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((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++

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// C++ implementation of th 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

1

Time Complexity: O(n)



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