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++ 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 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 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# 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

<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> 

3

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

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Article Tags :

• Strings
• Analysis of Algorithms
• Dynamic Programming
• Greedy
• Searching
• DSA
• Arrays
• subsequence
• frequency-counting

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• Dynamic Programming
• Greedy
• Searching
• Strings

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