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Max Xor Subarray of size K
Views: 4 | Author: prakhardoneria
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | class Solution: def maxSubarrayXOR(self, arr, k): n = len(arr) current_xor = 0 # 1. Calculate the XOR of the first window of size k for i in range(k): current_xor ^= arr[i] max_xor = current_xor # 2. Slide the window across the rest of the array for i in range(k, n): # XOR out the element leaving the window and XOR in the new element current_xor = (current_xor ^ arr[i-k]) ^ arr[i] # Update max_xor if the current window's XOR is higher if current_xor > max_xor: max_xor = current_xor return max_xor |