Count of Binary strings of length N having atmost M consecutive 1s or 0s alternatively exactly K times
Given three integers, N, K and M. The task is to find out the number of binary strings of length N which always starts with 1, in which there can be at most M consecutive 1’s or 0’s and they alternate exactly K times.
Examples:
Input: N = 5, K = 3, M = 2
Output: 3
The 3 configurations are:
11001
10011
11011
Explanation:
Notice that the groups of 1’s and 0’s alternate exactly K timesInput: N = 7, K = 4, M = 3
Output: 16
Approach: Since this problem involves both overlapping sub-problem and optimal substructure. So, this problem can be solved using dynamic programming.
- Sub-problem: DP[i][j] represents the number of binary strings upto length i having j alternating groups till now. So, to calculate dp[N][K] if we know the value of dp[n-j][k-1], then we can easily get the result by summing up the sub-problem value over j = 1 to m (DP[N][K] represents the final answer).
As shown below in the recursion tree diagram, it is observed many sub-problem overlaps. So, the result needs to be cached to avoid redundant calculations.
- Optimal substructure:
![$dp[i][j]=$$\sum_{j=1}^{M} f(N-j, K-1)](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-82f50b7780ce69c6fffb54bf5b2617dd_l3.png "Rendered by QuickLaTeX.com")
- By following the top-down DP approach:
As we can have a group which can be atmost of the length M, so we iterate on every possible length and recur with new N and decreasing K by 1, as a new group is formed. Solution to sub-problem is cached and summed up to give final result dp[N][K]. - Base Case:
- When N is 0 and K is 0, then return 1
- When N is 0 but K is not 0, then return 0
- When N is not 0 but K is 0, then return 0
- When both are negative, return 0
Below is the implementation of above approach:
C++
// C++ program to find the count // of Binary strings of length N // having atmost M consecutive 1s or 0s // alternatively exactly K times #include <bits/stdc++.h> using namespace std; // Array to contain the final result int dp[1000][1000]; // Function to get the number // of desirable binary strings int solve(int n, int k, int m) { // if we reach end of string // and groups are exhausted, // return 1 if (n == 0 && k == 0) return 1; // if length is exhausted but // groups are still to be made, // return 0 if (n == 0 && k != 0) return 0; // if length is not exhausted // but groups are exhausted, // return 0 if (n != 0 && k == 0) return 0; // if both are negative // just return 0 if (n < 0 || k < 0) return 0; // if already calculated, // return it if (dp[n][k]) return dp[n][k]; // initialise answer // for each state int ans = 0; // loop through every // possible m for (int j = 1; j <= m; j++) { ans += solve(n - j, k - 1, m); } return dp[n][k] = ans; } // Driver code int main() { int N = 7, K = 4, M = 3; cout << solve(N, K, M); } |
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Java
// Java program to find the count of // Binary Strings of length N having // atmost M consecutive 1s or 0s // alternatively exactly K times import java.util.*; class GFG{ // Array to contain the final result static int [][]dp = new int[1000][1000]; // Function to get the number // of desirable binary strings static int solve(int n, int k, int m) { // If we reach end of string // and groups are exhausted, // return 1 if (n == 0 && k == 0) return 1; // If length is exhausted but // groups are still to be made, // return 0 if (n == 0 && k != 0) return 0; // If length is not exhausted // but groups are exhausted, // return 0 if (n != 0 && k == 0) return 0; // If both are negative // just return 0 if (n < 0 || k < 0) return 0; // If already calculated, // return it if (dp[n][k] > 0) return dp[n][k]; // Initialise answer // for each state int ans = 0; // Loop through every // possible m for(int j = 1; j <= m; j++) { ans += solve(n - j, k - 1, m); } return dp[n][k] = ans; } // Driver code public static void main(String[] args) { int N = 7, K = 4, M = 3; System.out.print(solve(N, K, M)); } } // This code is contributed by Rajput-Ji |
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Python 3
# Python3 program to find the count # of Binary strings of length N # having atmost M consecutive 1s or # 0s alternatively exactly K times # List to contain the final result rows, cols = (1000, 1000) dp = [[0 for i in range(cols)] for j in range(rows)] # Function to get the number # of desirable binary strings def solve(n, k, m): # If we reach end of string # and groups are exhausted, # return 1 if n == 0 and k == 0: return 1 # If length is exhausted but # groups are still to be made, # return 0 if n == 0 and k != 0: return 0 # If length is not exhausted # but groups are exhausted, # return 0 if n != 0 and k == 0: return 0 # If both are negative # just return 0 if n < 0 or k < 0: return 0 # If already calculated, # return it if dp[n][k]: return dp[n][k] # Initialise answer # for each state ans = 0 # Loop through every # possible m for j in range(1, m + 1): ans = ans + solve(n - j, k - 1, m) dp[n][k] = ans return dp[n][k] # Driver code N = 7K = 4M = 3 print(solve(N, K, M)) # This code is contributed by ishayadav181 |
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C#
// C# program to find the count of // binary strings of length N having // atmost M consecutive 1s or 0s // alternatively exactly K times using System; class GFG{ // Array to contain the readonly result static int [,]dp = new int[1000, 1000]; // Function to get the number // of desirable binary strings static int solve(int n, int k, int m) { // If we reach end of string // and groups are exhausted, // return 1 if (n == 0 && k == 0) return 1; // If length is exhausted but // groups are still to be made, // return 0 if (n == 0 && k != 0) return 0; // If length is not exhausted // but groups are exhausted, // return 0 if (n != 0 && k == 0) return 0; // If both are negative // just return 0 if (n < 0 || k < 0) return 0; // If already calculated, // return it if (dp[n, k] > 0) return dp[n, k]; // Initialise answer // for each state int ans = 0; // Loop through every // possible m for(int j = 1; j <= m; j++) { ans += solve(n - j, k - 1, m); } return dp[n, k] = ans; } // Driver code public static void Main(String[] args) { int N = 7, K = 4, M = 3; Console.Write(solve(N, K, M)); } } // This code is contributed by gauravrajput1 |
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Output:
16
Time complexity: O(N*K*M)
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