# Count maximum occurrence of subsequence in string such that indices in subsequence is in A.P.

Given a string S, the task is to count the maximum occurrence of subsequence in the given string such that indices of the characters of the subsequence are in Arithmetic Progression.

Examples:

Input: S = “xxxyy”
Output: 6
Explanation:
There is a subsequence “xy”, where indices of each character of the subsequence are in A.P.
The indices of the different characters that form the subsequence “xy” –
{(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

Input: S = “pop”
Output: 2
Explanation:
There is a subsequence “p”, where indices of each character of the subsequence are in A.P.
The indices of the different characters that form the subsequence “p” –
{(1), (2)}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The key observation in the problem is if there are two characters in string whose collectively occurrence is greater than the occurrence of any single character, then these characters will form the maximum occurrence subsequence in the string with the character in Arithmetic progression because of every two integers will always form an arithmetic progression. Below is the illustration of the steps:

• Iterate over the string and count the frequency of the characters of the string. That is considering the subsequences of length 1.
• Iterate over the string and choose every two possible characters of the string and increment the frequency of the subsequence of the string.
• Finally, find the maximum frequency of the subsequence from length 1 and 2.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` `int` `maximumOccurrence(string s) ` `{ ` `    ``int` `n = s.length(); ` ` `  `    ``// Frequencies of subsequence ` `    ``map freq; ` ` `  `    ``// Loop to find the frequencies  ` `    ``// of subsequence of length 1 ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``string temp = ``""``; ` `        ``temp += s[i]; ` `        ``freq[temp]++; ` `    ``} ` `     `  `    ``// Loop to find the frequencies  ` `    ``// subsequence of length 2  ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``string temp = ``""``; ` `            ``temp += s[i]; ` `            ``temp += s[j]; ` `            ``freq[temp]++; ` `        ``} ` `    ``} ` ` `  `    ``int` `answer = INT_MIN; ` ` `  `    ``// Finding maximum frequency ` `    ``for` `(``auto` `it : freq) ` `        ``answer = max(answer, it.second); ` `    ``return` `answer; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s = ``"xxxyy"``; ` ` `  `    ``cout << maximumOccurrence(s); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `    ``// Function to find the  ` `    ``// maximum occurence of the subsequence ` `    ``// such that the indices of characters ` `    ``// are in arithmetic progression ` `    ``static` `int` `maximumOccurrence(String s) ` `    ``{ ` `        ``int` `n = s.length(); ` `      `  `        ``// Frequencies of subsequence ` `        ``HashMap freq = ``new` `HashMap(); ` `        ``int` `i, j; ` ` `  `        ``// Loop to find the frequencies  ` `        ``// of subsequence of length 1 ` `        ``for` `( i = ``0``; i < n; i++) { ` `            ``String temp = ``""``; ` `            ``temp += s.charAt(i); ` `            ``if` `(freq.containsKey(temp)){ ` `                ``freq.put(temp,freq.get(temp)+``1``);  ` `            ``} ` `            ``else``{ ` `                ``freq.put(temp, ``1``);  ` `            ``} ` `        ``} ` `          `  `        ``// Loop to find the frequencies  ` `        ``// subsequence of length 2  ` `        ``for` `(i = ``0``; i < n; i++) { ` `            ``for` `(j = i + ``1``; j < n; j++) { ` `                ``String temp = ``""``; ` `                ``temp += s.charAt(i); ` `                ``temp += s.charAt(j); ` `                ``if``(freq.containsKey(temp)) ` `                    ``freq.put(temp,freq.get(temp)+``1``); ` `                ``else`  `                    ``freq.put(temp,``1``); ` `            ``} ` `        ``} ` `        ``int` `answer = Integer.MIN_VALUE; ` `      `  `        ``// Finding maximum frequency ` `        ``for` `(``int` `it : freq.values()) ` `            ``answer = Math.max(answer, it); ` `        ``return` `answer; ` `    ``} ` `      `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `        ``String s = ``"xxxyy"``; ` `      `  `        ``System.out.print(maximumOccurrence(s)); ` `    ``} ` `} ` ` `  `// This code is contributed by chitranayal `

## Python3

 `# Python3 implementation to find the ` `# maximum occurence of the subsequence ` `# such that the indices of characters ` `# are in arithmetic progression ` ` `  `# Function to find the ` `# maximum occurence of the subsequence ` `# such that the indices of characters ` `# are in arithmetic progression ` `def` `maximumOccurrence(s): ` `    ``n ``=` `len``(s) ` ` `  `    ``# Frequencies of subsequence ` `    ``freq ``=` `{} ` ` `  `    ``# Loop to find the frequencies ` `    ``# of subsequence of length 1 ` `    ``for` `i ``in` `s: ` `        ``temp ``=` `"" ` `        ``temp ``+``=` `i ` `        ``freq[temp] ``=` `freq.get(temp, ``0``) ``+` `1` ` `  `    ``# Loop to find the frequencies ` `    ``# subsequence of length 2 ` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``temp ``=` `"" ` `            ``temp ``+``=` `s[i] ` `            ``temp ``+``=` `s[j] ` `            ``freq[temp] ``=` `freq.get(temp, ``0``) ``+` `1` ` `  `    ``answer ``=` `-``10``*``*``9` ` `  `    ``# Finding maximum frequency ` `    ``for` `it ``in` `freq: ` `        ``answer ``=` `max``(answer, freq[it]) ` `    ``return` `answer ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``s ``=` `"xxxyy"` ` `  `    ``print``(maximumOccurrence(s)) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```6
```

Time Complexity: O(N2)

Efficient Approach: The idea is to use the dynamic programming paradigm to compute the frequency of the subsequences of the length 1 and 2 in the string. Below is the illustration of the steps:

• Compute the frequency of the characters of the string in a frequency array.
• For subsequences of the string of length 2, the DP state will be
```dp[i][j] = Total number of times ith
character occured before jth character.
```

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` ` `  `#include ` ` `  `using` `namespace` `std; ` ` `  `// Function to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` `int` `maximumOccurrence(string s) ` `{ ` `    ``int` `n = s.length(); ` ` `  `    ``// Frequency for characters ` `    ``int` `freq = { 0 }; ` `    ``int` `dp = { 0 }; ` `     `  `    ``// Loop to count the occurence ` `    ``// of ith character before jth ` `    ``// character in the given string ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``int` `c = (s[i] - ``'a'``); ` ` `  `        ``for` `(``int` `j = 0; j < 26; j++) ` `            ``dp[j] += freq[j]; ` ` `  `        ``// Increase the frequency  ` `        ``// of s[i] or c of string ` `        ``freq++; ` `    ``} ` ` `  `    ``int` `answer = INT_MIN; ` `     `  `    ``// Maximum occurence of subsequence ` `    ``// of length 1 in given string ` `    ``for` `(``int` `i = 0; i < 26; i++) ` `        ``answer = max(answer, freq[i]); ` `         `  `    ``// Maximum occurence of subsequence ` `    ``// of length 2 in given string ` `    ``for` `(``int` `i = 0; i < 26; i++) { ` `        ``for` `(``int` `j = 0; j < 26; j++) { ` `            ``answer = max(answer, dp[i][j]); ` `        ``} ` `    ``} ` ` `  `    ``return` `answer; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string s = ``"xxxyy"``; ` ` `  `    ``cout << maximumOccurrence(s); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` ` `  ` `  `class` `GFG{ ` `  `  `// Function to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` `static` `int` `maximumOccurrence(String s) ` `{ ` `    ``int` `n = s.length(); ` `  `  `    ``// Frequency for characters ` `    ``int` `freq[] = ``new` `int``[``26``]; ` `    ``int` `dp[][] = ``new` `int``[``26``][``26``]; ` `      `  `    ``// Loop to count the occurence ` `    ``// of ith character before jth ` `    ``// character in the given String ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``int` `c = (s.charAt(i) - ``'a'``); ` `  `  `        ``for` `(``int` `j = ``0``; j < ``26``; j++) ` `            ``dp[j] += freq[j]; ` `  `  `        ``// Increase the frequency  ` `        ``// of s[i] or c of String ` `        ``freq++; ` `    ``} ` `  `  `    ``int` `answer = Integer.MIN_VALUE; ` `      `  `    ``// Maximum occurence of subsequence ` `    ``// of length 1 in given String ` `    ``for` `(``int` `i = ``0``; i < ``26``; i++) ` `        ``answer = Math.max(answer, freq[i]); ` `          `  `    ``// Maximum occurence of subsequence ` `    ``// of length 2 in given String ` `    ``for` `(``int` `i = ``0``; i < ``26``; i++) { ` `        ``for` `(``int` `j = ``0``; j < ``26``; j++) { ` `            ``answer = Math.max(answer, dp[i][j]); ` `        ``} ` `    ``} ` `  `  `    ``return` `answer; ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String s = ``"xxxyy"``; ` `  `  `    ``System.out.print(maximumOccurrence(s)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## C#

 `// C# implementation to find the  ` `// maximum occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` `using` `System; ` `class` `GFG{ ` `  `  `// Function to find the maximum  ` `// occurence of the subsequence ` `// such that the indices of characters ` `// are in arithmetic progression ` `static` `int` `maximumOccurrence(``string` `s) ` `{ ` `    ``int` `n = s.Length; ` `      `  `    ``// Frequency for characters ` `    ``int` `[]freq = ``new` `int``; ` `    ``int` `[,]dp = ``new` `int``[26, 26]; ` `          `  `    ``// Loop to count the occurence ` `    ``// of ith character before jth ` `    ``// character in the given String ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `       ``int` `x = (s[i] - ``'a'``); ` `        `  `       ``for``(``int` `j = 0; j < 26; j++) ` `          ``dp[x, j] += freq[j]; ` `           `  `       ``// Increase the frequency  ` `       ``// of s[i] or c of String ` `       ``freq[x]++; ` `    ``} ` `      `  `    ``int` `answer = ``int``.MinValue; ` `          `  `    ``// Maximum occurence of subsequence ` `    ``// of length 1 in given String ` `    ``for``(``int` `i = 0; i < 26; i++) ` `       ``answer = Math.Max(answer, freq[i]); ` `              `  `    ``// Maximum occurence of subsequence ` `    ``// of length 2 in given String ` `    ``for``(``int` `i = 0; i < 26; i++) ` `    ``{ ` `       ``for``(``int` `j = 0; j < 26; j++) ` `       ``{ ` `          ``answer = Math.Max(answer, dp[i, j]); ` `       ``} ` `    ``} ` `    ``return` `answer; ` `} ` `      `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``string` `s = ``"xxxyy"``; ` `      `  `    ``Console.Write(maximumOccurrence(s)); ` `} ` `} ` ` `  `// This code is contributed by Yash_R `

Output:

```6
```

Time complexity: O(26 * N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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