📅  最后修改于: 2023-12-03 15:42:04.203000             🧑  作者: Mango
计算两个二进制字符串重新排列后,能够形成不同XOR的数量。例如,对于字符串s1 = "1100"和s2 = "1010",可以将s1重新排列为"0011",两个字符串 XOR 后结果为"1001",因此只有一个不同的XOR 可以被形成。
我们可以通过首先将两个字符串中的字符出现次数进行比较,然后计算可以形成不同 XOR 的数量。具体步骤如下:
from collections import Counter
def count_bits(s):
counter = Counter(s)
return counter["0"], counter["1"]
s1 = "1100"
s2 = "1010"
zeros_1, ones_1 = count_bits(s1)
zeros_2, ones_2 = count_bits(s2)
diff_zeros = abs(zeros_1 - zeros_2)
diff_ones = abs(ones_1 - ones_2)
total_diff = 0
if diff_zeros % 2 != 0:
total_diff += 1
if diff_ones % 2 != 0:
total_diff += 1
print(f"There are {total_diff} different XORs can be formed.")
from collections import Counter
def count_bits(s):
counter = Counter(s)
return counter["0"], counter["1"]
s1 = "1100"
s2 = "1010"
zeros_1, ones_1 = count_bits(s1)
zeros_2, ones_2 = count_bits(s2)
diff_zeros = abs(zeros_1 - zeros_2)
diff_ones = abs(ones_1 - ones_2)
total_diff = 0
if diff_zeros % 2 != 0:
total_diff += 1
if diff_ones % 2 != 0:
total_diff += 1
print(f"There are {total_diff} different XORs can be formed.")
本文介绍了如何通过重新排列两个二进制字符串形成的不同XOR的计数。我们首先统计每个字符出现的次数,然后根据出现次数的奇偶性来确定可以形成不同XOR的数量。这个方法可以很容易地扩展到更多的二进制字符串。