Given a binary string, task is to erase exactly one integer in the array so that the XOR of the remaining numbers is zero. The task is to count number of ways to remove one element so that XOR of that string become ZERO.
Input : 10000 Output : 1 We only have 1 ways to Input : 10011 Output : 3 There are 3 ways to make XOR 0. We can remove any of the three 1's. Input : 100011100 Output : 5 There are 5 ways to make XOR 0. We can remove any of the give 0's
A simple solution is to one by one remove an element, then compute XOR of remaining string. And count occurrences where removing an element makes XOR 0.
An efficient solution is based on following fact. If count of 1s is odd, then we must remove a 1 to make count 0 and we can remove any of the 1s. If count of 1s is even, then XOR is 0, We can remove any of the 0s and XOR will remain 0.
# Python 3 program to count number of
# ways to remove an element so that
# XOR of remaining string becomes 0.
# Return number of ways in which XOR
# become ZERO by remove 1 element
one_count = 0
zero_count = 0
n = len(str)
# Counting number of 0 and 1
for i in range(0, n, 1):
if (str[i] == ‘1’):
one_count += 1
zero_count += 1
# If count of ones is even
# then return count of zero
# else count of one
if (one_count % 2 == 0):
# Driver Code
if __name__ == ‘__main__’:
str = “11111”
# This code is contributed by
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