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Taking equal number of elements from two arrays, such that the taken values have as few duplicates as possible

Consider we have 2 arrays of size N, with their values in the range [0, N-1]. For example:
a = np.array([0, 1, 2, 0])
b = np.array([2, 0, 3, 3])
I need to produce a new array c which contains exactly N/2 elements from a and b respectively, i.e. the values must be taken evenly/equally from both parent arrays.
(For odd length, this would be (N-1)/2 and (N+1)/2. Can also ignore odd length case, not important).
Taking equal number of elements from two arrays is pretty trivial, but there is an additional constraint: c should have as many unique numbers as possible / as few duplicates as possible.
For example, a solution to a and b above is:
c = np.array([b[0], a[1], b[2], a[3]])
>>> c
array([2, 1, 3, 0])
Note that the position/order is preserved. Each element of a and b that we took to form c is in same position. If element i in c is from a, c[i] == a[i], same for b.
A straightforward solution for this is simply a sort of path traversal, easy enough to implement recursively:
def traverse(i, a, b, path, n_a, n_b, best, best_path):
if n_a == 0 and n_b == 0:
score = len(set(path))
return (score, path.copy()) if score > best else (best, best_path)
if n_a > 0:
path.append(a[i])
best, best_path = traverse(i + 1, a, b, path, n_a - 1, n_b, best, best_path)
path.pop()
if n_b > 0:
path.append(b[i])
best, best_path = traverse(i + 1, a, b, path, n_a, n_b - 1, best, best_path)
path.pop()
return best, best_path
Here n_a and n_b are how many values we will take from a and b respectively, it's 2 and 2 as we want to evenly take 4 items.
>>> score, best_path = traverse(0, a, b, [], 2, 2, 0, None)
>>> score, best_path
(4, [2, 1, 3, 0])
Is there a way to implement the above in a more vectorized/efficient manner, possibly through numpy?

The algorithm is slow mainly because it runs in an exponential time. There is no straightforward way to vectorize this algorithm using only Numpy because of the recursion. Even if it would be possible, the huge number of combinations would cause most Numpy implementations to be inefficient (due to large Numpy arrays to compute). Additionally, there is AFAIK no vectorized operation to count the number of unique values of many rows efficiently (the usual way is to use np.unique which is not efficient in this case and cannot be use without a loop). As a result, there is two possible strategy to speed this up:
trying to find an algorithm with a reasonable complexity (eg. <= O(n^4));
using compilation methods, micro-optimizations and tricks to write a faster brute-force implementation.
Since finding a correct sub-exponential algorithm turns out not to be easy, I choose the other approach (though the first approach is the best).
The idea is to:
remove the recursion by generating all possible solutions using a loop iterating on integer;
write a fast way to count unique items of an array;
use the Numba JIT compiler so to optimize the code that is only efficient once compiled.
Here is the final code:
import numpy as np
import numba as nb
# Naive way to count unique items.
# This is a slow fallback implementation.
#nb.njit
def naive_count_unique(arr):
count = 0
for i in range(len(arr)):
val = arr[i]
found = False
for j in range(i):
if arr[j] == val:
found = True
break
if not found:
count += 1
return count
# Optimized way to count unique items on small arrays.
# Count items 2 by 2.
# Fast on small arrays.
#nb.njit
def optim_count_unique(arr):
count = 0
for i in range(0, len(arr), 2):
if arr[i] == arr[i+1]:
tmp = 1
for j in range(i):
if arr[j] == arr[i]: tmp = 0
count += tmp
else:
val1, val2 = arr[i], arr[i+1]
tmp1, tmp2 = 1, 1
for j in range(i):
val = arr[j]
if val == val1: tmp1 = 0
if val == val2: tmp2 = 0
count += tmp1 + tmp2
return count
#nb.njit
def count_unique(arr):
if len(arr) % 2 == 0:
return optim_count_unique(arr)
else:
# Odd case: not optimized yet
return naive_count_unique(arr)
# Count the number of bits in a 32-bit integer
# See https://stackoverflow.com/questions/71097470/msb-lsb-popcount-in-numba
#nb.njit('int_(uint32)', inline='always')
def popcount(v):
v = v - ((v >> 1) & 0x55555555)
v = (v & 0x33333333) + ((v >> 2) & 0x33333333)
c = np.uint32((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24
return c
# Count the number of bits in a 64-bit integer
#nb.njit(inline='always')
def bit_count(n):
if n < (1 << 30):
return popcount(np.uint32(n))
else:
return popcount(np.uint32(n)) + popcount(np.uint32(n >> 32))
# Mutate `out` so not to create an expensive new temporary array
#nb.njit
def int_to_path(n, out, a, b):
for i in range(len(out)):
out[i] = a[i] if ((n >> i) & 1) else b[i]
#nb.njit(['(int32[:], int32[:], int64, int64)', '(int64[:], int64[:], int64, int64)'])
def traverse_fast(a, b, n_a, n_b):
# This assertion is needed because the paths are encoded using 64-bit.
# This should not be a problem in practice since the number of solutions to
# test would be impracticably huge to test using this algorithm anyway.
assert n_a + n_b < 62
max_iter = 1 << (n_a + n_b)
path = np.empty(n_a + n_b, dtype=a.dtype)
score, best_score, best_i = 0, 0, 0
# Iterate over all cases (more than the set of possible solution)
for i in range(max_iter):
# Filter the possible solutions
if bit_count(i) != n_b:
continue
# Analyse the score of the solution
int_to_path(i, path, a, b)
score = count_unique(path)
# Store it if it better than the previous one
if score > best_score:
best_score = score
best_i = i
int_to_path(best_i, path, a, b)
return best_score, path
This implementation is about 30 times faster on arrays of size 8 on my machine. On could use several cores to speed this up even further. However, I think it is better to focus on finding a sub-exponential implementation so to avoid wasting more computing resources. Note that the path is different from the initial function but the score is the same on random arrays. It can help others to test their implementation on larger arrays without waiting for a long time.

Test this heavily.
import numpy as np
from numpy.random._generator import default_rng
rand = default_rng(seed=1)
n = 16
a = rand.integers(low=0, high=n, size=n)
b = rand.integers(low=0, high=n, size=n)
uniques = np.setxor1d(a, b)
print(a)
print(b)
print(uniques)
def limited_uniques(arr: np.ndarray) -> np.ndarray:
choose = np.zeros(shape=n, dtype=bool)
_, idx, _ = np.intersect1d(arr, uniques, return_indices=True)
idx = idx[:n//2]
choose[idx] = True
n_missing = n//2 - len(idx)
counts = choose.cumsum()
diffs = np.arange(n) - counts
at = np.searchsorted(diffs, n_missing)
choose[:at] = True
return arr[choose]
a_half = limited_uniques(a)
uniques = np.union1d(uniques, np.setdiff1d(a, a_half))
interleaved = np.empty_like(a)
interleaved[0::2] = a_half
interleaved[1::2] = limited_uniques(b)
print(interleaved)
[ 7 8 12 15 0 2 13 15 3 4 13 6 4 13 4 6]
[10 8 1 0 13 12 13 8 13 5 7 12 1 4 1 7]
[ 1 2 3 5 6 10 15]
[ 7 10 8 8 12 1 15 0 0 13 2 12 3 5 6 4]

Maximum Score from Two Arrays | Which Test Case is this approach missing?

Problem Statement
Given two integer arrays A and B of size N and M respectively. You begin with a score of 0. You want to perform exactly K operations. On the iᵗʰ operation (1-indexed), you will:
Choose one integer x from either the start or the end of any one array, A or B. Remove it from that array
Add x to score.
Return the maximum score after performing K operations.
Example
Input: A = [3,1,2], B = [2,8,1,9] and K=5
Output: 24
Explanation: An optimal solution is as follows:
Choose from end of B, add 9 to score. Remove 9 from B
Choose from start of A, add 3 to score. Remove 3 from A
Choose from start of B, add 2 to score. Remove 2 from B
Choose from start of B, add 8 to score. Remove 8 from B
Choose from end of A, add 2 to score. Remove 2 from A
The total score is 9+3+2+8+2 = 24
Constraints
1 ≤ N ≤ 6000
1 ≤ M ≤ 6000
1 ≤ A[i] ≤ 109
1 ≤ B[i] ≤ 109
1 ≤ K ≤ N+M
My Approach
Since, greedy [choosing maximum end from both array] approach is failing here [because it will produce conflict when maximum end of both array is same], it suggests we have to look for all possible combinations. There will be overlapping sub-problems, hence DP!
Here is the python reprex code for the same.
A = [3,1,2]
N = len(A)
B = [2,8,1,9]
M = len(B)
K = 5
memo = {}
def solve(i,j, AL, BL):
if (i,j,AL,BL) in memo:
return memo[(i,j,AL,BL)]
AR = (N-1)-(i-AL)
BR = (M-1)-(j-BL)
if AL>AR or BL>BR or i+j==K:
return 0
op1 = A[AL] + solve(i+1,j,AL+1,BL)
op2 = B[BL] + solve(i,j+1,AL,BL+1)
op3 = A[AR] + solve(i+1,j,AL,BL)
op4 = B[BR] + solve(i,j+1,AL,BL)
memo[(i,j,AL,BL)] = max(op1,op2,op3,op4)
return memo[(i,j,AL,BL)]
print(solve(0,0,0,0))
In brief,
i indicates that we have performed i operations from A
j indicates that we have performed j operations from B
Total operation is thus i+j
AL indicates index on left of which which all integers of A are used. Similarly AR indicates index on right of which all integers of A used for operation.
BL indicates index on left of which which all integers of B are used. Similarly BR indicates index on right of which all integers of B used for operation.
We are trying out all possible combination, and choosing maximum from them in each step. Also memoizing our answer.
Doubt
The code worked fine for several test cases, but also failed for few. The message was Wrong Answer means there was no Time Limit Exceed, Memory Limit Exceed, Syntax Error or Run Time Error. This means there is some logical error only.
Can anyone help in identifying those Test Cases? And, also in understanding intuition/reason behind why this approach failed in some case?

Examples were posted code gives the wrong answer:
Example 1.
A = [1, 1, 1]
N = len(A)
B = [1, 1]
M = len(B)
K = 5
print(print(solve(0,0,0,0))) # Output: 4 (which is incorrect)
# Correct answer is 5
Example 2.
A = [1, 1]
B = [1]
N = len(A)
M = len(B)
K = 3
print(print(solve(0,0,0,0))) # Output: 2 (which is incorrect)
# Correct answer is 3
Alternative Code
def solve(A, B, k):
def solve_(a_left, a_right, b_left, b_right, remaining_ops, sum_):
'''
a_left - left pointer into A
a_right - right pointer in A
b_left - left pointer into B
b_right - right pointer into B
remaining_ops - remaining operations
sum_ - sum from previous operations
'''
if remaining_ops == 0:
return sum_ # out of operations
if a_left > a_right and b_left > b_right:
return sum_ # both left and right are empty
if (a_left, a_right, b_left, b_right) in cache:
return cache[(a_left, a_right, b_left, b_right)]
max_ = sum_ # init to current sum
if a_left <= a_right: # A not empty
max_ = max(max_,
solve_(a_left + 1, a_right, b_left, b_right, remaining_ops - 1, sum_ + A[a_left]), # Draw from left of A
solve_(a_left, a_right - 1, b_left, b_right, remaining_ops - 1, sum_ + A[a_right])) # Draw from right of A
if b_left <= b_right: # B not empty
max_ = max(max_,
solve_(a_left, a_right, b_left + 1, b_right, remaining_ops - 1, sum_ + B[b_left]), # Draw from left of B
solve_(a_left, a_right, b_left, b_right - 1, remaining_ops - 1, sum_ + B[b_right])) # Draw from right of B
cache[(a_left, a_right, b_left, b_right)] = max_ # update cache
return cache[(a_left, a_right, b_left, b_right)]
cache = {}
return solve_(0, len(A) - 1, 0, len(B) - 1, k, 0)
Tests
print(solve([3,1,2], [2,8,1,9], 5) # Output 24
print(solve([1, 1, 1], [1, 1, 1], 5) # Output 5

The approach is failing because the Recursive Functions stops computing further sub-problems when either "AL exceeds AR" or "BL exceeds BR".
We should stop computing and return 0 only when both of them are True. If either of "AL exceeds AR" or "BL exceeds BR" evaluates to False, means we can solve that sub-problem.
Moreover, one quick optimization here is that when N+M==K, in this case we can get maximum score by choosing all elements from both the arrays.
Here is the correct code!
A = [3,1,2]
B = [2,8,1,9]
K = 5
N, M = len(A), len(B)
memo = {}
def solve(i,j, AL, BL):
if (i,j,AL,BL) in memo:
return memo[(i,j,AL,BL)]
AR = (N-1)-(i-AL)
BR = (M-1)-(j-BL)
if i+j==K or (AL>AR and BL>BR):
return 0
ans = -float('inf')
if AL<=AR:
ans = max(A[AL]+solve(i+1,j,AL+1,BL),A[AR]+solve(i+1,j,AL,BL),ans)
if BL<=BR:
ans = max(B[BL]+solve(i,j+1,AL,BL+1),B[BR]+solve(i,j+1,AL,BL),ans)
memo[(i,j,AL,BL)] = ans
return memo[(i,j,AL,BL)]
if N+M==K:
print(sum(A)+sum(B))
else:
print(solve(0,0,0,0))
[This answer was published taking help from DarryIG's Answer. The reason for publishing answer is to write code similar to code in question body. DarryIG's answer used different prototype for function]

Create random systems of linear equations - Python

Edit: more details
Hello I found this problem through one of my teachers but I still don't understand how to approach to it, and I would like to know if anyone had any ideas for it:
Create a program capable of generating systems of equations (randomly) that contain between 2 and 8 variables. The program will ask the user for a number of variables in the system of equations using the input function. The range of the coefficients must be between [-10,10], however, no coefficient should be 0. Both the coefficients and the solutions must be integers.
The goal is to print the system and show the solution to the variables (x,y,z,...). NumPy is allowed.
As far as I understand it should work this way:
Enter the number of variables: 2
x + y = 7
4x - y =3
x = 2
y = 5
I'm still learning arrays in python, but do they work the same as in matlab?
Thank you in advance :)!

For k variables, the lhs of the equations will be k number of unknowns and a kxk matrix for the coefficients. The dot product of those two should give you the rhs. Then it's a simple case of printing that however you want.
import numpy as np
def generate_linear_equations(k):
coeffs = [*range(-10, 0), *range(1, 11)]
rng = np.random.default_rng()
return rng.choice(coeffs, size=(k, k)), rng.integers(-10, 11, k)
k = int(input('Enter the number of variables: '))
if not 2 <= k <= 8:
raise ValueError('The number of variables must be between 2 and 8.')
coeffs, variables = generate_linear_equations(k)
solution = coeffs.dot(variables)
symbols = 'abcdefgh'[:k]
for row, sol in zip(coeffs, solution):
lhs = ' '.join(f'{r:+}{s}' for r, s in zip(row, symbols)).lstrip('+')
print(f'{lhs} = {sol}')
print()
for s, v in zip(symbols, variables):
print(f'{s} = {v}')
Which for example can give
Enter the number of variables: 3
8a +6b -4c = -108
9a -9b -4c = 3
10a +10b +9c = -197
a = -9
b = -8
c = -3
If you specifically want the formatting of the lhs to have a space between the sign and to not show a coefficient if it has a value of 1, then you need something more complex. Substitute lhs for the following:
def sign(n):
return '+' if n > 0 else '-'
lhs = ' '.join(f'{sign(r)} {abs(r)}{s}' if r not in (-1, 1) else f'{sign(r)} {s}' for r, s in zip(row, symbols))
lhs = lhs[2:] if lhs.startswith('+') else f'-{lhs[2:]}'

I did this by randomly generating the left hand side and the solution within your constraints, then plugging the solutions into the equations to generate the right hand side. Feel free to ask for clarification about any part of the code.
import numpy as np
num_variables = int(input('Number of variables:'))
valid_integers = np.asarray([x for x in range(-10,11) if x != 0])
lhs = np.random.choice(valid_integers, lhs_shape)
solution = np.random.randint(-10, 11, num_variables)
rhs = lhs.dot(solution)
for i in range(num_variables):
for j in range(num_variables):
symbol = '=' if j == num_variables-1 else '+'
print(f'{lhs[i, j]:3d}*x{j+1} {symbol} ', end='')
print(rhs[i])
for i in range(num_variables):
print(f'x{i+1} = {solution[i]}'
Example output:
Number of variables:2
2*x1 + -7*x2 = -84
-4*x1 + 1*x2 = 38
x1 = -7
x2 = 10

Recurrent sequence task

Given the sequence f0, f1, f2, ... given by the recurrence relations f0 = 0, f1 = 1, f2 = 2 and fk = f (k-1) + f (k-3)
Write a program that calculates the n elements of this sequence with the numbers k1, k2, ..., kn.
Input format
The first line of the input contains an integer n (1 <= n <= 1000)
The second line contains n non-negative integers ki (0 <= ki <= 16000), separated by spaces.
Output format
Output space-separated values ​​for fk1, fk2, ... fkn.
Memory Limit: 10MB
Time limit: 1 second
The problem is that the recursive function at large values ​​goes beyond the limit.
def f (a):
    if a <= 2:
        return a
    return f (a - 1) + f (a - 3)
n = int (input ())
nums = list (map (int, input (). split ()))
for i in range (len (nums)):
    if i <len (nums) - 1:
        print (f (nums [i]), end = '')
    else:
        print (f (nums [i]))
I also tried to solve through a cycle, but the task does not go through time (1 second):
fk1 = 0
fk2 = 0
fk3 = 0
n = int (input ())
nums = list (map (int, input (). split ()))
a = []
for i in range (len (nums)):
    itog = 0
    for j in range (1, nums [i] + 1):
        if j <= 2:
            itog = j
        else:
            if j == 3:
                itog = 0 + 2
                fk1 = itog
                fk2 = 2
                fk3 = 1
            else:
                itog = fk1 + fk3
                fk1, fk2, fk3 = itog, fk1, fk2
    if i <len (nums) - 1:
        print (itog, end = '')
    else:
        print (itog)
How else can you solve this problem so that it is optimal in time and memory?

Concerning the memory, the best solution probably is the iterative one. I think you are not far from the answer. The idea would be to first check for the simple cases f(k) = k (ie, k <= 2), for all other cases k > 2 you can simply compute fi using (fi-3, fi-2, fi-1) until i = k. What you need to do during this process is indeed to keep track of the last three values (similar to what you did in the line fk1, fk2, fk3 = itog, fk1, fk2).
On the other hand, there is one thing that you need to do here. If you just perform computations of fk1, fk2, ... fkn independently, then you are screwed (unless you use a super fast machine or a Cython implementation). On the other hand, there is no reason to perform n independent computations, you can just compute fx for x = max(k1, k2, ..., kn) and on the way you'll store every answer for fk1, fk2, ..., fkn (this will slow down the computation of fx by a little bit, but instead of doing this n times you'll do it only once). This way it can be solved under 1s even for n = 1000.
On my machine, independent calculations for f15000, f15001, ..., f16000 takes roughly 30s, the "all at once" solution takes roughly 0.035s.
Honestly, that's not such an easy exercise, it would be interesting to show your solution on a site like code review to get some feedback on your solution once you found one :).

First, you have to sort the numbers. Then calculate values of the sequence one by one:
while True:
a3 = a2 + a0
a0 = a3 + a1
a1 = a0 + a2
a2 = a1 + a3
Lastly, return values in beginning order. To do that you have to remember position of every number. From [45, 22, 14, 33] make [[45,0], [22,1], [14,2], [33,3]] and then sort, calculate values and change them with argument [[f45,0], [f22,1], [f14,2], [f33,3]], then sort by second value.

Need to reduce run time on creating a list of numbers based on a formula and a number n

Need a better way to create a list of numbers, so that the run time is less. Or probably figure out a better approach to my problem.
I'm running a code to create a series of numbers based on 2 formulas. Starting from 1, the formulas create the following numbers. The idea is to return the number n from the list that is created at the end. Even tough the formulas create the same number in some cases, only unique values remain, and the list is sorted to match. I use a while loop to create the list, and I believe that reducing the number of repetitions can help with my problem, but I can't figure out a way to effectively reduce it, without ruining the purpose of my code.
def dbl_linear(n):
x = 1
y = 0
z = 0
i = 0
u = []
u.append(x)
while i <= n:
x = (u)[i]
y = 2 * x + 1
u.append(y)
z = 3 * x + 1
u.append(z)
i = i + 1
u.sort()
uFix = set(u)
uFix = list(uFix)
uFix.sort()
return uFix[n]
print(dbl_linear(50))
These are the expected results. Which I get, but it takes too long.
dbl_linear(10), 22)
dbl_linear(20), 57)
dbl_linear(30), 91)
dbl_linear(50), 175)

Your function can be considerably simplified to:
Code:
def dbl_linear(n):
u = [1]
for i in range(n):
x = u[i]
u.extend((2 * x + 1, 3 * x + 1))
return sorted(set(u))[n]
Test Code:
assert dbl_linear(10) == 22
assert dbl_linear(20) == 57
assert dbl_linear(30) == 91
assert dbl_linear(50) == 175