# Longest sub-sequence of a binary string divisible by 3

Given a binary string S of length N, the task is to find the length of the longest sub-sequence in it which is divisible by 3. Leading zeros in the sub-sequences are allowed.

Examples:

Input: S = “1001”
Output: 4
The longest sub-sequence divisible by 3 is “1001”.
1001 = 9 which is divisible by 3.

Input: S = “1011”
Output: 3

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

Naive approach: Generate all the possible sub-sequences and check if they are divisible by 3. The time complexity for this will be O((2N) * N).

Efficient approach: Dynamic programming can be used to solve this problem. Let’s look at the states of DP.
DP[i][r] will store the longest sub-sequence of the substring S[i…N-1] such that it gives a remainder of (3 – r) % 3 when divided by 3.
Let’s write the recurrence relation now.

DP[i][r] = max(1 + DP[i + 1][(r * 2 + s[i]) % 3], DP[i + 1][r])

The recurrence is derived because of the following two choices:

1. Include the current index i in the sub-sequence. Thus, the r will be updated as r = (r * 2 + s[i]) % 3.
2. Don’t include the current index in the sub-sequence.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define N 100 ` ` `  `int` `dp[N]; ` `bool` `v[N]; ` ` `  `// Function to return the length of the ` `// largest sub-string divisible by 3 ` `int` `findLargestString(string& s, ``int` `i, ``int` `r) ` `{ ` `    ``// Base-case ` `    ``if` `(i == s.size()) { ` `        ``if` `(r == 0) ` `            ``return` `0; ` `        ``else` `            ``return` `INT_MIN; ` `    ``} ` ` `  `    ``// If the state has been solved ` `    ``// before then return its value ` `    ``if` `(v[i][r]) ` `        ``return` `dp[i][r]; ` ` `  `    ``// Marking the state as solved ` `    ``v[i][r] = 1; ` ` `  `    ``// Recurrence relation ` `    ``dp[i][r] ` `        ``= max(1 + findLargestString(s, i + 1, ` `                                    ``(r * 2 + (s[i] - ``'0'``)) % 3), ` `              ``findLargestString(s, i + 1, r)); ` `    ``return` `dp[i][r]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"101"``; ` ` `  `    ``cout << findLargestString(s, 0, 0); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of th approach  ` `class` `GFG  ` `{ ` ` `  `    ``final` `static` `int` `N = ``100` `; ` `    ``final` `static` `int` `INT_MIN = Integer.MIN_VALUE; ` `     `  `    ``static` `int` `dp[][] = ``new` `int``[N][``3``];  ` `    ``static` `int` `v[][] = ``new` `int``[N][``3``];  ` `     `  `     `  `    ``// Function to return the length of the  ` `    ``// largest sub-string divisible by 3  ` `    ``static` `int` `findLargestString(String s, ``int` `i, ``int` `r)  ` `    ``{  ` `        ``// Base-case  ` `        ``if` `(i == s.length()) ` `        ``{  ` `            ``if` `(r == ``0``)  ` `                ``return` `0``;  ` `            ``else` `                ``return` `INT_MIN;  ` `        ``}  ` `     `  `        ``// If the state has been solved  ` `        ``// before then return its value  ` `        ``if` `(v[i][r] == ``1``)  ` `            ``return` `dp[i][r];  ` `     `  `        ``// Marking the state as solved  ` `        ``v[i][r] = ``1``;  ` `     `  `        ``// Recurrence relation  ` `        ``dp[i][r] = Math.max(``1` `+ findLargestString(s, i + ``1``,  ` `                          ``(r * ``2` `+ (s.charAt(i) - ``'0'``)) % ``3``),  ` `                            ``findLargestString(s, i + ``1``, r));  ` `        ``return` `dp[i][r]; ` `    ``} ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``String s = ``"101"``;  ` `     `  `        ``System.out.print(findLargestString(s, ``0``, ``0``));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach  ` `import` `numpy as np ` `import` `sys ` ` `  `N ``=` `100` `INT_MIN ``=` `-``(sys.maxsize ``-` `1``) ` ` `  `dp ``=` `np.zeros((N, ``3``));  ` `v ``=` `np.zeros((N, ``3``));  ` ` `  `# Function to return the length of the  ` `# largest sub-string divisible by 3  ` `def` `findLargestString(s, i, r) :  ` ` `  `    ``# Base-case  ` `    ``if` `(i ``=``=` `len``(s)) : ` `        ``if` `(r ``=``=` `0``) : ` `            ``return` `0``;  ` `        ``else` `: ` `            ``return` `INT_MIN;  ` ` `  `    ``# If the state has been solved  ` `    ``# before then return its value  ` `    ``if` `(v[i][r]) : ` `        ``return` `dp[i][r];  ` ` `  `    ``# Marking the state as solved  ` `    ``v[i][r] ``=` `1``;  ` ` `  `    ``# Recurrence relation  ` `    ``dp[i][r] ``=` `max``(``1` `+` `findLargestString(s, i ``+` `1``,  ` `                  ``(r ``*` `2` `+` `(``ord``(s[i]) ``-` `ord``(``'0'``))) ``%` `3``), ` `                       ``findLargestString(s, i ``+` `1``, r));  ` `                 `  `    ``return` `dp[i][r];  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``s ``=` `"101"``;  ` ` `  `    ``print``(findLargestString(s, ``0``, ``0``));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of th approach  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``readonly` `static` `int` `N = 100 ; ` `    ``readonly` `static` `int` `INT_MIN = ``int``.MinValue; ` `     `  `    ``static` `int` `[,]dp = ``new` `int``[N, 3];  ` `    ``static` `int` `[,]v = ``new` `int``[N, 3];  ` `     `  `    ``// Function to return the length of the  ` `    ``// largest sub-string divisible by 3  ` `    ``static` `int` `findLargestString(String s, ``int` `i, ``int` `r)  ` `    ``{  ` `        ``// Base-case  ` `        ``if` `(i == s.Length) ` `        ``{  ` `            ``if` `(r == 0)  ` `                ``return` `0;  ` `            ``else` `                ``return` `INT_MIN;  ` `        ``}  ` `     `  `        ``// If the state has been solved  ` `        ``// before then return its value  ` `        ``if` `(v[i, r] == 1)  ` `            ``return` `dp[i, r];  ` `     `  `        ``// Marking the state as solved  ` `        ``v[i, r] = 1;  ` `     `  `        ``// Recurrence relation  ` `        ``dp[i, r] = Math.Max(1 + findLargestString(s, i + 1,  ` `                                ``(r * 2 + (s[i] - ``'0'``)) % 3),  ` `                            ``findLargestString(s, i + 1, r));  ` `        ``return` `dp[i, r]; ` `    ``} ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `        ``String s = ``"101"``;  ` `     `  `        ``Console.Write(findLargestString(s, 0, 0));  ` `    ``}  ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```2
```

Time Complexity: O(n)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : AnkitRai01, 29AjayKumar