# 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 subsequences in the given string such that the indices of the characters of the subsequence are Arithmetic Progression.

**Examples:**

Input:S = “xxxyy”Output:6Explanation:

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:2Explanation:

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)}

**Approach:** The key observation in the problem is if there are two characters in a string whose collective 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 every two integers will always form an arithmetic progression. Below is an 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 lengths 1 and 2.

Below is the implementation of the above approach:

## C++

`// C++ implementation to find the ` `// maximum occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the ` `// maximum occurrence 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<string, ` `int` `> 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 occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `import` `java.util.*;` `class` `GFG ` `{` ` ` `// Function to find the ` ` ` `// maximum occurrence 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<String, Integer> freq = ` `new` `HashMap<String,Integer>();` ` ` `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 occurrence of the subsequence` `# such that the indices of characters` `# are in arithmetic progression` `# Function to find the` `# maximum occurrence 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` |

## C#

`// C# implementation to find the ` `// maximum occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG ` `{` `// Function to find the ` `// maximum occurrence 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` ` ` `Dictionary<` `string` `, ` ` ` `int` `> freq = ` `new` `Dictionary<` `string` `, ` ` ` `int` `>();` ` ` `int` `i, j;` ` ` `// Loop to find the frequencies ` ` ` `// of subsequence of length 1` ` ` `for` `( i = 0; i < n; i++) ` ` ` `{` ` ` `string` `temp = ` `""` `;` ` ` `temp += s[i];` ` ` `if` `(freq.ContainsKey(temp))` ` ` `{` ` ` `freq[temp]++;` ` ` `}` ` ` `else` ` ` `{` ` ` `freq[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[i];` ` ` `temp += s[j];` ` ` `if` `(freq.ContainsKey(temp))` ` ` `freq[temp]++;` ` ` `else` ` ` `freq[temp] = 1;` ` ` `}` ` ` `}` ` ` `int` `answer =` `int` `.MinValue;` ` ` `// Finding maximum frequency` ` ` `foreach` `(KeyValuePair<` `string` `,` ` ` `int` `> it ` `in` `freq)` ` ` `answer = Math.Max(answer, it.Value);` ` ` `return` `answer;` `}` ` ` `// Driver Code` `public` `static` `void` `Main(` `string` `[]args)` `{` ` ` `string` `s = ` `"xxxyy"` `;` ` ` `Console.Write(maximumOccurrence(s));` `}` `}` `// This code is contributed by Rutvik_56` |

## Javascript

`<script>` `// Javascript implementation to find the ` `// maximum occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `// Function to find the ` `// maximum occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `function` `maximumOccurrence(s)` `{` ` ` `var` `n = s.length;` ` ` `// Frequencies of subsequence` ` ` `var` `freq = ` `new` `Map();` ` ` `// Loop to find the frequencies ` ` ` `// of subsequence of length 1` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `var` `temp = ` `""` `;` ` ` `temp += s[i];` ` ` `if` `(freq.has(temp))` ` ` `freq.set(temp, freq.get(temp)+1)` ` ` `else` ` ` `freq.set(temp, 1)` ` ` `}` ` ` ` ` `// Loop to find the frequencies ` ` ` `// subsequence of length 2 ` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `for` `(` `var` `j = i + 1; j < n; j++) {` ` ` `var` `temp = ` `""` `;` ` ` `temp += s[i];` ` ` `temp += s[j];` ` ` `if` `(freq.has(temp))` ` ` `freq.set(temp, freq.get(temp)+1)` ` ` `else` ` ` `freq.set(temp, 1)` ` ` `}` ` ` `}` ` ` `var` `answer = -1000000000;` ` ` `// Finding maximum frequency` ` ` `freq.forEach((value, key) => {` ` ` `answer = Math.max(answer, value);` ` ` `});` ` ` `return` `answer;` `}` `// Driver Code` `var` `s = ` `"xxxyy"` `;` `document.write( maximumOccurrence(s));` `</script>` |

**Output:**

6

**Time Complexity:** O(N^{2}), for using two nested loops.**Auxiliary Space: **O(N), where N is the size of the given string.

**Efficient Approach:** The idea is to use the dynamic programming paradigm to compute the frequency of the subsequences of lengths 1 and 2 in the string. Below is an 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 i^{th}character occurred before j^{th}character.

Below is the implementation of the above approach:

## C++

`// C++ implementation to find the ` `// maximum occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the ` `// maximum occurrence 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[26] = { 0 };` ` ` `int` `dp[26][26] = { 0 };` ` ` ` ` `// Loop to count the occurrence` ` ` `// 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 occurrence of subsequence` ` ` `// of length 1 in given string` ` ` `for` `(` `int` `i = 0; i < 26; i++)` ` ` `answer = max(answer, freq[i]);` ` ` ` ` `// Maximum occurrence 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 occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `class` `GFG{` ` ` `// Function to find the ` `// maximum occurrence 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 occurrence` ` ` `// 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 occurrence of subsequence` ` ` `// of length 1 in given String` ` ` `for` `(` `int` `i = ` `0` `; i < ` `26` `; i++)` ` ` `answer = Math.max(answer, freq[i]);` ` ` ` ` `// Maximum occurrence 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` |

## Python3

`# Python3 implementation to find the` `# maximum occurrence of the subsequence` `# such that the indices of characters` `# are in arithmetic progression` `import` `sys` `# Function to find the maximum occurrence` `# of the subsequence such that the ` `# indices of characters are in ` `# arithmetic progression` `def` `maximumOccurrence(s):` ` ` ` ` `n ` `=` `len` `(s)` ` ` `# Frequency for characters` ` ` `freq ` `=` `[` `0` `] ` `*` `(` `26` `)` ` ` `dp ` `=` `[[` `0` `for` `i ` `in` `range` `(` `26` `)] ` ` ` `for` `j ` `in` `range` `(` `26` `)]` ` ` `# Loop to count the occurrence` ` ` `# of ith character before jth` ` ` `# character in the given String` ` ` `for` `i ` `in` `range` `(n):` ` ` `c ` `=` `(` `ord` `(s[i]) ` `-` `ord` `(` `'a'` `))` ` ` `for` `j ` `in` `range` `(` `26` `):` ` ` `dp[j] ` `+` `=` `freq[j]` ` ` `# Increase the frequency` ` ` `# of s[i] or c of String` ` ` `freq ` `+` `=` `1` ` ` `answer ` `=` `-` `sys.maxsize` ` ` `# Maximum occurrence of subsequence` ` ` `# of length 1 in given String` ` ` `for` `i ` `in` `range` `(` `26` `):` ` ` `answer ` `=` `max` `(answer, freq[i])` ` ` `# Maximum occurrence of subsequence` ` ` `# of length 2 in given String` ` ` `for` `i ` `in` `range` `(` `26` `):` ` ` `for` `j ` `in` `range` `(` `26` `):` ` ` `answer ` `=` `max` `(answer, dp[i][j])` ` ` `return` `answer` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `s ` `=` `"xxxyy"` ` ` `print` `(maximumOccurrence(s))` `# This code is contributed by Princi Singh` |

## C#

`// C# implementation to find the ` `// maximum occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` `using` `System;` `class` `GFG{` ` ` `// Function to find the maximum ` `// occurrence 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 occurrence` ` ` `// 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 occurrence of subsequence` ` ` `// of length 1 in given String` ` ` `for` `(` `int` `i = 0; i < 26; i++)` ` ` `answer = Math.Max(answer, freq[i]);` ` ` ` ` `// Maximum occurrence 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` |

## Javascript

`<script>` `// javascript implementation to find the ` `// maximum occurrence of the subsequence` `// such that the indices of characters` `// are in arithmetic progression` ` ` `// Function to find the` ` ` `// maximum occurrence of the subsequence` ` ` `// such that the indices of characters` ` ` `// are in arithmetic progression` ` ` `function` `maximumOccurrence(s) {` ` ` `var` `n = s.length;` ` ` `// Frequency for characters` ` ` `var` `freq = Array(26).fill(0);` ` ` `var` `dp = Array(26).fill().map(()=>Array(26).fill(0));` ` ` `// Loop to count the occurrence` ` ` `// of ith character before jth` ` ` `// character in the given String` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` `var` `c = (s.charCodeAt(i) - ` `'a'` `.charCodeAt(0));` ` ` `for` `(` `var` `j = 0; j < 26; j++)` ` ` `dp[j] += freq[j];` ` ` `// Increase the frequency` ` ` `// of s[i] or c of String` ` ` `freq++;` ` ` `}` ` ` `var` `answer = Number.MIN_VALUE;` ` ` `// Maximum occurrence of subsequence` ` ` `// of length 1 in given String` ` ` `for` `(` `var` `i = 0; i < 26; i++)` ` ` `answer = Math.max(answer, freq[i]);` ` ` `// Maximum occurrence of subsequence` ` ` `// of length 2 in given String` ` ` `for` `(` `var` `i = 0; i < 26; i++) {` ` ` `for` `(` `var` `j = 0; j < 26; j++) {` ` ` `answer = Math.max(answer, dp[i][j]);` ` ` `}` ` ` `}` ` ` `return` `answer;` ` ` `}` ` ` `// Driver Code` ` ` ` ` `var` `s = ` `"xxxyy"` `;` ` ` `document.write(maximumOccurrence(s));` `// This code contributed by Princi Singh ` `</script>` |

**Output:**

6

**Time complexity:** O(26 * N)**Auxiliary space**: O(1) as constant space is required by the algorithm.