
import itertools as it
K, M = map(int, input().strip().split())
# reading the K lines and appending lists to 'L'
L = []
for i in range(K):
lst = list(map(int, input().strip().split()))
L.append(lst[1:])
# Looping through Cartesian product
MAX = -1
for i in it.product(*L):
MAX = max(sum(map(lambda x: x**2, i))% M, MAX)
print(MAX)
I have a problem,
where I would like to copy values of X a ( n x n ) array into smaller array A, B , C, D
here is my current code.
X = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
n = len(X)
n_by_2 = n // 2
A = [[]]
B = [[]]
C = [[]]
D = [[]]
"""
X = [
A<-- B<--
[1,2, | 3,4],
[5,6, | 7,8],
-------|-------
D<-- | C<--
[9,10, | 11,12],
[13,14,| 15,16]
]
"""
for i in range(len(X)):
for j in range(len(X[i])):
if (i >= 0 and i < n_by_2) and (j >= 0 and j < n_by_2):
#copy 1st quadrent values to A
#A[i][j] = X[i][j]
if (i >= 0 and i < n_by_2) and (j >= n_by_2 and j < n):
#copy 2nd quadrent values to B
#B[i][j] = X[i][j]
if (i >= n_by_2 and i < n) and (j >= 0 and j < n_by_2):
#copy 3rd quadrent values to C
#C[i][j] = X[i][j]
if (i >= n_by_2 and i < n) and (j >= n_by_2 and j < n):
#copy 4th quadrent values to D
#D[i][j] = X[i][j]
Be careful that you cannot add a new item x to an empty list l by doing.
l[0] = x. This only works if the list already has a length greater than 0. In that situation, you can use the append method instead.
For your problem, this code will do the job:
X = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
n = len(X)
n_by_2 = n // 2
A = []
B = []
C = []
D = []
for i in range(n_by_2):
A.append(X[i][:n_by_2])
B.append(X[i][n_by_2-1:-1])
for i in range(n_by_2):
C.append(X[i + n_by_2][:n_by_2])
D.append(X[i + n_by_2][n_by_2-1:-1])
print(A)
print(B)
print(C)
print(D)
This question was asked in a challenge in HackerEarth:
Mark is solving an interesting question. He needs to find out number
of distinct ways such that
(i + 2*j+ k) % (x + y + 2*z) = 0, where 1 <= i,j,k,x,y,z <= N
Help him find it.
Constraints:
1<= T<= 10
1<=N<= 1000
Input Format:
First line contains T, the number of test cases. Each of the test case
contains a single integer,N in a separate line.
Output Format:
For each test case , output in a separate line, the number of distinct
ways.
Sample Input
2
1
2
Sample Output
1
15
Explanation
In the first case, the only possible way is i = j = k = x =y = z = 1
I am not getting any way how to solve this problem, I have tried one and I know it's not even close to the question.
import random
def CountWays (N):
# Write your code here
i = random.uniform(1,N)
j = random.uniform(1,N)
k = random.uniform(1,N)
x = random.uniform(1,N)
y = random.uniform(1,N)
z = random.uniform(1,N)
d = 0
for i in range(N):
if (i+2*j+k)%(x+y+2*z)==0:
d += 1
return d
T = int(input())
for _ in range(T):
N = int(input())
out_ = CountWays(N)
print (out_)
My Output
0
0
Instead it should give the output
1
15
The value of the numerator (num) can range from 4 to 4N. The value of the denominator (dom) can range from 4 to num. You can split your problem into two smaller problems: 1) How many values of the denominator is a given value of the numerator divisible by? 2) How many ways can a given denominator and numerator be constructed?
To answer 1) we can simply loop through all the possible values of the numerator, then loop over all the values of the denominator where numerator % denominator == 0. To answer 2) we can find all the partitions of the numerator and denominator that satisfies the equality and constraints. The number of ways to construct a given numerator and denominator will be the product of the number of partitions of each.
import itertools
def divisible_numbers(n):
"""
Get all numbers with which n is divisible.
"""
for i in range(1,n+1):
if n % i == 0:
yield i
if i >= n:
break
def get_partitions(n):
"""
Generate ALL ways n can be partitioned into 3 integers.
Modified from http://code.activestate.com/recipes/218332-generator-for-integer-partitions/#c9
"""
a = [1]*n
y = -1
v = n
while v > 0:
v -= 1
x = a[v] + 1
while y >= 2 * x:
a[v] = x
y -= x
v += 1
w = v + 1
while x <= y:
a[v] = x
a[w] = y
if w == 2:
yield a[:w + 1]
x += 1
y -= 1
a[v] = x + y
y = a[v] - 1
if w == 3:
yield a[:w]
def get_number_of_valid_partitions(num, N):
"""
Get number of valid partitions of num, given that
num = i + j + 2k, and that 1<=i,j,k<=N
"""
n = 0
for partition in get_partitions(num):
# This can be done a bit more cleverly, but makes
# the code extremely complicated to read, so
# instead we just brute force the 6 combinations,
# ignoring non-unique permutations using a set
for i,j,k in set(itertools.permutations(partition)):
if i <= N and j <= N and k <= 2*N and k % 2 == 0:
n += 1
return n
def get_number_of_combinations(N):
"""
Get number of ways the equality can be solved under the given constraints
"""
out = 0
# Create a dictionary of number of valid partitions
# for all numerator values we can encounter
n_valid_partitions = {i: get_number_of_valid_partitions(i, N) for i in range(1,4*N+1)}
for numerator in range(4,4*N+1):
numerator_permutations = n_valid_partitions[numerator]
for denominator in divisible_numbers(numerator):
denominator_permutations = n_valid_partitions[denominator]
if denominator < 4:
continue
out += numerator_permutations * denominator_permutations
return out
N = 2
out = get_number_of_combinations(N)
print(out)
The scaling of the code right now is very poor due to the way the get_partitions and the get_number_of_valid_partitions functions interact.
EDIT
The following code is much faster. There's a small improvement to divisible_numbers, but the main speedup lies in get_number_of_valid_partitions not creating a needless amount of temporary lists as it has now been joined with get_partitions in a single function. Other big speedups comes from using numba. The code of get_number_of_valid_partitions is all but unreadable now, so I've added a much simpler but slightly slower version named get_number_of_valid_partitions_simple so you can understand what is going on in the complicated function.
import numba
#numba.njit
def divisible_numbers(n):
"""
Get all numbers with which n is divisible.
Modified from·
"""
# We can save some time by only looking at
# values up to n/2
for i in range(4,n//2+1):
if n % i == 0:
yield i
yield n
def get_number_of_combinations(N):
"""
Get number of ways the equality can be solved under the given constraints
"""
out = 0
# Create a dictionary of number of valid partitions
# for all numerator values we can encounter
n_valid_partitions = {i: get_number_of_valid_partitions(i, N) for i in range(4,4*N+1)}
for numerator in range(4,4*N+1):
numerator_permutations = n_valid_partitions[numerator]
for denominator in divisible_numbers(numerator):
if denominator < 4:
continue
denominator_permutations = n_valid_partitions[denominator]
out += numerator_permutations * denominator_permutations
return out
#numba.njit
def get_number_of_valid_partitions(num, N):
"""
Get number of valid partitions of num, given that
num = i + j + 2l, and that 1<=i,j,l<=N.
"""
count = 0
# In the following, k = 2*l
#There's different cases for i,j,k that we can treat separately
# to give some speedup due to symmetry.
#i,j can be even or odd. k <= N or N < k <= 2N.
# Some combinations only possible if num is even/odd
# num is even
if num % 2 == 0:
# i,j odd, k <= 2N
k_min = max(2, num - 2 * (N - (N + 1) % 2))
k_max = min(2 * N, num - 2)
for k in range(k_min, k_max + 1, 2):
# only look at i<=j
i_min = max(1, num - k - N + (N + 1) % 2)
i_max = min(N, (num - k)//2)
for i in range(i_min, i_max + 1, 2):
j = num - i - k
# if i == j, only one permutations
# otherwise two due to symmetry
if i == j:
count += 1
else:
count += 2
# i,j even, k <= N
# only look at k<=i<=j
k_min = max(2, num - 2 * (N - N % 2))
k_max = min(N, num // 3)
for k in range(k_min, k_max + 1, 2):
i_min = max(k, num - k - N + N % 2)
i_max = min(N, (num - k) // 2)
for i in range(i_min, i_max + 1, 2):
j = num - i - k
if i == j == k:
# if i == j == k, only one permutation
count += 1
elif i == j or i == k or j == k:
# if only two of i,j,k are the same there are 3 permutations
count += 3
else:
# if all differ, there are six permutations
count += 6
# i,j even, N < k <= 2N
k_min = max(N + 1 + (N + 1) % 2, num - 2 * N)
k_max = min(2 * N, num - 4)
for k in range(k_min, k_max + 1, 2):
# only look for i<=j
i_min = max(2, num - k - N + 1 - (N + 1) % 2)
i_max = min(N, (num - k) // 2)
for i in range(i_min, i_max + 1, 2):
j = num - i - k
if i == j:
# if i == j, only one permutation
count += 1
else:
# if all differ, there are two permutations
count += 2
# num is odd
else:
# one of i,j is even, the other is odd. k <= N
# We assume that j is odd, k<=i and correct the symmetry in the counts
k_min = max(2, num - 2 * N + 1)
k_max = min(N, (num - 1) // 2)
for k in range(k_min, k_max + 1, 2):
i_min = max(k, num - k - N + 1 - N % 2)
i_max = min(N, num - k - 1)
for i in range(i_min, i_max + 1, 2):
j = num - i - k
if i == k:
# if i == j, two permutations
count += 2
else:
# if i and k differ, there are four permutations
count += 4
# one of i,j is even, the other is odd. N < k <= 2N
# We assume that j is odd and correct the symmetry in the counts
k_min = max(N + 1 + (N + 1) % 2, num - 2 * N + 1)
k_max = min(2 * N, num - 3)
for k in range(k_min, k_max + 1, 2):
i_min = max(2, num - k - N + (N + 1) % 2)
i_max = min(N, num - k - 1)
for i in range(i_min, i_max + 1, 2):
j = num - i - k
count += 2
return count
#numba.njit
def get_number_of_valid_partitions_simple(num, N):
"""
Simpler but slower version of 'get_number_of_valid_partitions'
"""
count = 0
for k in range(2, 2 * N + 1, 2):
for i in range(1, N + 1):
j = num - i - k
if 1 <= j <= N:
count += 1
return count
if __name__ == "__main__":
N = int(sys.argv[1])
out = get_number_of_combinations(N)
print(out)
The current issue with your code is that you've picked random numbers once, then calculate the same equation N times.
I assume you wanted to generate 1..N for each individual variable, which would require 6 nested loops from 1..N, for each variable
Now, that's the brute force solution, which probably fails on large N values, so as I commented, there's some trick to find the multiples of the right side of the modulo, then check if the left side portion is contained in that list. That would only require two triple nested lists, I think
(i + 2*j+ k) % (x + y + 2*z) = 0, where 1 <= i,j,k,x,y,z <= N
(2*j + i + k) is a multiple of (2*z + x + y)
N = 2
min(2*j + i + k) = 4
max(2*j + i + k) = 8
ways to make 4: 1 * 1 = 1
ways to make 5: 2 * 2 = 4
ways to make 6: 2 * 2 = 4
ways to make 7: 2 * 2 = 4
ways to make 8: 1 * 1 = 1
Total = 14
But 8 is a multiple of 4 so we add one more instance for a total of 15.
Here's the Problem Statement:
Given an array nums of n integers,
are there elements a, b, c in nums such that a + b + c = 0?
Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[[-1, 0, 1],[-1, -1, 2]]
I'm solving Leetcode 3Sum problem right now and getting Time Limit Exceeded error for the below Code:
class Solution:
def threeSum(self, nums):
triplets=[]
nums.sort()
for i in range(len(nums)-1):
l=i+1
r=len(nums)-1
while l<r:
sum=nums[i]+nums[l]+nums[r]
if sum==0:
if not [nums[i],nums[l],nums[r]] in triplets:
triplets+=[[nums[i],nums[l],nums[r]]]
if sum<0:
l+=1
else:
r-=1
return triplets
Can anyone tell me where can I optimize this code?
Your algorithm looks optimal in general (slightly better complexity exists but approach perhaps is too complex for practical purposes).
But seems that searching of sublist in list is rather slow operation (probably linear for unsorted)
Use dictionary instead and extract triplets at the end.
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
result = []
N = len(nums)
for i in range(N):
if i > 0 and nums[i] == nums[i-1]:
continue
target = -nums[i]
l = i + 1
r = N - 1
while l < r:
if nums[l] + nums[r] == target:
result.append([nums[l],nums[r],nums[i]])
l = l + 1
while l <= r and nums[l] == nums[l - 1]:
l = l + 1
elif nums[l] + nums[r] > target:
r = r - 1
else:
l = l + 1
return result