Can someone explain how the if statement is working or the meaning of the code in the following code?
I have two lists, A and B, and I need to see if there exists a pair of elements, one from A the other from B, such that swapping them will make the sum of both lists equal.
My method, O(n^2) is to find the sumOfA and sumOfB.
Find the halfdiff = (sumOfA - sumOfB)/2
For each element in A, see if there's a B[i] so that (A[j] - B[i]) = halfdiff.
But the following code does it in O(n+m). And I don't understand the meaning of "if" statement (LINE 11) here. Does it guarantee that if it is true we have the required pair?
1 def fast_solution(A, B, m):
2 n = len(A)
3 sum_a = sum(A)
4 sum_b = sum(B)
5 d = sum_b - sum_a
6 if d % 2 == 1:
7 return False
8 d //= 2
9 count = counting(A, m)
10 for i in xrange(n):
11 if 0 <= B[i] - d and B[i] - d <= m and count[B[i] - d] > 0:
12 return True
13 return False
You have to find i, j such that sum(A) - a[i] + b[j] = sum(B) - b[j] + a[i], or equivalently, sum(A) - 2*a[i] = sum(B) - 2*b[j].
You can do this by calculating all possible results of the right-hand-side, and then searching through possible i values.
def exists_swap(A, B):
sumA = sum(A)
sumB = sum(B)
bVals = set(sumB - 2 * bj for bj in B)
return any(sumA - 2 * ai in bVals for ai in A)
The partial code in your question is doing a similar thing, except d = (sum(B)-sum(A))/2 and count is itertools.Counter(A) (that is, it's a dict that maps any x to the number of times it appears in A). Then count[B[i] - d] > 0 is equivalent to there being a j such that B[i] - d = A[j], or B[i] - A[j] = (sum(B) - sum(A))/2.
It may be that instead of using sets or dicts, the value m is the maximum value allowed in A and B. Then counting could be defined like this:
def counting(xs, m):
r = [0] * (m+1)
for x in xs:
r[x] += 1
return r
This is a simple but inefficient way to represent a set of integers, but it makes sense of the missing parts of your question and explains the bounds checking 0 <= B[i] - d and B[i] - d <= m which is unnecessary if you use a set or dict, but necessary if counting returns an array.
Actually, it's not O(n+m). Linear estimation is just amortized because of hashmap count usage. This knowledge may help you to understand that your code is an obfuscated version of
bool solve(A,B) {
sum_a = sum(A)
sum_b = sum(B)
sort(B)
for(val in A)
if( binary_search(B, val - (sum_b - sum_a)/2 ) )
return true
return false
}
As Paul pointed out, 0 <= B[i] - d and B[i] - d <= m is just a validation of count argument. BTW his solution is purely linear, well implemented and much simplier to understand.
Related
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]
I have a function that takes three arrays and runs a few lines of code to interpolate the data.
It looks something like this:
def interpolate(a, b, c):
# Avoid crashing if a > max(b)
xm_ = b[-1]
a[a > xm_] = xm_
i = np.arange(a.size)
j = np.searchsorted(b, a) - 1
d = (a - b[j]) / (b[j + 1] - b[j])
return (1 - d) * c[i, j] + c[i, j + 1] * d
Here are the shapes of the arrays:
a is (163080,)
b is (71,)
c is (162080, 71)
I keep running into the IndexError. I have tried to extend the b with higher values to try to not hit max values. I have also tried to use up to fourth last index when assigning xm_.
Any suggestions to what might be happening here? And potentially how to solve it?
Cheers!
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]
The rating for Alice's challenge is the triplet a = (a[0], a[1], a[2]), and the rating for Bob's challenge is the triplet b = (b[0], b[1], b[2]).
The task is to find their comparison points by comparing a[0] with b[0], a[1] with b[1], and a[2] with b[2].
If a[i] > b[i], then Alice is awarded 1 point.
If a[i] < b[i], then Bob is awarded 1 point.
If a[i] = b[i], then neither person receives a point.
Below is the function body:
# The function is expected to return an INTEGER_ARRAY.
# The function accepts following parameters:
# 1. INTEGER_ARRAY a
# 2. INTEGER_ARRAY b
#
def compareTriplets(a, b):
# Write your code here
c=0
d=0
for i in range(0,len(a)-1):
if(a[i]>b[i]):
c+=1
elif(a[i]<b[i]):
d+=1
return array.array("i",[c,d])
First of all, not sure why you're iterating with range(0,len(a)-1) - you are skipping the last element. It should be range(len(a)).
Second, it is more "pythonic" in these cases to use the zip function instead of using indexes. So you could do:
def compareTriplets(alice, bob):
alice_score = 0
bob_score = 0
for a, b in zip(alice, bob):
if a > b:
alice_score += 1
elif a < b:
bob_score += 1
return array.array("i", [alice_score, bob_score])
If you're looking for shorter code (note that it doesn't mean better or faster code!), you could use the sum function:
def compareTriplets(alice, bob):
alice_score = sum(a > b for a, b in zip(alice, bob))
bob_score = sum(a < b for a, b in zip(alice, bob))
return array.array("i", [alice_score, bob_score])
As you can see this loops over the zip twice. For two tuples of three elements this is insignificant, but it is important to take in count that once you're dealing with big data, this approach is not ideal.
Here is the comprehension for ya:
def compareTriplets(a, b):
return [sum(tot) for tot in zip(*((aa > bb, aa < bb) for aa, bb in zip(a, b)))]
e.g.
a = 'abc123def'
b = 'abcdef'
I want a function which can judge whether b in a.
contains(a,b)=True
p.s. gap is also allowed in the represention of b, e.g.
b='abc_def'
but regular expressions are not allowed.
If what you want to do is to check whether b is a subsequence of a, you can write:
def contains(a, b):
n, m = len(a), len(b)
j = 0
for i in range(n):
if j < m and a[i] == b[j]:
j += 1
return j == m
Try using list comprehension:
def contains(main_string, sub_string):
return all([i in main_string for i in sub_string])
NOTE: 'all' is a builtin function which takes an iterable of booleans and returns try if all are True.
def new_contained(a,b):
boo = False
c = [c for c in a]
d = [i for i in b]
if len(c)<=len(d):
for i in c:
if i in d:
boo = True
return boo