Count substrings with different first and last characters

Given a string S, the task is to print the count of substrings from a given string whose first and last characters are different.

Examples:

Input: S = “abcab”
Output: 8
Explanation:
There are 8 substrings having first and last characters different {ab, abc, abcab, bc, bca, ca, cab, ab}.

Input: S = “aba”
Output: 2
Explanation:
There are 2 substrings having first and last characters different {ab, ba}.

Naive Approach: The idea is to generate all possible substrings of a given string and for each substring, check if the first and the last characters are different or not. If found to be true, then increment the count by 1 and check for the next substring. Print the count after traversal of all the substring.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the substrings
// having different first and last
// characters
int countSubstring(string s, int n)
{
    // Store the final count
    int ans = 0;

    // Loop to traverse the string
    for (int i = 0; i < n; i++) {

        // Counter for each iteration
        int cnt = 0;

        // Iterate over substrings
        for (int j = i + 1; j < n; j++) {

            // Compare the characters
            if (s[j] != s[i])

                // Increase count
                cnt++;
        }

        // Adding count of substrings
        // to the final count
        ans += cnt;
    }

    // Print the final count
    cout << ans;
}

// Driver Code
int main()
{
    // Given string
    string S = "abcab";

    // Length of the string
    int N = 5;

    // Function Call
    countSubstring(S, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG{

// Function to count the substrings
// having different first and last
// characters
static void countSubstring(String s, int n)
{

    // Store the final count
    int ans = 0;

    // Loop to traverse the string
    for(int i = 0; i < n; i++)
    {

        // Counter for each iteration
        int cnt = 0;

        // Iterate over substrings
        for(int j = i + 1; j < n; j++)
        {

            // Compare the characters
            if (s.charAt(j) != s.charAt(i))

                // Increase count
                cnt++;
        }

        // Adding count of substrings
        // to the final count
        ans += cnt;
    }

    // Print the final count
    System.out.print(ans);
}

// Driver Code
public static void main(String[] args)
{

    // Given string
    String S = "abcab";

    // Length of the string
    int N = 5;

    // Function call
    countSubstring(S, N);
}
}

// This code is contributed by code_hunt

Python3

# Python3 program for the above approach

# Function to count the substrings
# having different first and last
# characters
def countSubstring(s, n):

    # Store the final count
    ans = 0

    # Loop to traverse the string
    for i in range(n):

        # Counter for each iteration
        cnt = 0

        # Iterate over substrings
        for j in range(i + 1, n):

            # Compare the characters
            if (s[j] != s[i]):

                # Increase count
                cnt += 1

        # Adding count of substrings
        # to the final count
        ans += cnt

    # Print the final count
    print(ans)

# Driver Code

# Given string
S = "abcab"

# Length of the string
N = 5

# Function call
countSubstring(S, N)

# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;

class GFG{

// Function to count the substrings
// having different first and last
// characters
static void countSubstring(string s, int n)
{

    // Store the final count
    int ans = 0;

    // Loop to traverse the string
    for(int i = 0; i < n; i++)
    {

        // Counter for each iteration
        int cnt = 0;

        // Iterate over substrings
        for(int j = i + 1; j < n; j++)
        {

            // Compare the characters
            if (s[j] != s[i])

                // Increase count
                cnt++;
        }

        // Adding count of substrings
        // to the final count
        ans += cnt;
    }

    // Print the final count
    Console.Write(ans);
}

// Driver Code
public static void Main(string[] args)
{

    // Given string
    string S = "abcab";

    // Length of the string
    int N = 5;

    // Function call
    countSubstring(S, N);
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// JavaScript program for
// the above approach

// Function to count the substrings
// having different first and last
// characters
function countSubstring(s, n)
{

    // Store the final count
    let ans = 0;

    // Loop to traverse the string
    for(let i = 0; i < n; i++)
    {

        // Counter for each iteration
        let cnt = 0;

        // Iterate over substrings
        for(let j = i + 1; j < n; j++)
        {

            // Compare the characters
            if (s[j] != s[i])

                // Increase count
                cnt++;
        }

        // Adding count of substrings
        // to the final count
        ans += cnt;
    }

    // Print the final count
    document.write(ans);
}


// Driver code

    // Given string
    let S = "abcab";

    // Length of the string
    let N = 5;

    // Function call
    countSubstring(S, N);

     // This code is contributed by splevel62.
</script>

Output:

8

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Map by to store the frequency of the characters of the string. Follow the steps below to solve the problem:

1. Initialize two variables, one for counting different characters for every iteration (say cur) and one to store the final count of substrings (say ans).
2. Initialize a map M to store the frequency of all characters in it.
3. Traverse the given string and for each character, follow the steps below:
   • Iterate the map M.
   • If the first element i.e., the key of the map is not the same as the current character, then proceed.
   • Otherwise, add the value corresponding to the current character.
4. After traversal of the Map, add cur to the final result i.e., ans += cur.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the substrings
// having different first & last character
int countSubstring(string s, int n)
{
    // Stores frequency of each char
    map<char, int> m;

    // Loop to store frequency of
    // the characters in a Map
    for (int i = 0; i < n; i++)
        m[s[i]]++;

    // To store final result
    int ans = 0;

    // Traversal of string
    for (int i = 0; i < n; i++) {

        // Store answer for every
        // iteration
        int cnt = 0;
        m[s[i]]--;

        // Map traversal
        for (auto value : m) {

            // Compare current char
            if (value.first == s[i]) {
                continue;
            }
            else {
                cnt += value.second;
            }
        }

        ans += cnt;
    }

    // Print the final count
    cout << ans;
}

// Driver Code
int main()
{
    // Given string
    string S = "abcab";

    // Length of the string
    int N = 5;

    // Function Call
    countSubstring(S, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to count the subStrings
// having different first & last character
static void countSubString(char []s, int n)
{

    // Stores frequency of each char
    HashMap<Character,
            Integer> mp = new HashMap<Character,
                                      Integer>();

    // Loop to store frequency of
    // the characters in a Map
    for(int i = 0; i < n; i++)

        if (mp.containsKey(s[i]))
        {
            mp.put(s[i], mp.get(s[i]) + 1);
        }
        else
        {
            mp.put(s[i], 1);
        }

    // To store final result
    int ans = 0;

    // Traversal of String
    for(int i = 0; i < n; i++)
    {

        // Store answer for every
        // iteration
        int cnt = 0;
        if (mp.containsKey(s[i]))
        {
            mp.put(s[i], mp.get(s[i]) - 1);

            // Map traversal
            for(Map.Entry<Character,
                          Integer> value : mp.entrySet())
            {

                // Compare current char
                if (value.getKey() == s[i])
                {
                    continue;
                }
                else
                {
                    cnt += value.getValue();
                }
            }
            ans += cnt;
        }
    }

    // Print the final count
    System.out.print(ans);
}

// Driver Code
public static void main(String[] args)
{

    // Given String
    String S = "abcab";

    // Length of the String
    int N = 5;

    // Function call
    countSubString(S.toCharArray(), N);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach

# Function to count the substrings
# having different first & last character
def countSubstring(s, n):

    # Stores frequency of each char
    m = {}

    # Loop to store frequency of
    # the characters in a Map
    for i in range(n):
        if s[i] in m:
            m[s[i]] += 1
        else:
            m[s[i]] = 1

    # To store final result
    ans = 0

    # Traversal of string
    for i in range(n):

        # Store answer for every
        # iteration
        cnt = 0

        if s[i] in m:
            m[s[i]] -= 1
        else:
            m[s[i]] = -1

        # Map traversal
        for value in m:

            # Compare current char
            if (value == s[i]):
                continue
            else:
                cnt += m[value]

        ans += cnt

    # Print the final count
    print(ans)

# Driver code

# Given string
S = "abcab"

# Length of the string
N = 5

# Function Call
countSubstring(S, N)

# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to count the subStrings
// having different first & last character
static void countSubString(char []s, int n)
{

    // Stores frequency of each char
    Dictionary<char,
               int> mp = new Dictionary<char,
                                        int>();

    // Loop to store frequency of
    // the characters in a Map
    for(int i = 0; i < n; i++)

        if (mp.ContainsKey(s[i]))
        {
            mp[s[i]] = mp[s[i]] + 1;
        }
        else
        {
            mp.Add(s[i], 1);
        }

    // To store readonly result
    int ans = 0;

    // Traversal of String
    for(int i = 0; i < n; i++)
    {

        // Store answer for every
        // iteration
        int cnt = 0;
        if (mp.ContainsKey(s[i]))
        {
            mp[s[i]] = mp[s[i]] - 1;

            // Map traversal
            foreach(KeyValuePair<char,
                                 int> value in mp)
            {

                // Compare current char
                if (value.Key == s[i])
                {
                    continue;
                }
                else
                {
                    cnt += value.Value;
                }
            }
            ans += cnt;
        }
    }

    // Print the readonly count
    Console.Write(ans);
}

// Driver Code
public static void Main(String[] args)
{

    // Given String
    String S = "abcab";

    // Length of the String
    int N = 5;

    // Function call
    countSubString(S.ToCharArray(), N);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// Javascript program for the above approach

// Function to count the substrings
// having different first & last character
function countSubstring(s, n)
{

    // Stores frequency of each char
    var m = new Map();

    // Loop to store frequency of
    // the characters in a Map
    for (var i = 0; i < n; i++)
    {
        if(m.has(s[i]))
            m.set(s[i], m.get(s[i])+1)
        else
            m.set(s[i], 1)
    }

    // To store final result
    var ans = 0;

    // Traversal of string
    for (var i = 0; i < n; i++) {

        // Store answer for every
        // iteration
        var cnt = 0;
        if(m.has(s[i]))
            m.set(s[i], m.get(s[i])-1)


        // Map traversal
        m.forEach((value, key) => {

            // Compare current char
            if (key != s[i]) {
                cnt += value;
            }

        });

        ans += cnt;
    }

    // Print the final count
    document.write( ans);
}

// Driver Code

// Given string
var S = "abcab";

// Length of the string
var N = 5;

// Function Call
countSubstring(S, N);

// This code is contributed by itsok.
</script>

Output:

8

Time Complexity: O(N*26)
Auxiliary Space: O(N)