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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 <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 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




<script>
 
// Javascript implementation of th 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)
 

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