Frequency of maximum occurring subsequence in given string

Given a string str of lowercase English alphabets, our task is to find the frequency of occurrence a subsequence of the string which occurs the maximum times.

Examples:

Input: s = “aba” 
Output: 2 
Explanation:
For “aba”, subsequence “ab” occurs maximum times in subsequence ‘ab’ and ‘aba’.

Input: s = “acbab” 
Output: 3 
Explanation:
For “acbab”, “ab” occurs 3 times which is the maximum.

Approach: The problem can be solved using Dynamic Programming. To solve the problem mentioned above the key observation is that the resultant subsequence will be of length 1 or 2 because frequency of any subsequence of length > 2 will be lower than the subsequence of length 1 or 2 as they are also present in higher length subsequences. So we need to check for the subsequence of length 1 or 2 only. Below are the steps:

• For length 1 count the frequency of each alphabet in the string.
• For length 2 form a 2D array dp[26][26], where dp[i][j] tells frequency of string of char(‘a’ + i) + char(‘a’ + j).
• The recurrence relation is used in the step 2 is given by:

 dp[i][j] = dp[i][j] + freq[i] 
where, 
freq[i] = frequency of character char(‘a’ + i) 
dp[i][j] = frequency of string formed by current_character + char(‘a’ + i).

• The maximum of frequency array and array dp[][] gives the maximum count of any subsequence in the given string.

Below is the implementation of the above approach:

C++

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

// Function to find the frequency
ll findCount(string s)
{
    // freq stores frequency of each
    // english lowercase character
    ll freq[26];

    // dp[i][j] stores the count of
    // subsequence with 'a' + i
    // and 'a' + j character
    ll dp[26][26];

    memset(freq, 0, sizeof freq);

    // Initialize dp to 0
    memset(dp, 0, sizeof dp);

    for (int i = 0; i < s.size(); ++i) {

        for (int j = 0; j < 26; j++) {

            // Increment the count of
            // subsequence j and s[i]
            dp[j][s[i] - 'a'] += freq[j];
        }

        // Update the frequency array
        freq[s[i] - 'a']++;
    }

    ll ans = 0;

    // For 1 length subsequence
    for (int i = 0; i < 26; i++)
        ans = max(freq[i], ans);

    // For 2 length subsequence
    for (int i = 0; i < 26; i++) {
        for (int j = 0; j < 26; j++) {

            ans = max(dp[i][j], ans);
        }
    }

    // Return the final result
    return ans;
}

// Driver Code
int main()
{
    // Given string str
    string str = "acbab";

    // Function Call
    cout << findCount(str);

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to find the frequency
static int findCount(String s)
{

    // freq stores frequency of each
    // english lowercase character
    int []freq = new int[26];

    // dp[i][j] stores the count of
    // subsequence with 'a' + i
    // and 'a' + j character
    int [][]dp = new int[26][26];

    for(int i = 0; i < s.length(); ++i)
    {
        for(int j = 0; j < 26; j++)
        {

            // Increment the count of
            // subsequence j and s[i]
            dp[j][s.charAt(i) - 'a'] += freq[j];
        }

        // Update the frequency array
        freq[s.charAt(i) - 'a']++;
    }

    int ans = 0;

    // For 1 length subsequence
    for(int i = 0; i < 26; i++)
        ans = Math.max(freq[i], ans);

    // For 2 length subsequence
    for(int i = 0; i < 26; i++)
    {
        for(int j = 0; j < 26; j++)
        {
            ans = Math.max(dp[i][j], ans);
        }
    }

    // Return the final result
    return ans;
}

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

    // Given String str
    String str = "acbab";

    // Function call
    System.out.print(findCount(str));
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 program for the above approach
import numpy

# Function to find the frequency
def findCount(s):

    # freq stores frequency of each
    # english lowercase character
    freq = [0] * 26

    # dp[i][j] stores the count of
    # subsequence with 'a' + i
    # and 'a' + j character
    dp = [[0] * 26] * 26

    freq = numpy.zeros(26)
    dp = numpy.zeros([26, 26])

    for i in range(0, len(s)):
        for j in range(26):

            # Increment the count of
            # subsequence j and s[i]
            dp[j][ord(s[i]) - ord('a')] += freq[j]

        # Update the frequency array
        freq[ord(s[i]) - ord('a')] += 1

    ans = 0

    # For 1 length subsequence
    for i in range(26):
        ans = max(freq[i], ans)

    # For 2 length subsequence
    for i in range(0, 26):
        for j in range(0, 26):
            ans = max(dp[i][j], ans)

    # Return the final result
    return int(ans)

# Driver Code

# Given string str
str = "acbab"

# Function call
print(findCount(str))

# This code is contributed by sanjoy_62

C#

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

class GFG{

// Function to find the frequency
static int findCount(String s)
{

    // freq stores frequency of each
    // english lowercase character
    int []freq = new int[26];

    // dp[i,j] stores the count of
    // subsequence with 'a' + i
    // and 'a' + j character
    int [,]dp = new int[26, 26];

    for(int i = 0; i < s.Length; ++i)
    {
        for(int j = 0; j < 26; j++)
        {

            // Increment the count of
            // subsequence j and s[i]
            dp[j, s[i] - 'a'] += freq[j];
        }

        // Update the frequency array
        freq[s[i] - 'a']++;
    }

    int ans = 0;

    // For 1 length subsequence
    for(int i = 0; i < 26; i++)
        ans = Math.Max(freq[i], ans);

    // For 2 length subsequence
    for(int i = 0; i < 26; i++)
    {
        for(int j = 0; j < 26; j++)
        {
            ans = Math.Max(dp[i, j], ans);
        }
    }

    // Return the readonly result
    return ans;
}

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

    // Given String str
    String str = "acbab";

    // Function call
    Console.Write(findCount(str));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program for the above approach

// Function to find the frequency
function findCount(s)
{
    // freq stores frequency of each
    // english lowercase character
    var freq = Array(26).fill(0);

    // dp[i][j] stores the count of
    // subsequence with 'a' + i
    // and 'a' + j character
    var dp = Array.from(Array(26), ()=>Array(26).fill(0));

    for (var i = 0; i < s.length; ++i) {

        for (var j = 0; j < 26; j++) {

            // Increment the count of
            // subsequence j and s[i]
            dp[j][s[i].charCodeAt(0) -
            'a'.charCodeAt(0)] += freq[j];
        }

        // Update the frequency array
        freq[s[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
    }

    var ans = 0;

    // For 1 length subsequence
    for (var i = 0; i < 26; i++)
        ans = Math.max(freq[i], ans);

    // For 2 length subsequence
    for (var i = 0; i < 26; i++) {
        for (var j = 0; j < 26; j++) {

            ans = Math.max(dp[i][j], ans);
        }
    }

    // Return the final result
    return ans;
}

// Driver Code

// Given string str
var str = "acbab";

// Function Call
document.write( findCount(str));

</script>

Output:

3

Time Complexity: O(26*N), where N is the length of the given string.
Auxiliary Space: O(M), where M = 26*26