给定两个二进制字符串s和t 。任务是以这样的方式重新排列字符串s ，使得字符串t作为s 中的子字符串的出现次数最多。
例子：
Input: s = “101101”, t = “110”
Output: 110110
Input: s = “10”, t = “11100”
Output: 10
Input: s = “11000100”, t = “101”
Output: 10101000
处理方法：如果我们不能使字符串T发生任何字符串s，然后输出S中的任意排列。否则，以t开始重新排列的字符串。现在我们将在字符串s中使字符串t的下一次出现尽可能靠左，以最大化出现次数。为了实现这一目标，我们会发现，字符串T前缀匹配字符串T最大的后缀
相同的长度。它可以通过 Prefix 函数、 Z-Function 或 Hashes 找到。
下面是上述方法的实现：
# Python3 implementation of the approach
from collections import defaultdict
# Function to return the length of maximum
# proper suffix which is also proper prefix
def Prefix_Array(t):
m = len(t)
arr =[-1]*m
k =-1
for i in range(1, m):
while k>-1 and t[k + 1]!= t[i]:
k = arr[k]
if t[k + 1]== t[i]:
k+= 1
arr[i]= k
return arr[-1]
# Function to return the rearranged string
def Rearranged(ds, dt):
check = Prefix_Array(t)
# If there is no proper suffix which is
# also a proper prefix
if check ==-1:
if ds['1']0:
temp+='1'
ds['1']-= 1
while ds['0']>0:
temp+='0'
ds['0']-= 1
# Return the rearranged string
return temp
# If count of 0's in string t is 0
if dt['0']== 0 and ds['1']!= 0:
n = ds['1']//dt['1']
temp = t * n
ds['1']-= n * dt['1']
while ds['1']>0:
temp+='1'
ds['1']-= 1
while ds['0']>0:
temp+='0'
ds['0']-= 1
# Return the rearranged string
return temp
# If both 1's and 0's are present in
# string t
m1 = ds['1']//dt['1']
m2 = ds['0']//dt['0']
n = min(m1, m2)
temp = t * n
ds['1']-= n * dt['1']
ds['0']-= n * dt['0']
while ds['1']>0:
temp+='1'
ds['1']-= 1
while ds['0']>0:
temp+='0'
ds['0']-= 1
return temp
# If there is a suffix which is
# also a prefix in string t
else:
if ds['1']0:
temp+='1'
ds['1']-= 1
while ds['0']>0:
temp+='0'
ds['0']-= 1
return temp
# If count of 1's in string r is 0
if dr['1']== 0 and ds['0']!= 0:
n = ds['0']//dr['0']
temp+= r * n
ds['0']-= n * dr['0']
while ds['1']>0:
temp+='1'
ds['1']-= 1
while ds['0']>0:
temp+='0'
ds['0']-= 1
return temp
# If count of 0's in string r is 0
if dr['0']== 0 and ds['1']!= 0:
n = ds['1']//dr['1']
temp+= r * n
ds['1']-= n * dr['1']
while ds['1']>0:
temp+='1'
ds['1']-= 1
while ds['0']>0:
temp+='0'
ds['0']-= 1
return temp
# If string r have both 0's and 1's
m1 = ds['1']//dr['1']
m2 = ds['0']//dr['0']
n = min(m1, m2)
temp+= r * n
ds['1']-= n * dr['1']
ds['0']-= n * dr['0']
while ds['1']>0:
temp+='1'
ds['1']-= 1
while ds['0']>0:
temp+='0'
ds['0']-= 1
return temp
# Driver code
if __name__=="__main__":
s ="10101000"
t ="101"
# Count of 0's and 1's in string s
ds = defaultdict(int)
# Count of 0's and 1's in string t
dt = defaultdict(int)
for i in s:
ds[i]= ds[i]+1
for i in t:
dt[i]= dt[i]+1
print(Rearranged(ds, dt))
输出：
10101000
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