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Longest sub-sequence of a binary string divisible by 3
• Difficulty Level : Easy
• Last Updated : 14 Jan, 2020

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)

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