递归解决方案，以计数具有相同的首尾字符的子字符串

给定一个字符串S，我们需要找到所有以相同字符开始和结束的连续子字符串的计数。

例子 ：

Input  : S = "abcab"
Output : 7
There are 15 substrings of "abcab"
a, ab, abc, abca, abcab, b, bc, bca
bcab, c, ca, cab, a, ab, b
Out of the above substrings, there 
are 7 substrings : a, abca, b, bcab, 
c, a and b.

Input  : S = "aba"
Output : 4
The substrings are a, b, a and aba

我们在下面的文章中讨论了不同的解决方案。

计算具有相同首尾字符的子字符串

在本文中，讨论了一个简单的递归解决方案。

C++

// c++ program to count substrings with same
// first and last characters
#include 
#include 
using namespace std;
  
/* Function to count subtrings with same first and 
  last characters*/
int countSubstrs(string str, int i, int j, int n)
{
    // base cases
    if (n == 1)
        return 1;
    if (n <= 0)
        return 0;
  
    int res =  countSubstrs(str, i + 1, j, n - 1) +  
               countSubstrs(str, i, j - 1, n - 1) -
               countSubstrs(str, i + 1, j - 1, n - 2);            
  
    if (str[i] == str[j])
        res++; 
  
    return res;
}
  
// driver code
int main()
{
    string str = "abcab";
    int n = str.length();
    cout << countSubstrs(str, 0, n - 1, n);
}

Java

// Java program to count substrings 
// with same first and last characters
  
class GFG
{
    // Function to count subtrings
    // with same first and 
    // last characters
    static int countSubstrs(String str, int i, 
                                         int j, int n)
    {
        // base cases
        if (n == 1)
            return 1;
        if (n <= 0)
            return 0;
      
        int res = countSubstrs(str, i + 1, j, n - 1) + 
                countSubstrs(str, i, j - 1, n - 1) -
                countSubstrs(str, i + 1, j - 1, n - 2);         
      
        if (str.charAt(i) == str.charAt(j))
            res++; 
      
        return res;
    }
      
    // Driver code
    public static void main (String[] args)
    {
        String str = "abcab";
        int n = str.length();
        System.out.print(countSubstrs(str, 0, n - 1, n));
    }
}
  
// This code is contributed by Anant Agarwal.

Python 3

# Python 3 program to count substrings with same
# first and last characters
  
# Function to count subtrings with same first and 
# last characters
def countSubstrs(str, i, j, n):
  
    # base cases
    if (n == 1):
        return 1
    if (n <= 0):
        return 0
  
    res = (countSubstrs(str, i + 1, j, n - 1) 
        + countSubstrs(str, i, j - 1, n - 1) 
        - countSubstrs(str, i + 1, j - 1, n - 2))     
  
    if (str[i] == str[j]):
        res += 1
  
    return res
  
# driver code
str = "abcab"
n = len(str)
print(countSubstrs(str, 0, n - 1, n))
  
# This code is contributed by Smitha

C#

// C# program to count substrings 
// with same first and last characters
using System;
  
class GFG {
      
    // Function to count subtrings
    // with same first and 
    // last characters
    static int countSubstrs(string str, int i, 
                                 int j, int n)
    {
          
        // base cases
        if (n == 1)
            return 1;
        if (n <= 0)
            return 0;
      
        int res = countSubstrs(str, i + 1, j, n - 1)
                + countSubstrs(str, i, j - 1, n - 1)
            - countSubstrs(str, i + 1, j - 1, n - 2);     
      
        if (str[i] == str[j])
            res++; 
      
        return res;
    }
      
    // Driver code
    public static void Main ()
    {
        string str = "abcab";
        int n = str.Length;
          
        Console.WriteLine(
                countSubstrs(str, 0, n - 1, n));
    }
}
  
// This code is contributed by vt_m.

输出：

7

上述解决方案的时间复杂度是指数的。在最坏的情况下，我们可能最终会执行O(3 ⁿ )运算。