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# Frequency of maximum occurring subsequence in given string

• Difficulty Level : Hard
• Last Updated : 14 Jun, 2021

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:

• 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 ` `#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

 ``

Output:

`3`

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

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