包含另一个给定字符串作为子字符串的子字符串计数|套装2

给定分别长度为N和M的两个字符串S和T ，任务是将其中包含字符串T的S的子字符串的数目计为一个子字符串。

例子：

Input: S = “dabc”, T = “ab”
Output: 4
Explanation:
Substrings of S containing T as a substring are:

1. S[0, 2] = “dab”
2. S[1, 2] = “ab”
3. S[1, 3] = “abc”
4. S[0, 3] = “dabc”

Input: S = “hshshshs” T = “hs”
Output: 25

天真的方法：有关解决问题的最简单方法，请参阅本文的上一篇文章。

时间复杂度： O(N ² )
辅助空间： O(N ² )

高效的方法：为了优化上述方法，我们的想法是找出S中所有T的出现。只要在S中找到T ，就添加所有包含此T出现的子串，但不包括在先前出现中已计算出的子串。请按照以下步骤解决问题：

• 初始化一个变量，例如answer ，以存储子字符串的计数。
• 初始化一个变量，表示最后，存储的T S中的最后出现的起始索引。
• 使用变量i遍历[0，N – M]范围。
  • 检查子字符串S [i，i + M]是否等于T。如果发现是真的，则加(i + 1 –最后)*(N –(i + M – 1))回答，最后更新为(i + 1)。
  • 否则，请继续进行下一次迭代。
• 完成上述步骤后，打印答案的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count the substrings of
// string containing another given
// string as a substring
void findOccurrences(string S, string T)
{
    // Store length of string S
    int n1 = S.size();
 
    // Store length of string T
    int n2 = T.size();
 
    // Store the required count of
    // substrings
    int ans = 0;
 
    // Store the starting index of
    // last occurence of T in S
    int last = 0;
 
    // Iterate in range [0, n1-n2]
    for (int i = 0; i <= n1 - n2; i++) {
 
        // Check if substring from i
        // to i + n2 is equal to T
        bool chk = true;
 
        // Check if substring from i
        // to i + n2 is equal to T
        for (int j = 0; j < n2; j++) {
 
            // Mark chk as false and
            // break the loop
            if (T[j] != S[i + j]) {
                chk = false;
                break;
            }
        }
 
        // If chk is true
        if (chk) {
 
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
 
            // Update the last to i + 1
            last = i + 1;
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver code
int main()
{
    string S = "dabc", T = "ab";
 
    // Function Call
    findOccurrences(S, T);
}

Java

// Java program for the above approach
class GFG{
   
// Function to count the substrings of
// string containing another given
// string as a substring
static void findOccurrences(String S, String T)
{
   
    // Store length of string S
    int n1 = S.length();
 
    // Store length of string T
    int n2 = T.length();
 
    // Store the required count of
    // substrings
    int ans = 0;
 
    // Store the starting index of
    // last occurence of T in S
    int last = 0;
 
    // Iterate in range [0, n1-n2]
    for (int i = 0; i <= n1 - n2; i++)
    {
 
        // Check if substring from i
        // to i + n2 is equal to T
        boolean chk = true;
 
        // Check if substring from i
        // to i + n2 is equal to T
        for (int j = 0; j < n2; j++)
        {
 
            // Mark chk as false and
            // break the loop
            if (T.charAt(j) != S.charAt(i + j))
            {
                chk = false;
                break;
            }
        }
 
        // If chk is true
        if (chk)
        {
 
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
 
            // Update the last to i + 1
            last = i + 1;
        }
    }
 
    // Print the answer
    System.out.println(ans);
}
 
  // Driver code
  public static void main (String[] args)
  {
    String S = "dabc", T = "ab";
 
    // Function Call
    findOccurrences(S, T);
  }
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
 
# Function to count the substrings of
# containing another given
# as a sub
def findOccurrences(S, T):
   
    # Store length of S
    n1 = len(S)
 
    # Store length of T
    n2 = len(T)
 
    # Store the required count of
    # substrings
    ans = 0
 
    # Store the starting index of
    # last occurence of T in S
    last = 0
 
    # Iterate in range [0, n1-n2]
    for i in range(n1 - n2 + 1):
 
        # Check if subfrom i
        # to i + n2 is equal to T
        chk = True
 
        # Check if subfrom i
        # to i + n2 is equal to T
        for j in range(n2):
 
            # Mark chk as false and
            # break the loop
            if (T[j] != S[i + j]):
                chk = False
                break
 
        # If chk is true
        if (chk):
 
            # Add (i + 1 - last) *
            # (n1 - (i + n2 - 1))
            # to answer
            ans += (i + 1 - last) * (n1 - (i + n2 - 1))
 
            # Update the last to i + 1
            last = i + 1
 
    # Prthe answer
    print(ans)
 
# Driver code
if __name__ == '__main__':
    S,T = "dabc","ab"
 
    # Function Call
    findOccurrences(S, T)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG
{
   
// Function to count the substrings of
// string containing another given
// string as a substring
static void findOccurrences(String S, String T)
{
   
    // Store length of string S
    int n1 = S.Length;
 
    // Store length of string T
    int n2 = T.Length;
 
    // Store the required count of
    // substrings
    int ans = 0;
 
    // Store the starting index of
    // last occurence of T in S
    int last = 0;
 
    // Iterate in range [0, n1-n2]
    for (int i = 0; i <= n1 - n2; i++)
    {
 
        // Check if substring from i
        // to i + n2 is equal to T
        bool chk = true;
 
        // Check if substring from i
        // to i + n2 is equal to T
        for (int j = 0; j < n2; j++)
        {
 
            // Mark chk as false and
            // break the loop
            if (T[j] != S[i + j])
            {
                chk = false;
                break;
            }
        }
 
        // If chk is true
        if (chk)
        {
 
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
 
            // Update the last to i + 1
            last = i + 1;
        }
    }
 
    // Print the answer
    Console.WriteLine(ans);
}
 
  // Driver code
  public static void Main(String[] args)
  {
    String S = "dabc", T = "ab";
 
    // Function Call
    findOccurrences(S, T);
  }
}
 
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript program for above approach
 
// Function to count the substrings of
// string containing another given
// string as a substring
function findOccurrences(S, T)
{
    
    // Store length of string S
    let n1 = S.length;
  
    // Store length of string T
    let n2 = T.length;
  
    // Store the required count of
    // substrings
    let ans = 0;
  
    // Store the starting index of
    // last occurence of T in S
    let last = 0;
  
    // Iterate in range [0, n1-n2]
    for (let i = 0; i <= n1 - n2; i++)
    {
  
        // Check if substring from i
        // to i + n2 is equal to T
        let chk = true;
  
        // Check if substring from i
        // to i + n2 is equal to T
        for (let j = 0; j < n2; j++)
        {
  
            // Mark chk as false and
            // break the loop
            if (T[j] != S[i + j])
            {
                chk = false;
                break;
            }
        }
  
        // If chk is true
        if (chk)
        {
  
            // Add (i + 1 - last) *
            // (n1 - (i + n2 - 1))
            // to answer
            ans += (i + 1 - last)
                   * (n1 - (i + n2 - 1));
  
            // Update the last to i + 1
            last = i + 1;
        }
    }
  
    // Prlet the answer
    document.write(ans);
}
 
// Driver Code
 
     let S = "dabc", T = "ab";
  
    // Function Call
    findOccurrences(S, T);
     

输出：

4

时间复杂度： O(N * M)
辅助空间： O(1)