# Number of sub-sequences of non-zero length of a binary string divisible by 3

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

Examples:

Input: S = “1001”
Output: 5
“11”, “1001”, “0”, “0” and “00” are
the only subsequences divisible by 3.

Input: S = “1”
Output: 0

## 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. Time complexity for this will be O((2N) * N).

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

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

The recurrence is derived because of the two choices below:

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 current index in the sub-sequence.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of th approach ` `#include ` `using` `namespace` `std; ` `#define N 100 ` ` `  `int` `dp[N]; ` `bool` `v[N]; ` ` `  `// Function to return the number of ` `// sub-sequences divisible by 3 ` `int` `findCnt(string& s, ``int` `i, ``int` `r) ` `{ ` `    ``// Base-cases ` `    ``if` `(i == s.size()) { ` `        ``if` `(r == 0) ` `            ``return` `1; ` `        ``else` `            ``return` `0; ` `    ``} ` ` `  `    ``// 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] ` `        ``= findCnt(s, i + 1, (r * 2 + (s[i] - ``'0'``)) % 3) ` `          ``+ findCnt(s, i + 1, r); ` ` `  `    ``return` `dp[i][r]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"11"``; ` ` `  `    ``cout << (findCnt(s, 0, 0) - 1); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of th approach  ` `class` `GFG  ` `{ ` ` `  `    ``static` `final` `int` `N = ``100``;  ` `     `  `    ``static` `int` `dp[][] = ``new` `int``[N][``3``];  ` `    ``static` `int` `v[][] = ``new` `int``[N][``3``];  ` `     `  `    ``// Function to return the number of  ` `    ``// sub-sequences divisible by 3  ` `    ``static` `int` `findCnt(String s, ``int` `i, ``int` `r)  ` `    ``{  ` `        ``// Base-cases  ` `        ``if` `(i == s.length())  ` `        ``{  ` `            ``if` `(r == ``0``)  ` `                ``return` `1``;  ` `            ``else` `                ``return` `0``;  ` `        ``}  ` `     `  `        ``// 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] = findCnt(s, i + ``1``, (r * ``2` `+ (s.charAt(i) - ``'0'``)) % ``3``)  ` `                    ``+ findCnt(s, i + ``1``, r);  ` `     `  `        ``return` `dp[i][r];  ` `    ``} ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``String s = ``"11"``;  ` `     `  `        ``System.out.print(findCnt(s, ``0``, ``0``) - ``1``);  ` `     `  `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of th approach  ` `import` `numpy as np ` `N ``=` `100` ` `  `dp ``=` `np.zeros((N, ``3``));  ` `v ``=` `np.zeros((N, ``3``)); ` ` `  `# Function to return the number of  ` `# sub-sequences divisible by 3  ` `def` `findCnt(s, i, r) : ` ` `  `    ``# Base-cases  ` `    ``if` `(i ``=``=` `len``(s)) :  ` `         `  `        ``if` `(r ``=``=` `0``) : ` `            ``return` `1``;  ` `        ``else` `: ` `            ``return` `0``;  ` ` `  `    ``# 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] ``=` `findCnt(s, i ``+` `1``, (r ``*` `2` `+`  `                      ``(``ord``(s[i]) ``-` `ord``(``'0'``))) ``%` `3``) ``+` `\ ` `               ``findCnt(s, i ``+` `1``, r);  ` ` `  `    ``return` `dp[i][r];  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``s ``=` `"11"``;  ` ` `  `    ``print``(findCnt(s, ``0``, ``0``) ``-` `1``);  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of th approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``static` `readonly` `int` `N = 100;  ` `     `  `    ``static` `int` `[,]dp = ``new` `int``[N, 3];  ` `    ``static` `int` `[,]v = ``new` `int``[N, 3];  ` `     `  `    ``// Function to return the number of  ` `    ``// sub-sequences divisible by 3  ` `    ``static` `int` `findCnt(String s, ``int` `i, ``int` `r)  ` `    ``{  ` `        ``// Base-cases  ` `        ``if` `(i == s.Length)  ` `        ``{  ` `            ``if` `(r == 0)  ` `                ``return` `1;  ` `            ``else` `                ``return` `0;  ` `        ``}  ` `     `  `        ``// 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] = findCnt(s, i + 1, (r * 2 + (s[i] - ``'0'``)) % 3)  ` `                    ``+ findCnt(s, i + 1, r);  ` `     `  `        ``return` `dp[i, r];  ` `    ``} ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``String s = ``"11"``;  ` `     `  `        ``Console.Write(findCnt(s, 0, 0) - 1);  ` `     `  `    ``}  ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```1
```

Time Complexity: O(n)

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