Skip to content
Related Articles
Number of sub-sequences of non-zero length of a binary string divisible by 3
• Last Updated : 13 May, 2021

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:
“11”, “1001”, “0”, “0” and “00” are
the only subsequences divisible by 3.
Input: S = “1”
Output:

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 a 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`

## Javascript

 ``
Output:
`1`

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up