包含给定字符K 次的子字符串的计数
给定一个由数字字母、字符C和整数K组成的字符串，任务是找到恰好包含K次字符C的可能子字符串的数量。
例子：
Input : str = "212", c = '2', k = 1
Output : 4
Possible substrings are {"2", "21", "12", "2"}
that contains 2 exactly 1 time.
Input : str = "55555", c = '5', k = 4
Output : 2
Possible substrings are {"5555", "5555"}
that contains 5 exactly 4 times朴素方法：一个简单的解决方案是生成字符串的所有子字符串，并计算给定字符恰好出现 k 次的子字符串的数量。
该解决方案的时间复杂度为 O(N 2 )，其中 N 是字符串的长度。
有效方法：一种有效的解决方案是使用滑动窗口技术。从字符C 开始，找到恰好包含字符C 的子串 K 次。计算子串两边的字符数。将计数相乘以获得具有给定子字符串的可能子字符串的数量
下面是上述方法的实现：
C++
// C++ program to count the number of substrings
// which contains the character C exactly K times
#include
using namespace std;
// Function to count the number of substrings
// which contains the character C exactly K times
int countSubString(string s, char c, int k)
{
// left and right counters for characters on
// both sides of substring window
int leftCount = 0, rightCount = 0;
// left and right pointer on both sides
// of substring window
int left = 0, right = 0;
// initialize the frequency
int freq = 0;
// result and length of string
int result = 0, len = s.length();
// initialize the left pointer
while (s[left] != c && left < len) {
left++;
leftCount++;
}
// initialize the right pointer
right = left + 1;
while (freq != (k - 1) && (right - 1) < len) {
if (s[right] == c)
freq++;
right++;
}
// traverse all the window substrings
while (left < len && (right - 1) < len) {
// counting the characters on leftSide
// of substring window
while (s[left] != c && left < len) {
left++;
leftCount++;
}
// counting the characters on rightSide of
// substring window
while (right < len && s[right] != c) {
if (s[right] == c)
freq++;
right++;
rightCount++;
}
// Add the possible substrings on both
// sides to result
result = result + (leftCount + 1) * (rightCount + 1);
// Setting the frequency for next
// substring window
freq = k - 1;
// reset the left, right counters
leftCount = 0;
rightCount = 0;
left++;
right++;
}
return result;
}
// Driver code
int main()
{
string s = "3123231";
char c = '3';
int k = 2;
cout << countSubString(s, c, k);
return 0;
} Java
// Java program to count the number of substrings
// which contains the character C exactly K times
class GFG
{
// Function to count the number of substrings
// which contains the character C exactly K times
static int countSubString(String s, char c, int k)
{
// left and right counters for characters on
// both sides of substring window
int leftCount = 0, rightCount = 0;
// left and right pointer on both sides
// of substring window
int left = 0, right = 0;
// initialize the frequency
int freq = 0;
// result and length of string
int result = 0, len = s.length();
// initialize the left pointer
while (s.charAt(left) != c && left < len)
{
left++;
leftCount++;
}
// initialize the right pointer
right = left + 1;
while (freq != (k - 1) && (right - 1) < len)
{
if (s.charAt(right) == c)
{
freq++;
}
right++;
}
// traverse all the window substrings
while (left < len && (right - 1) < len)
{
// counting the characters on leftSide
// of substring window
while (s.charAt(left) != c && left < len)
{
left++;
leftCount++;
}
// counting the characters on rightSide of
// substring window
while (right < len && s.charAt(right) != c)
{
if (s.charAt(right) == c)
{
freq++;
}
right++;
rightCount++;
}
// Add the possible substrings on both
// sides to result
result = result + (leftCount + 1) * (rightCount + 1);
// Setting the frequency for next
// substring window
freq = k - 1;
// reset the left, right counters
leftCount = 0;
rightCount = 0;
left++;
right++;
}
return result;
}
// Driver code
public static void main(String args[])
{
String s = "3123231";
char c = '3';
int k = 2;
System.out.println(countSubString(s, c, k));
}
}
/* This code is contributed by PrinciRaj1992 */Python3
# Python3 program to count the number of substrings
# which contains the character C exactly K times
# Function to count the number of substrings
# which contains the character C exactly K times
def countSubString(s, c, k):
# left and right counters for characters
# on both sides of subwindow
leftCount = 0
rightCount = 0
# left and right pointer on both sides
# of subwindow
left = 0
right = 0
# Initialize the frequency
freq = 0
# result and Length of string
result = 0
Len = len(s)
# initialize the left pointer
while (s[left] != c and left < Len):
left += 1
leftCount += 1
# initialize the right pointer
right = left + 1
while (freq != (k - 1) and (right - 1) < Len):
if (s[right] == c):
freq += 1
right += 1
# traverse all the window substrings
while (left < Len and (right - 1) < Len):
# counting the characters on leftSide
# of subwindow
while (s[left] != c and left < Len):
left += 1
leftCount += 1
# counting the characters on rightSide of
# subwindow
while (right < Len and s[right] != c):
if (s[right] == c):
freq += 1
right += 1
rightCount += 1
# Add the possible substrings on both
# sides to result
result = (result + (leftCount + 1) *
(rightCount + 1))
# Setting the frequency for next
# subwindow
freq = k - 1
# reset the left, right counters
leftCount = 0
rightCount = 0
left += 1
right += 1
return result
# Driver code
s = "3123231"
c = '3'
k = 2
print(countSubString(s, c, k))
# This code is contributed by Mohit KumarC#
// C# program to count the number of substrings
// which contains the character C exactly K times
using System;
class GFG
{
// Function to count the number of substrings
// which contains the character C exactly K times
static int countSubString(string s, char c, int k)
{
// left and right counters for characters on
// both sides of substring window
int leftCount = 0, rightCount = 0;
// left and right pointer on both sides
// of substring window
int left = 0, right = 0;
// initialize the frequency
int freq = 0;
// result and length of string
int result = 0, len = s.Length;
// initialize the left pointer
while (s[left] != c && left < len)
{
left++;
leftCount++;
}
// initialize the right pointer
right = left + 1;
while (freq != (k - 1) && (right - 1) < len)
{
if (s[right] == c)
freq++;
right++;
}
// traverse all the window substrings
while (left < len && (right - 1) < len)
{
// counting the characters on leftSide
// of substring window
while (s[left] != c && left < len)
{
left++;
leftCount++;
}
// counting the characters on rightSide of
// substring window
while (right < len && s[right] != c)
{
if (s[right] == c)
freq++;
right++;
rightCount++;
}
// Add the possible substrings on both
// sides to result
result = result + (leftCount + 1) * (rightCount + 1);
// Setting the frequency for next
// substring window
freq = k - 1;
// reset the left, right counters
leftCount = 0;
rightCount = 0;
left++;
right++;
}
return result;
}
// Driver code
static public void Main ()
{
string s = "3123231";
char c = '3';
int k = 2;
Console.WriteLine(countSubString(s, c, k));
}
}
// This code is contributed by AnkitRai01Javascript
输出：
8时间复杂度：O(N)