恰好包含给定字符k 次的子字符串的数量

给定字符串str 、一个字符c和一个整数k > 0 。任务是找出恰好包含k次字符c的子串的数量。

例子：

Input: str = “abada”, c = ‘a’, K = 2 
Output: 4 
All possible sub-strings are “aba”, “abad”, “bada” and “ada”.

Input: str = “55555”, c = ‘5’, k = 4 
Output: 2 

天真的方法：一个简单的解决方案是生成所有子字符串并检查给定字符的计数是否恰好是 k 次。
这种方法的时间复杂度是O(n ² ) 。

高效方法：一种有效的解决方案是使用滑动窗口技术。找到恰好包含给定字符k 次的子字符串，从字符c 开始。计算子字符串两边的字符数。将计数相乘以获得可能的子字符串的数量。这种方法的时间复杂度是O(n) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count of required sub-strings
int countSubString(string s, char c, int k)
{
 
    // Left and right counters for characters on
    // both sides of sub-string window
    int leftCount = 0, rightCount = 0;
 
    // Left and right pointer on both
    // sides of sub-string 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 sub-strings
    while (left < len && (right - 1) < len) {
 
        // Counting the characters on left side
        // of the sub-string window
        while (s[left] != c && left < len) {
            left++;
            leftCount++;
        }
 
        // Counting the characters on right side
        // of the sub-string window
        while (right < len && s[right] != c) {
            if (s[right] == c)
                freq++;
            right++;
            rightCount++;
        }
 
        // Add the possible sub-strings
        // on both sides to result
        result = result + (leftCount + 1) * (rightCount + 1);
 
        // Setting the frequency for next
        // sub-string window
        freq = k - 1;
 
        // Reset the left and right counters
        leftCount = 0;
        rightCount = 0;
 
        left++;
        right++;
    }
    return result;
}
 
// Driver code
int main()
{
    string s = "abada";
    char c = 'a';
    int k = 2;
 
    cout << countSubString(s, c, k) << "\n";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to return the count of required sub-strings
static int countSubString(char[] s, char c, int k)
{
 
    // Left and right counters for characters on
    // both sides of sub-string window
    int leftCount = 0, rightCount = 0;
 
    // Left and right pointer on both
    // sides of sub-string 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 sub-strings
    while (left < len && (right - 1) < len)
    {
 
        // Counting the characters on left side
        // of the sub-string window
        while (s[left] != c && left < len)
        {
            left++;
            leftCount++;
        }
 
        // Counting the characters on right side
        // of the sub-string window
        while (right < len && s[right] != c)
        {
            if (s[right] == c)
                freq++;
            right++;
            rightCount++;
        }
 
        // Add the possible sub-strings
        // on both sides to result
        result = result + (leftCount + 1) * (rightCount + 1);
 
        // Setting the frequency for next
        // sub-string window
        freq = k - 1;
 
        // Reset the left and right counters
        leftCount = 0;
        rightCount = 0;
 
        left++;
        right++;
    }
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "abada";
    char c = 'a';
    int k = 2;
 
    System.out.println(countSubString(s.toCharArray(), c, k));
}
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python 3 implementation of the approach
 
# Function to return the count
# of required sub-strings
def countSubString(s, c, k):
     
    # Left and right counters for characters
    # on both sides of sub-string window
    leftCount = 0
    rightCount = 0
 
    # Left and right pointer on both
    # sides of sub-string window
    left = 0
    right = 0
 
    # Initialize the frequency
    freq = 0
 
    # Result and length of string
    result = 0
    len1 = len(s)
 
    # Initialize the left pointer
    while (s[left] != c and left < len1):
        left += 1
        leftCount += 1
 
    # Initialize the right pointer
    right = left + 1
    while (freq != (k - 1) and
          (right - 1) < len1):
        if (s[right] == c):
            freq += 1
        right += 1
 
    # Traverse all the window sub-strings
    while (left < len1 and
          (right - 1) < len1):
         
        # Counting the characters on left side
        # of the sub-string window
        while (s[left] != c and left < len1):
            left += 1
            leftCount += 1
 
        # Counting the characters on right side
        # of the sub-string window
        while (right < len1 and
             s[right] != c):
            if (s[right] == c):
                freq += 1
            right += 1
            rightCount += 1
 
        # Add the possible sub-strings
        # on both sides to result
        result = (result + (leftCount + 1) *
                           (rightCount + 1))
 
        # Setting the frequency for next
        # sub-string window
        freq = k - 1
 
        # Reset the left and right counters
        leftCount = 0
        rightCount = 0
 
        left += 1
        right += 1
 
    return result
 
# Driver code
if __name__ == '__main__':
    s = "abada"
    c = 'a'
    k = 2
 
    print(countSubString(s, c, k))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the count of required sub-strings
static int countSubString(char[] s, char c, int k)
{
 
    // Left and right counters for characters on
    // both sides of sub-string window
    int leftCount = 0, rightCount = 0;
 
    // Left and right pointer on both
    // sides of sub-string 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 sub-strings
    while (left < len && (right - 1) < len)
    {
 
        // Counting the characters on left side
        // of the sub-string window
        while (s[left] != c && left < len)
        {
            left++;
            leftCount++;
        }
 
        // Counting the characters on right side
        // of the sub-string window
        while (right < len && s[right] != c)
        {
            if (s[right] == c)
                freq++;
            right++;
            rightCount++;
        }
 
        // Add the possible sub-strings
        // on both sides to result
        result = result + (leftCount + 1) * (rightCount + 1);
 
        // Setting the frequency for next
        // sub-string window
        freq = k - 1;
 
        // Reset the left and right counters
        leftCount = 0;
        rightCount = 0;
 
        left++;
        right++;
    }
    return result;
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "abada";
    char c = 'a';
    int k = 2;
 
    Console.WriteLine(countSubString(s.ToCharArray(), c, k));
}
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the approach
 
// Function to return the count of required sub-strings
function countSubString(s, c, k)
{
 
    // Left and right counters for characters on
    // both sides of sub-string window
    var leftCount = 0, rightCount = 0;
 
    // Left and right pointer on both
    // sides of sub-string window
    var left = 0, right = 0;
 
    // Initialize the frequency
    var freq = 0;
 
    // Result and length of string
    var 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 sub-strings
    while (left < len && (right - 1) < len) {
 
        // Counting the characters on left side
        // of the sub-string window
        while (s[left] != c && left < len) {
            left++;
            leftCount++;
        }
 
        // Counting the characters on right side
        // of the sub-string window
        while (right < len && s[right] != c) {
            if (s[right] == c)
                freq++;
            right++;
            rightCount++;
        }
 
        // Add the possible sub-strings
        // on both sides to result
        result = result + (leftCount + 1) * (rightCount + 1);
 
        // Setting the frequency for next
        // sub-string window
        freq = k - 1;
 
        // Reset the left and right counters
        leftCount = 0;
        rightCount = 0;
 
        left++;
        right++;
    }
    return result;
}
 
// Driver code
var s = "abada";
var c = 'a';
var k = 2;
document.write( countSubString(s, c, k) + "<br>");
 

输出：

4