Program to make the XOR of all segments equal to zero in Python

Suppose we have an array called nums and another value k. The XOR of a segment [left, right] (left

We have to find the minimum number of elements to change in the array such that the XOR of all segments of size k is same as zero.

So, if the input is like nums = [3,4,5,2,1,7,3,4,7], k = 3, then the output will be 3 because we can modify elements at indices 2, 3, 4 to make the array [3,4,7,3,4,7,3,4,7].

To solve this, we will follow these steps −

• LIMIT := 1024

• temp := make an array whose size is LIMIT x k, and fill with 0

• for each index i and value x in nums, do

  • temp[i mod k, x] := temp[i mod k, x] + 1

• dp := an array of size LIMIT and fill with -2000

• dp[0] := 0

• for each row in temp, do

  • maxprev := maximum of dp

  • new_dp := an array of size LIMIT and fill with maxprev

  • for each index i and value cnt row, do

    • if cnt > 0, then

      • for each index j and value prev in dp, do

        • new_dp[i XOR j] := maximum of new_dp[i XOR j] and prev+cnt

  • dp := new_dp

• return size of nums - new_dp[0]

Example

Let us see the following implementation to get better understanding

def solve(nums, k):
   LIMIT = 2**10
   temp = [[0 for _ in range(LIMIT)] for _ in range(k)]
   for i,x in enumerate(nums):
      temp[i%k][x] += 1

   dp = [-2000 for _ in range(LIMIT)]
   dp[0] = 0
   for row in temp:
      maxprev = max(dp)
      new_dp = [maxprev for _ in range(LIMIT)]
      for i,cnt in enumerate(row):
         if cnt > 0:
            for j,prev in enumerate(dp):
               new_dp[i^j] = max(new_dp[i^j], prev+cnt)
         dp = new_dp
   return len(nums) - new_dp[0]

nums = [3,4,5,2,1,7,3,4,7]
k = 3
print(solve(nums, k))

Input

[3,4,5,2,1,7,3,4,7], 3

Output

-9