给定由小英文字母和26位比特字符串Q组成的字符串P ，其中1表示特殊字符， 0表示26个英文字母的普通字符。任务是找到最多包含K个正常字符的最长子字符串的长度。
例子：
Input : P = “normal”, Q = “00000000000000000000000000”, K=1
Output : 1
Explanation : In string Q all characters are normal.
Hence we can select any substring of length 1.
Input : P = “giraffe”, Q = “01111001111111111011111111”, K=2
Output : 3
Explanation : Normal characters in P from Q are {a, f, g, r}.
Therefore, possible substrings with at most 2 normal characters are {gir, ira, ffe}.
The maximum length of all substring is 3.
方法：
为了解决上述问题，我们将使用两个指针的概念。因此，请保持子字符串的左右指针以及正常字符。递增右索引，直到正常字符数最多为K。然后使用直到现在为止遇到的最大子串长度来更新答案。递增左索引和递减计数，直到大于K。
下面是上述方法的实现：
C++
// C++ implementation to Find
// length of longest substring
// with at most K normal characters
#include
using namespace std;
// Function to find maximum
// length of normal substrings
int maxNormalSubstring(string& P, string& Q,
int K, int N)
{
if (K == 0)
return 0;
// keeps count of normal characters
int count = 0;
// indexes of substring
int left = 0, right = 0;
// maintain length of longest substring
// with at most K normal characters
int ans = 0;
while (right < N) {
while (right < N && count <= K) {
// get position of character
int pos = P[right] - 'a';
// check if current character is normal
if (Q[pos] == '0') {
// check if normal characters
// count exceeds K
if (count + 1 > K)
break;
else
count++;
}
right++;
// update answer with substring length
if (count <= K)
ans = max(ans, right - left);
}
while (left < right) {
// get position of character
int pos = P[left] - 'a';
left++;
// check if character is
// normal then decrement count
if (Q[pos] == '0')
count--;
if (count < K)
break;
}
}
return ans;
}
// Driver code
int main()
{
// initialise the string
string P = "giraffe", Q = "01111001111111111011111111";
int K = 2;
int N = P.length();
cout << maxNormalSubstring(P, Q, K, N);
return 0;
} Java
// Java implementation to Find
// length of longest subString
// with at most K normal characters
class GFG{
// Function to find maximum
// length of normal subStrings
static int maxNormalSubString(char []P, char []Q,
int K, int N)
{
if (K == 0)
return 0;
// keeps count of normal characters
int count = 0;
// indexes of subString
int left = 0, right = 0;
// maintain length of longest subString
// with at most K normal characters
int ans = 0;
while (right < N) {
while (right < N && count <= K) {
// get position of character
int pos = P[right] - 'a';
// check if current character is normal
if (Q[pos] == '0') {
// check if normal characters
// count exceeds K
if (count + 1 > K)
break;
else
count++;
}
right++;
// update answer with subString length
if (count <= K)
ans = Math.max(ans, right - left);
}
while (left < right) {
// get position of character
int pos = P[left] - 'a';
left++;
// check if character is
// normal then decrement count
if (Q[pos] == '0')
count--;
if (count < K)
break;
}
}
return ans;
}
// Driver code
public static void main(String[] args)
{
// initialise the String
String P = "giraffe", Q = "01111001111111111011111111";
int K = 2;
int N = P.length();
System.out.print(maxNormalSubString(P.toCharArray(), Q.toCharArray(), K, N));
}
}
// This code is contributed by Princi SinghPython3
# Function to find maximum
# length of normal substrings
def maxNormalSubstring(P, Q, K, N):
if (K == 0):
return 0
# keeps count of normal characters
count = 0
# indexes of substring
left, right = 0, 0
# maintain length of longest substring
# with at most K normal characters
ans = 0
while (right < N):
while (right < N and count <= K):
# get position of character
pos = ord(P[right]) - ord('a')
# check if current character is normal
if (Q[pos] == '0'):
# check if normal characters
# count exceeds K
if (count + 1 > K):
break
else:
count += 1
right += 1
# update answer with substring length
if (count <= K):
ans = max(ans, right - left)
while (left < right):
# get position of character
pos = ord(P[left]) - ord('a')
left += 1
# check if character is
# normal then decrement count
if (Q[pos] == '0'):
count -= 1
if (count < K):
break
return ans
# Driver code
if(__name__ == "__main__"):
# initialise the string
P = "giraffe"
Q = "01111001111111111011111111"
K = 2
N = len(P)
print(maxNormalSubstring(P, Q, K, N))
# This code is contributed by skylagsC#
// C# implementation to Find
// length of longest subString
// with at most K normal characters
using System;
public class GFG{
// Function to find maximum
// length of normal subStrings
static int maxNormalSubString(char []P, char []Q,
int K, int N)
{
if (K == 0)
return 0;
// keeps count of normal characters
int count = 0;
// indexes of subString
int left = 0, right = 0;
// maintain length of longest subString
// with at most K normal characters
int ans = 0;
while (right < N) {
while (right < N && count <= K) {
// get position of character
int pos = P[right] - 'a';
// check if current character is normal
if (Q[pos] == '0') {
// check if normal characters
// count exceeds K
if (count + 1 > K)
break;
else
count++;
}
right++;
// update answer with subString length
if (count <= K)
ans = Math.Max(ans, right - left);
}
while (left < right) {
// get position of character
int pos = P[left] - 'a';
left++;
// check if character is
// normal then decrement count
if (Q[pos] == '0')
count--;
if (count < K)
break;
}
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
// initialise the String
String P = "giraffe", Q = "01111001111111111011111111";
int K = 2;
int N = P.Length;
Console.Write(maxNormalSubString(P.ToCharArray(),
Q.ToCharArray(), K, N));
}
}
// This code contributed by Princi Singh输出：
3
时间复杂度：上述方法需要O(N)时间。