Skip to content
Related Articles
Queries to count distinct Binary Strings of all lengths from N to M satisfying given properties
• Difficulty Level : Medium
• Last Updated : 25 Mar, 2021

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] to Q[i] 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 = 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] <= j <= Q[i].

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]] – dp[Q[i] – 1] and print as result.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``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 > Q)``{``    ``// Initialize dp``    ``dp = 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]] - dp[Q[i] - 1];` `        ``if` `(ans < 0)``            ``ans = ans + MOD;` `        ``cout << ans << endl;``    ``}``}` `// Driver Code``int` `main()``{` `    ``int` `K = 3;` `    ``vector > Q``        ``= { { 1, 4 }, { 3, 7 } };` `    ``countStrings(K, Q);` `    ``return` `0;``}`

## Java

 `// 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``    ``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`

## Python3

 `# 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``    ``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`

## C#

 `// 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``    ``dp = 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`

## Javascript

 ``
Output:
```7
24```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up