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 

Naive approach: Generate all the possible sub-sequences and check if they are divisible by 3. Time complexity for this will be O((2^N) * 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 the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100

int dp[N][3];
bool v[N][3];

// 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 the 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 the 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 the 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

<script>

// Javascript implementation of the approach
var N = 100

var dp = Array.from(Array(N), ()=> Array(3));
var v = Array.from(Array(N), ()=> Array(3));

// Function to return the number of
// sub-sequences divisible by 3
function findCnt(s, i, 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])
        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
var s = "11";
document.write( (findCnt(s, 0, 0) - 1));

</script>

Output:

1

Time Complexity: O(n)
Auxiliary Space: O(n * 3) ⇒ O(n), where n is the length of the given string.