Count substrings with same first and last characters

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

Method 1 (Simple): In this approach, we use brute force and find all the sub-strings and pass them through our function checkEquality to see if starting and ending characters are same.

Implementation:

C++

// C++ program to count all substrings with same
// first and last characters.
#include <bits/stdc++.h>
using namespace std;

// Returns true if first and last characters
// of s are same.
int checkEquality(string s)
{
    return (s[0] == s[s.size() - 1]);
}

int countSubstringWithEqualEnds(string s)
{
    int result = 0;
    int n = s.length();

    // Starting point of substring
    for (int i = 0; i < n; i++)

       // Length of substring
       for (int len = 1; len <= n-i; len++)

          // Check if current substring has same
          // starting and ending characters.
          if (checkEquality(s.substr(i, len)))
            result++;

    return result;
}

// Driver function
int main()
{
    string s("abcab");
    cout << countSubstringWithEqualEnds(s);
    return 0;
}

Java

// Java program to count all substrings with same
// first and last characters.
public class GFG {

    // Returns true if first and last characters
    // of s are same.
    static boolean checkEquality(String s)
    {
        return (s.charAt(0) == s.charAt(s.length() - 1));
    }

    static int countSubstringWithEqualEnds(String s)
    {
        int result = 0;
        int n = s.length();

        // Starting point of substring
        for (int i = 0; i < n; i++)

           // Length of substring
           for (int len = 1; len <= n-i; len++)

              // Check if current substring has same
              // starting and ending characters.
              if (checkEquality(s.substring(i, i + len)))
                result++;

        return result;
    }

    // Driver function
    public static void main(String args[])
    {
        String s = "abcab";
        System.out.println(countSubstringWithEqualEnds(s));
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python program to count all substrings with same
# first and last characters.

# Returns true if first and last characters
# of s are same.
def checkEquality(s):
    return (ord(s[0]) == ord(s[len(s) - 1]));

def countSubstringWithEqualEnds(s):
    result = 0;
    n = len(s);

    # Starting point of substring
    for i in range(n):

    # Length of substring
        for j in range(1,n-i+1):

        # Check if current substring has same
        # starting and ending characters.
            if (checkEquality(s[i:i+j])):
                result+=1;

    return result;

# Driver code
s = "abcab";
print(countSubstringWithEqualEnds(s));

# This code contributed by PrinciRaj1992

C#

// C# program to count all
// substrings with same
// first and last characters.
using System;

class GFG
{

    // Returns true if first and
    //  last characters of s are same.
    static bool checkEquality(string s)
    {
        return (s[0] == s[s.Length - 1]);
    }

    static int countSubstringWithEqualEnds(string s)
    {
        int result = 0;
        int n = s.Length;

        // Starting point of substring
        for (int i = 0; i < n; i++)

        // Length of substring
        for (int len = 1; len <= n-i; len++)

            // Check if current substring has same
            // starting and ending characters.
            if (checkEquality(s.Substring(i, len)))
                result++;

        return result;
    }

    // Driver code
    public static void Main()
    {
        string s = "abcab";
        Console.WriteLine(countSubstringWithEqualEnds(s));
    }
}

// This code is contributed by Code_Mech

PHP

<?php
// PHP program to count all substrings
// with same first and last characters.

// Returns true if first and last
// characters of s are same.
function checkEquality($s)
{
    return ($s[0] == $s[strlen($s) - 1]);
}

function countSubstringWithEqualEnds($s)
{
    $result = 0;
    $n = strlen($s);

    // Starting point of substring
    for ($i = 0; $i < $n; $i++)

    // Length of substring
    for ($len = 1; $len <= $n - $i; $len++)

        // Check if current substring has same
        // starting and ending characters.
        if (checkEquality(substr($s, $i, $len)))
            $result++;

    return $result;
}

// Driver Code
$s = "abcab";
print(countSubstringWithEqualEnds($s));

// This code is contributed by chandan_jnu
?>

Javascript

<script>

// JavaScript program to count all substrings
// with same first and last characters.


// Returns true if first and last characters
// of s are same.
function checkEquality(s)
{
    return (s.charAt(0) == s.charAt(s.length - 1));
}

function countSubstringWithEqualEnds(s)
{
    var result = 0;
    var n = s.length;

    // Starting point of substring
    for (var i = 0; i < n; i++)

       // Length of substring
       for (var len = 1; len <= n-i; len++)

          // Check if current substring has same
          // starting and ending characters.
          if (checkEquality(s.substring(i, i + len)))
            result++;

    return result;
}

// Driver function

    var s = "abcab";
    document.write(countSubstringWithEqualEnds(s));


// This code contributed by shikhasingrajput

</script>

Output

7

Although the above code works fine, it’s not efficient as its time complexity is O(n²). Note that there are n*(n+1)/2 substrings of a string of length n. This solution also requires O(n) extra space as we one by one create all substrings.

Method 2 (Space Efficient): In this approach we don’t actually generate substrings rather we traverse the string in such a manner so that we can easily compare first and last characters. 

Implementation:

C++

// Space efficient C++ program to count all
// substrings with same first and last characters.
#include <bits/stdc++.h>
using namespace std;

int countSubstringWithEqualEnds(string s)
{
    int result = 0;
    int n = s.length();

    // Iterating through all substrings in
    // way so that we can find first and last
    // character easily
    for (int i=0; i<n; i++)
        for (int j=i; j<n; j++)
            if (s[i] == s[j])
                result++;

    return result;
}

// Driver function
int main()
{
    string s("abcab");
    cout << countSubstringWithEqualEnds(s);
    return 0;
}

Java

// Space efficient Java program to count all
// substrings with same first and last characters.
public class GFG {

    static int countSubstringWithEqualEnds(String s)
    {
        int result = 0;
        int n = s.length();

        // Iterating through all substrings in
        // way so that we can find first and last
        // character easily
        for (int i = 0; i < n; i++)
            for (int j = i; j < n; j++)
                if (s.charAt(i) == s.charAt(j))
                    result++;

        return result;
    }

    // Driver function
    public static void main(String args[])
    {
        String s = "abcab";
        System.out.println(countSubstringWithEqualEnds(s));
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Space efficient Python3 program to count all
# substrings with same first and last characters.
def countSubstringWithEqualEnds(s):

    result = 0;
    n = len(s);

    # Iterating through all substrings in
    # way so that we can find first and
    # last character easily
    for i in range(n):
        for j in range(i, n):
            if (s[i] == s[j]):
                result = result + 1

    return result

# Driver Code
s = "abcab";
print(countSubstringWithEqualEnds(s))

# This code is contributed
# by Akanksha Rai

C#

// Space efficient C# program to count all
// substrings with same first and last characters.
using System;

public class GFG {

    static int countSubstringWithEqualEnds(string s)
    {
        int result = 0;
        int n = s.Length;

        // Iterating through all substrings in
        // way so that we can find first and last
        // character easily
        for (int i = 0; i < n; i++)
            for (int j = i; j < n; j++)
                if (s[i] == s[j])
                    result++;

        return result;
    }

    // Driver function
    public static void Main()
    {
        string s = "abcab";
        Console.Write(countSubstringWithEqualEnds(s));
    }
}

// This code is contributed by nitin mittal

PHP

<?php
// Space efficient PHP program to count all
// substrings with same first and last characters.

function countSubstringWithEqualEnds($s)
{
    $result = 0;
    $n = strlen($s);

    // Iterating through all substrings in
    // way so that we can find first and last
    // character easily
    for ($i = 0; $i < $n; $i++)
        for ($j = $i; $j < $n; $j++)
            if ($s[$i] == $s[$j])
                $result++;

    return $result;
}

// Driver Code
$s = "abcab";
echo countSubstringWithEqualEnds($s);

// This code is contributed
// by Akanksha Rai

Javascript

<script>

// Space efficient javascript program to count all
// substrings with same first and last characters.
function countSubstringWithEqualEnds(s)
{
    var result = 0;
    var n = s.length;

    // Iterating through all substrings in
    // way so that we can find first and last
    // character easily
    for(i = 0; i < n; i++)
        for(j = i; j < n; j++)
            if (s.charAt(i) == s.charAt(j))
                result++;

    return result;
}

// Driver code
var s = "abcab";
document.write(countSubstringWithEqualEnds(s));

// This code is contributed by Princi Singh

</script>

Output

7

In the above code although we have reduced the extra space to O(1) time complexity is still O(n^2).

Method 3 (Best approach) : Now if we carefully observe then we can realize that the answer just depends on frequencies of characters in the original string. For example in string abcab, frequency of ‘a’ is 2, and substrings contributing to answering are a, abca and a respectively, a total of 3, which is calculated by ^{(frequency of ‘a’+1)}C₂. 

Implementation:

C++

// Most efficient C++ program to count all
// substrings with same first and last characters.
#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;  // assuming lower case only

int countSubstringWithEqualEnds(string s)
{
    int result = 0;
    int n = s.length();

    // Calculating frequency of each character
    // in the string.
    int count[MAX_CHAR] = {0};
    for (int i=0; i<n; i++)
        count[s[i]-'a']++;

    // Computing result using counts
    for (int i=0; i<MAX_CHAR; i++)
        result += (count[i]*(count[i]+1)/2);

    return result;
}

// Driver function
int main()
{
    string s("abcab");
    cout << countSubstringWithEqualEnds(s);
    return 0;
}

Java

// Most efficient Java program to count all
// substrings with same first and last characters.
public class GFG {

      // assuming lower case only
    static final int MAX_CHAR = 26;

    static int countSubstringWithEqualEnds(String s)
    {
        int result = 0;
        int n = s.length();

        // Calculating frequency of each character
        // in the string.
        int[] count =  new int[MAX_CHAR];
        for (int i = 0; i < n; i++)
            count[s.charAt(i)-'a']++;

        // Computing result using counts
        for (int i = 0; i < MAX_CHAR; i++)
            result += (count[i] * (count[i] + 1) / 2);

        return result;
    }

    // Driver function
    public static void main(String args[])
    {
        String s = "abcab";
        System.out.println(countSubstringWithEqualEnds(s));
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Most efficient Python program to count all
# substrings with same first and last characters.

MAX_CHAR = 26; # assuming lower case only

def countSubstringWithEqualEnds(s):
    result = 0;
    n = len(s);

    # Calculating frequency of each character
    # in the string.
    count = [0]*MAX_CHAR;
    for i in range(n):
        count[ord(s[i])-ord('a')]+=1;

    # Computing result using counts
    for i in range(MAX_CHAR):
        result += (count[i]*(count[i]+1)/2);

    return result;

# Driver code
s = "abcab";
print(countSubstringWithEqualEnds(s));


# This code is contributed by 29AjayKumar

C#

// Most efficient C# program to count all
// substrings with same first and last characters.
using System;

class GFG {

    // assuming lower case only
    static readonly int MAX_CHAR = 26;

    static int countSubstringWithEqualEnds(String s)
    {
        int result = 0;
        int n = s.Length;

        // Calculating frequency of each character
        // in the string.
        int[] count = new int[MAX_CHAR];
        for (int i = 0; i < n; i++)
            count[s[i] - 'a']++;

        // Computing result using counts
        for (int i = 0; i < MAX_CHAR; i++)
            result += (count[i] * (count[i] + 1) / 2);

        return result;
    }

    // Driver code
    public static void Main()
    {
        String s = "abcab";
        Console.Write(countSubstringWithEqualEnds(s));
    }
}

// This code is contributed by 29AjayKumar

PHP

<?php
// Most efficient PHP program to count all
// substrings with same first and last characters.
$MAX_CHAR = 26; // assuming lower case only

function countSubstringWithEqualEnds($s)
{
    global $MAX_CHAR;
    $result = 0;
    $n = strlen($s);

    // Calculating frequency of each
    // character in the string.
    $count = array_fill(0, $MAX_CHAR, 0);
    for ($i = 0; $i < $n; $i++)
        $count[ord($s[$i]) - ord('a')]++;

    // Computing result using counts
    for ($i = 0; $i < $MAX_CHAR; $i++)
        $result += ($count[$i] *
                   ($count[$i] + 1) / 2);

    return $result;
}

// Driver Code
$s = "abcab";
echo countSubstringWithEqualEnds($s);

// This code is contributed by mits
?>

Javascript

<script>
// Most efficient javascript program to count all
// substrings with same first and last characters.

      // assuming lower case only
    var MAX_CHAR = 26;

    function countSubstringWithEqualEnds(s)
    {
        var result = 0;
        var n = s.length;

        // Calculating frequency of each character
        // in the string.
        var count =  Array.from({length: MAX_CHAR}, (_, i) => 0);
        for (var i = 0; i < n; i++)
            count[s.charAt(i).charCodeAt(0)-'a'.charCodeAt(0)]++;

        // Computing result using counts
        for (var i = 0; i < MAX_CHAR; i++)
            result += (count[i] * (count[i] + 1) / 2);

        return result;
    }

    // Driver function
    var s = "abcab";
    document.write(countSubstringWithEqualEnds(s));

// This code is contributed by 29AjayKumar
</script>

Output

7

The above code has time complexity of O(n) and requires O(1) extra space.
Recursive solution to count substrings with same first and last characters

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