Queries to count distinct Binary Strings of all lengths from N to M satisfying given properties

Given a K and a matrix Q[][] consisting of queries of the form {N, M}, the task for each query is to count the number of strings possible of all lengths from Q[i][0] to Q[i][1] satisfying the following properties:  

  • The frequency of 0‘s is equal to a multiple of K.
  • Two strings are said to be different only if the frequencies of 0‘s and 1‘s are different

Since the answer can be quite large, compute the answer by mod 109 + 7.
Examples:  

Input: K = 3, Q[][] = {{1, 3}} 
Output: 4 
Explanation: 
All possible strings of length 1 : {“1”} 
All possible strings of length 2 : {“11”} 
All possible strings of length 3 : {“111”, “000”} 
Therefore, a total of 4 strings can be generated.
Input: K = 3, Q[][] = {{1, 4}, {3, 7}} 
Output: 

24 
 

Naive Approach: 
Follow the steps below to solve the problem:  

  • Initialize an array dp[] such that dp[i] denotes the number of strings possible of length i.
  • Initialize dp[0] = 1.
  • For every ith Length, at most two possibilities arise: 
    • Appending ‘1’ to the strings of length i – 1.
    • Add K 0‘s to all possible strings of length i-K.
  • Finally, for each query Q[i], print the sum of all dp[j] for Q[i][0] <= j <= Q[i][1].

Time Complexity: O(N*Q) 
Auxiliary Space: O(N)
Efficient Approach: 
The above approach can be optimized using Prefix Sum Array. Follow the steps below:  



  • Update the dp[] array by following the steps in the above approach.
  • Compute prefix sum array of the dp[] array.
  • Finally, for each query Q[i], calculate dp[Q[i][1]] – dp[Q[i][0] – 1] and print as result.

Below is the implementation of the above approach: 

C++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
const int N = 1e5 + 5;
 
const int MOD = 1000000007;
 
long int dp[N];
 
// Function to calculate the
// count of possible strings
void countStrings(int K,
                  vector<vector<int> > Q)
{
    // Initialize dp[0]
    dp[0] = 1;
 
    // dp[i] represents count of
    // strings of length i
    for (int i = 1; i < N; i++) {
 
        dp[i] = dp[i - 1];
 
        // Add dp[i-k] if i>=k
        if (i >= K)
            dp[i]
                = (dp[i] + dp[i - K]) % MOD;
    }
 
    // Update Prefix Sum Array
    for (int i = 1; i < N; i++) {
        dp[i] = (dp[i] + dp[i - 1]) % MOD;
    }
 
    for (int i = 0; i < Q.size(); i++) {
        long int ans
            = dp[Q[i][1]] - dp[Q[i][0] - 1];
 
        if (ans < 0)
            ans = ans + MOD;
 
        cout << ans << endl;
    }
}
 
// Driver Code
int main()
{
 
    int K = 3;
 
    vector<vector<int> > Q
        = { { 1, 4 }, { 3, 7 } };
 
    countStrings(K, Q);
 
    return 0;
}

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Java

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// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
static int N = (int)(1e5 + 5);
static int MOD = 1000000007;
static int []dp = new int[N];
 
// Function to calculate the
// count of possible Strings
static void countStrings(int K, int[][] Q)
{
     
    // Initialize dp[0]
    dp[0] = 1;
 
    // dp[i] represents count of
    // Strings of length i
    for(int i = 1; i < N; i++)
    {
        dp[i] = dp[i - 1];
 
        // Add dp[i-k] if i>=k
        if (i >= K)
            dp[i] = (dp[i] + dp[i - K]) % MOD;
    }
 
    // Update Prefix Sum Array
    for(int i = 1; i < N; i++)
    {
        dp[i] = (dp[i] + dp[i - 1]) % MOD;
    }
 
    for(int i = 0; i < Q.length; i++)
    {
        int ans = dp[Q[i][1]] - dp[Q[i][0] - 1];
 
        if (ans < 0)
            ans = ans + MOD;
 
        System.out.print(ans + "\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int K = 3;
 
    int [][]Q = { { 1, 4 }, { 3, 7 } };
 
    countStrings(K, Q);
}
}
 
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program to implement
# the above approach
N = int(1e5 + 5)
MOD = 1000000007
dp = [0] * N
 
# Function to calculate the
# count of possible strings
def countStrings(K, Q):
 
    # Initialize dp[0]
    dp[0] = 1
 
    # dp[i] represents count of
    # strings of length i
    for i in range(1, N):
        dp[i] = dp[i - 1]
 
        # Add dp[i-k] if i>=k
        if(i >= K):
            dp[i] = (dp[i] + dp[i - K]) % MOD
 
    # Update Prefix Sum Array
    for i in range(1, N):
        dp[i] = (dp[i] + dp[i - 1]) % MOD
 
    for i in range(len(Q)):
        ans = dp[Q[i][1]] - dp[Q[i][0] - 1]
 
        if (ans < 0):
            ans += MOD
             
        print(ans)
 
# Driver Code
K = 3
 
Q = [ [ 1, 4 ], [ 3, 7 ] ]
 
countStrings(K, Q)
 
# This code is contributed by Shivam Singh

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C#

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// C# program to implement
// the above approach
using System;
class GFG{
  
static int N = (int)(1e5 + 5);
static int MOD = 1000000007;
static int []dp = new int[N];
  
// Function to calculate the
// count of possible Strings
static void countStrings(int K,
                         int[,] Q)
{    
    // Initialize dp[0]
    dp[0] = 1;
  
    // dp[i] represents count of
    // Strings of length i
    for(int i = 1; i < N; i++)
    {
        dp[i] = dp[i - 1];
  
        // Add dp[i-k] if i>=k
        if (i >= K)
            dp[i] = (dp[i] +
                     dp[i - K]) % MOD;
    }
  
    // Update Prefix Sum Array
    for(int i = 1; i < N; i++)
    {
        dp[i] = (dp[i] +
                 dp[i - 1]) % MOD;
    }
  
    for(int i = 0; i < Q.GetLength(0); i++)
    {
        int ans = dp[Q[i, 1]] -
                  dp[Q[i, 0] - 1];
  
        if (ans < 0)
            ans = ans + MOD;
  
        Console.Write(ans + "\n");
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    int K = 3;
    int [,]Q = {{1, 4}, {3, 7}};
    countStrings(K, Q);
}
}
  
// This code is contributed by 29AjayKumar

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Output: 

7
24


 

Time Complexity: O(N + Q) 
Auxiliary Space: O(N) 

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