Count of substrings having the most frequent character in the string as first character

Given a string S consisting of lowercase alphabets of size N, the task is to count all substrings which contain the most frequent character in the string as the first character. 

Note: If more than one character has a maximum frequency, consider the lexicographically smallest among them.

Examples:

Input: S = “abcab”
Output: 7
Explanation:
There are two characters a and b occurring maximum times i.e., 2 times.
Selecting the lexicographically smaller character i.e. ‘a’.
Substrings starts with ‘a’ are: “a”, “ab”, “abc”, “abca”, “abcab”, “a”, “ab”.
Therefore the count is 7.

Input: S= “cccc”
Output: 10

Approach: The idea is to first find the character that occurs the maximum number of times and then count the substring starting with that character in the string. Follow the steps below to solve the problem:

• Initialize the count as 0 that will store the total count of strings.
• Find the maximum occurring character in the string S. Let that character be ch.
• Traverse the string using the variable i and if the character at i^th index is the same as ch, increment the count by (N – i).
• After the above steps, print the value of count as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find all substrings
// whose first character occurs
// maximum number of times
int substringCount(string s)
{

    // Stores frequency of characters
    vector<int> freq(26, 0);

    // Stores character that appears
    // maximum number of times
    char max_char = '#';

    // Stores max frequency of character
    int maxfreq = INT_MIN;

    // Updates frequency of characters
    for(int i = 0; i < s.size(); i++)
    {
        freq[s[i] - 'a']++;

        // Update maxfreq
        if (maxfreq < freq[s[i] - 'a'])
            maxfreq = freq[s[i] - 'a'];
    }

    // Character that occurs
    // maximum number of times
    for(int i = 0; i < 26; i++)
    {

        // Update the maximum frequency
        // character
        if (maxfreq == freq[i])
        {
            max_char = (char)(i + 'a');
            break;
        }
    }

    // Stores all count of substrings
    int ans = 0;

    // Traverse over string
    for(int i = 0; i < s.size(); i++)
    {

        // Get the current character
        char ch = s[i];

        // Update count of substrings
        if (max_char == ch)
        {
            ans += (s.size() - i);
        }
    }

    // Return the count of all
    // valid substrings
    return ans;
}

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

    // Function Call
    cout << (substringCount(S));
}

// This code is contributed by mohit kumar 29

Java

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

class GFG {

    // Function to find all substrings
    // whose first character occurs
    // maximum number of times
    static int substringCount(String s)
    {

        // Stores frequency of characters
        int[] freq = new int[26];

        // Stores character that appears
        // maximum number of times
        char max_char = '#';

        // Stores max frequency of character
        int maxfreq = Integer.MIN_VALUE;

        // Updates frequency of characters
        for (int i = 0;
             i < s.length(); i++) {
            freq[s.charAt(i) - 'a']++;

            // Update maxfreq
            if (maxfreq
                < freq[s.charAt(i) - 'a'])
                maxfreq
                    = freq[s.charAt(i) - 'a'];
        }

        // Character that occurs
        // maximum number of times
        for (int i = 0; i < 26; i++) {

            // Update the maximum frequency
            // character
            if (maxfreq == freq[i]) {
                max_char = (char)(i + 'a');
                break;
            }
        }

        // Stores all count of substrings
        int ans = 0;

        // Traverse over string
        for (int i = 0;
             i < s.length(); i++) {

            // Get the current character
            char ch = s.charAt(i);

            // Update count of substrings
            if (max_char == ch) {
                ans += (s.length() - i);
            }
        }

        // Return the count of all
        // valid substrings
        return ans;
    }

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

        String S = "abcab";

        // Function Call
        System.out.println(substringCount(S));
    }
}

Python3

# Python3 program for the above approach
import sys

# Function to find all substrings
# whose first character occurs
# maximum number of times
def substringCount(s):

    # Stores frequency of characters
    freq = [0 for i in range(26)]

    # Stores character that appears
    # maximum number of times
    max_char = '#'

    # Stores max frequency of character
    maxfreq = -sys.maxsize - 1

    # Updates frequency of characters
    for i in range(len(s)):
        freq[ord(s[i]) - ord('a')] += 1

        # Update maxfreq
        if (maxfreq < freq[ord(s[i]) - ord('a')]):
            maxfreq = freq[ord(s[i]) - ord('a')]

    # Character that occurs
    # maximum number of times
    for i in range(26):

        # Update the maximum frequency
        # character
        if (maxfreq == freq[i]):
            max_char = chr(i + ord('a'))
            break

    # Stores all count of substrings
    ans = 0

    # Traverse over string
    for i in range(len(s)):

        # Get the current character
        ch = s[i]

        # Update count of substrings
        if (max_char == ch):
            ans += (len(s) - i)

    # Return the count of all
    # valid substrings
    return ans

# Driver Code
if __name__ == '__main__':

    S = "abcab"

    # Function Call
    print(substringCount(S))

# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;

class GFG{

// Function to find all substrings
// whose first character occurs
// maximum number of times
static int substringCount(string s)
{

    // Stores frequency of characters
    int[] freq = new int[26];

    // Stores character that appears
    // maximum number of times
    char max_char = '#';

    // Stores max frequency of character
    int maxfreq = Int32.MinValue;

    // Updates frequency of characters
    for(int i = 0; i < s.Length; i++)
    {
        freq[s[i] - 'a']++;

        // Update maxfreq
        if (maxfreq < freq[s[i] - 'a'])
            maxfreq = freq[s[i] - 'a'];
    }

    // Character that occurs
    // maximum number of times
    for(int i = 0; i < 26; i++)
    {

        // Update the maximum frequency
        // character
        if (maxfreq == freq[i])
        {
            max_char = (char)(i + 'a');
            break;
        }
    }

    // Stores all count of substrings
    int ans = 0;

    // Traverse over string
    for(int i = 0; i < s.Length; i++)
    {

        // Get the current character
        char ch = s[i];

        // Update count of substrings
        if (max_char == ch)
        {
            ans += (s.Length - i);
        }
    }

    // Return the count of all
    // valid substrings
    return ans;
}

// Driver Code
public static void Main()
{
    string S = "abcab";

    // Function Call
    Console.WriteLine(substringCount(S));
}
}

// This code is contributed by susmitakundugoaldanga

Javascript

<script>

// JavaScript program for the above approach

// Function to find all substrings
// whose first character occurs
// maximum number of times
function substringCount(s)
{

    // Stores frequency of characters
    var freq = new Array(26).fill(0);

    // Stores character that appears
    // maximum number of times
    var max_char = "#";

    // Stores max frequency of character
    var maxfreq = -21474836487;

    // Updates frequency of characters
    for(var i = 0; i < s.length; i++)
    {
        freq[s[i].charCodeAt(0) -
              "a".charCodeAt(0)]++;

        // Update maxfreq
        if (maxfreq < freq[s[i].charCodeAt(0) -
                            "a".charCodeAt(0)])
            maxfreq = freq[s[i].charCodeAt(0) -
                            "a".charCodeAt(0)];
    }

    // Character that occurs
    // maximum number of times
    for(var i = 0; i < 26; i++)
    {

        // Update the maximum frequency
        // character
        if (maxfreq === freq[i])
        {
            max_char = String.fromCharCode(
                i + "a".charCodeAt(0));
            break;
        }
    }

    // Stores all count of substrings
    var ans = 0;

    // Traverse over string
    for(var i = 0; i < s.length; i++)
    {

        // Get the current character
        var ch = s[i];

        // Update count of substrings
        if (max_char === ch)
        {
            ans += s.length - i;
        }
    }

    // Return the count of all
    // valid substrings
    return ans;
}

// Driver Code
var S = "abcab";

// Function Call
document.write(substringCount(S));

// This code is contributed by rdtank

</script>

Output:

7

Time Complexity: O(N), as we are using a loop to traverse the string.
Auxiliary Space: O(1), as we are using freq array of size 26 which is constant.