Number of ways to make binary string of length N such that 0s always occur together in groups of size K

Given two integers N and K, the task is to count the number of ways to make a binary string of length N such that 0s always occur together in a group of size K.

Examples:

Input: N = 3, K = 2
Output : 3
No of binary strings:
111
100
001

Input : N = 4, K = 2
Output : 5

This problem can easily be solved using dynamic programming. Let dp[i] be the number of binary strings of length i satisfying the condition.

From the condition it can be deduced that:

  • dp[i] = 1 for 1 <= i < k.
  • Also dp[k] = 2 since a binary string of length K will either be a string of only zeros or only ones.
  • Now if we consider for i > k. If we decide the ith character to be ‘1’, then dp[i] = dp[i-1] since the number of binary strings would not change. However if we decide the ith character to be ‘0’, then we require that previous k-1 characters should also be ‘0’ and hence dp[i] = dp[i-k]. Therefore dp[i] will be the sum of these 2 values.

Thus,

dp[i] = dp[i - 1] + dp[i - k]

Below is the implementation of the above approach:

C++

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// C++ implementation of the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
const int mod = 1000000007;
  
// Function to return no of ways to build a binary 
// string of length N such that 0s always occur 
// in groups of size K
int noOfBinaryStrings(int N, int k)
{
    int dp[100002];
    for (int i = 1; i <= k - 1; i++) {
        dp[i] = 1;
    }
  
    dp[k] = 2;
  
    for (int i = k + 1; i <= N; i++) {
        dp[i] = (dp[i - 1] + dp[i - k]) % mod;
    }
  
    return dp[N];
}
  
// Driver Code
int main()
{
    int N = 4;
    int K = 2;
    cout << noOfBinaryStrings(N, K);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG 
{
      
static int mod = 1000000007;
  
// Function to return no of ways to build a binary 
// string of length N such that 0s always occur 
// in groups of size K
static int noOfBinaryStrings(int N, int k)
{
    int dp[] = new int[100002];
    for (int i = 1; i <= k - 1; i++) 
    {
        dp[i] = 1;
    }
  
    dp[k] = 2;
  
    for (int i = k + 1; i <= N; i++) 
    {
        dp[i] = (dp[i - 1] + dp[i - k]) % mod;
    }
  
    return dp[N];
}
  
// Driver Code
public static void main(String[] args) 
{
    int N = 4;
    int K = 2;
    System.out.println(noOfBinaryStrings(N, K));
}
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 iimplementation of the
# above approach 
mod = 1000000007
  
# Function to return no of ways to 
# build a binary string of length N
# such that 0s always occur in 
# groups of size K 
def noOfBinaryStrings(N, k) :
    dp = [0] * 100002
    for i in range(1, K) : 
        dp[i] = 1
      
    dp[k] = 2
  
    for i in range(k + 1, N + 1) :
        dp[i] = (dp[i - 1] + dp[i - k]) % mod; 
  
    return dp[N]; 
  
# Driver Code 
if __name__ == "__main__"
  
    N = 4
    K = 2
      
    print(noOfBinaryStrings(N, K)); 
  
# This code is contributed by Ryuga

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C#

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// C# implementation of the approach
using System;
  
class GFG 
{
      
static int mod = 1000000007;
  
// Function to return no of ways to build 
// a binary string of length N such that 
// 0s always occur in groups of size K
static int noOfBinaryStrings(int N, int k)
{
    int []dp = new int[100002];
    for (int i = 1; i <= k - 1; i++) 
    {
        dp[i] = 1;
    }
  
    dp[k] = 2;
  
    for (int i = k + 1; i <= N; i++) 
    {
        dp[i] = (dp[i - 1] + dp[i - k]) % mod;
    }
  
    return dp[N];
}
  
// Driver Code
public static void Main() 
{
    int N = 4;
    int K = 2;
    Console.WriteLine(noOfBinaryStrings(N, K));
}
}
  
/* This code contributed by PrinciRaj1992 */

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PHP

Output:

5


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