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Frequency of maximum occurring subsequence in given string
• Difficulty Level : Hard
• Last Updated : 14 Sep, 2020

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:
Explanation:
For “aba”, subsequence “ab” occurs maximum times in subsequence ‘ab’ and ‘aba’.

Input: s = “acbab”
Output:
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:

1. For length 1 count the frequency of each alphabet in the string.
2. For length 2 form a 2D array dp, where dp[i][j] tells frequency of string of char(‘a’ + i) + char(‘a’ + j).
3. 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).

4. 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  ``#define ll long long ``using` `namespace` `std; `` ` `// Function to find the frequency ``ll findCount(string s) ``{ ``    ``// freq stores frequnecy of each ``    ``// english lowercase character ``    ``ll freq; `` ` `    ``// dp[i][j] stores the count of ``    ``// subsequnce with 'a' + i ``    ``// and 'a' + j character ``    ``ll dp; `` ` `    ``memset``(freq, 0, ``sizeof` `freq); `` ` `    ``// Intialize 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 frequnecy of each``    ``// english lowercase character``    ``int` `[]freq = ``new` `int``[``26``];`` ` `    ``// dp[i][j] stores the count of``    ``// subsequnce 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 frequnecy of each ``    ``# english lowercase character ``    ``freq ``=` `[``0``] ``*` `26`` ` `    ``# dp[i][j] stores the count of ``    ``# subsequnce 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 frequnecy of each``    ``// english lowercase character``    ``int` `[]freq = ``new` `int``;`` ` `    ``// dp[i,j] stores the count of``    ``// subsequnce 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`
Output:
```3
```

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

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