I believe the online judge for this task has expired. Given my proposed solution below, is it logically sound? Can we do better in terms of time or space complexity? What would a practical brute-force method look like for comparison?
Task:
Given an array of length n, find the minimum number of elements that would need to be changed so that the XOR of every k consecutive elements is 0.
Constraints:
1 ≤ k ≤ n ≤ 10^4
0 ≤ A[i] < 1024
Proposed solution:
Say we have an optimal selection for the first k elements. To update the current window to the next contiguous k elements, we remove the contribution of the first element and XOR with the next proposed element. To remove the contribution of the first element, we XOR with it, which means the only option to make the next window XOR to zero is to XOR with the element we just removed. This means the optimal first k elements must continue to repeat throughout.
e1, e2, e3,...ek, e1, e2, e3,...ek, etc.
Lets call each sequence of elements A[i], A[i+k], A[i+2*k]... that must be equal to one another, seq(i). We can calculate a minimal upper bound on the number of changes needed by noting that if just one seq(i) is allowed to have its elements set to any one arbitrary element, we can incur that cost and make any selection for the remaining seq(i)s a viable solution, including the selection with least cost for each.
To try and do better than the minimal upper bound, we rule out using any arbitrary assignment, so all target options for each seq(i) must come from the set seq(i) itself. Furthermore, as we iterate, we can use the minimal upper bound to rule out any XOR prefix that costs as much or more.
Time complexity O(k * n/k * 1024) = O(n). Space complexity O(n).
Python 3 example:
from collections import defaultdict
from math import ceil
A = [1, 2, 3, 1, 4]
k = 3
n = len(A)
seqs = [None] * k
for i in range(k):
seqs[i] = defaultdict(lambda: 0)
for j in range(i, n, k):
seqs[i][A[j]] += 1
def cost(i, e):
return ceil((n - i) / k) - seqs[i][e]
def min_cost(i):
return min([cost(i, e) for e in seqs[i]])
total_min_cost = sum([min_cost(i) for i in range(k)])
upper_bound = total_min_cost + min([ceil((n - i) / k) - min_cost(i) for i in range(k)])
dp = {0: 0}
for i in range(k):
new_dp = defaultdict(lambda: float('inf'))
for e in seqs[i]:
for xor_pfx in dp:
new_cost = cost(i, e) + dp[xor_pfx]
if new_cost < upper_bound:
new_pfx = xor_pfx ^ e
new_dp[new_pfx] = min(new_dp[new_pfx], new_cost)
dp = new_dp
result = dp[0] if 0 in dp else upper_bound
print(result)