连接二进制字符串中的最大连续零
你得到一个长度为n的二进制字符串str 。假设您通过将 str 的k个副本连接在一起来创建另一个大小为 n * k 的字符串。仅由 0 组成的连接字符串的子字符串的最大大小是多少？鉴于 k > 1。
例子：
Input : str = “110010”, k = 2
Output : 2
String becomes 110010110010 after two concatenations. This string has two zeroes.
Input : str = “00100110”, k = 4
Output : 3
如果给定字符串全为零，则答案为 n * k。如果 S 包含 1，则答案要么是 str 的仅包含零的子串的最大长度，要么是 S 的仅包含零的最大前缀的长度与 str 的仅包含零的最大后缀的长度之和。只有当 k > 1 时才必须计算最后一个。
C++
// C++ program to find maximum number
// of consecutive zeroes after
// concatenating a binary string
#include
using namespace std;
// returns the maximum size of a
// substring consisting only of
// zeroes after k concatenation
int max_length_substring(string st,
int n, int k)
{
// stores the maximum length
// of the required substring
int max_len = 0;
int len = 0;
for (int i = 0; i < n; ++i)
{
// if the current character is 0
if (st[i] == '0')
len++;
else
len = 0;
// stores maximum length of current
// substrings with zeroes
max_len = max(max_len, len);
}
// if the whole string is
// filled with zero
if (max_len == n)
return n * k;
int pref = 0, suff = 0;
// computes the length of the maximal
// prefix which contains only zeroes
for (int i = 0; st[i] == '0';
++i, ++pref);
// computes the length of the maximal
// suffix which contains only zeroes
for (int i = n - 1; st[i] == '0';
--i, ++suff);
// if more than 1 concatenations
// are to be made
if (k > 1)
max_len = max(max_len,
pref + suff);
return max_len;
}
// Driver code
int main()
{
int n = 6;
int k = 3;
string st = "110010";
int ans = max_length_substring(st, n, k);
cout << ans;
}
// This code is contributed by ihritik Java
// Java program to find maximum number of
// consecutive zeroes after concatenating
// a binary string
class GFG {
// returns the maximum size of a substring
// consisting only of zeroes
// after k concatenation
static int max_length_substring(String st,
int n, int k)
{
// stores the maximum length of the
// required substring
int max_len = 0;
int len = 0;
for (int i = 0; i < n; ++i) {
// if the current character is 0
if (st.charAt(i) == '0')
len++;
else
len = 0;
// stores maximum length of current
// substrings with zeroes
max_len = Math.max(max_len, len);
}
// if the whole string is filled with zero
if (max_len == n)
return n * k;
int pref = 0, suff = 0;
// computes the length of the maximal
// prefix which contains only zeroes
for (int i = 0; st.charAt(i) == '0'; ++i, ++pref)
;
// computes the length of the maximal
// suffix which contains only zeroes
for (int i = n - 1; st.charAt(i) == '0'; --i, ++suff)
;
// if more than 1 concatenations are to be made
if (k > 1)
max_len = Math.max(max_len, pref + suff);
return max_len;
}
// Driver code
public static void main(String[] args)
{
int n = 6;
int k = 3;
String st = "110010";
int ans = max_length_substring(st, n, k);
System.out.println(ans);
}
}Python3
# Python3 program to find maximum
# number of consecutive zeroes
# after concatenating a binary string
# returns the maximum size of a
# substring consisting only of
# zeroes after k concatenation
def max_length_substring(st, n, k):
# stores the maximum length
# of the required substring
max_len = 0
len = 0
for i in range(0, n):
# if the current character is 0
if (st[i] == '0'):
len = len + 1;
else:
len = 0
# stores maximum length of
# current substrings with zeroes
max_len = max(max_len, len)
# if the whole is filled
# with zero
if (max_len == n):
return n * k
pref = 0
suff = 0
# computes the length of the maximal
# prefix which contains only zeroes
i = 0
while(st[i] == '0'):
i = i + 1
pref = pref + 1
# computes the length of the maximal
# suffix which contains only zeroes
i = n - 1
while(st[i] == '0'):
i = i - 1
suff = suff + 1
# if more than 1 concatenations
# are to be made
if (k > 1):
max_len = max(max_len,
pref + suff)
return max_len
# Driver code
n = 6
k = 3
st = "110010"
ans = max_length_substring(st, n, k)
print(ans)
# This code is contributed by ihritikC#
// C# program to find maximum number
// of consecutive zeroes after
// concatenating a binary string
using System;
class GFG
{
// returns the maximum size of
// a substring consisting only
// of zeroes after k concatenation
static int max_length_substring(string st,
int n, int k)
{
// stores the maximum length
// of the required substring
int max_len = 0;
int len = 0;
for (int i = 0; i < n; ++i)
{
// if the current character is 0
if (st[i] == '0')
len++;
else
len = 0;
// stores maximum length of current
// substrings with zeroes
max_len = Math.Max(max_len, len);
}
// if the whole string is
// filled with zero
if (max_len == n)
return n * k;
int pref = 0, suff = 0;
// computes the length of the maximal
// prefix which contains only zeroes
for (int i = 0; st[i] == '0';
++i, ++pref);
// computes the length of the maximal
// suffix which contains only zeroes
for (int i = n - 1; st[i] == '0';
--i, ++suff);
// if more than 1 concatenations
// are to be made
if (k > 1)
max_len = Math.Max(max_len,
pref + suff);
return max_len;
}
// Driver code
public static void Main(string[] args)
{
int n = 6;
int k = 3;
string st = "110010";
int ans = max_length_substring(st, n, k);
Console.WriteLine(ans);
}
}
// This code is contributed by ihritikPHP
1)
$max_len = max($max_len,
$pref + $suff);
return $max_len;
}
// Driver code
$n = 6;
$k = 3;
$st = "110010";
$ans = max_length_substring($st, $n, $k);
echo $ans;
// This code is contributed by ihritik
?>输出：
2