We are given a string S, we need to find count of all contiguous substrings starting and ending with same character.
Examples :
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
We have discussed different solutions in below post.
Count substrings with same first and last characters
In this article, a simple recursive solution is discussed.
C++
// c++ program to count substrings with same // first and last characters #include <iostream> #include <string> 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. |
PHP
<?php // PHP program to count // substrings with same // first and last characters //Function to count subtrings // with same first and // last characters function 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++; return $res; } // Driver Code $str = "abcab"; $n = strlen($str); echo(countSubstrs($str, 0, $n - 1, $n)); // This code is contributed by Ajit. ?> |
Output:
7
The time complexity of above solution is exponential. In Worst case, we may end up doing O(3n) operations.
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