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# Count of Binary strings of length N having atmost M consecutive 1s or 0s alternatively exactly K times

• Difficulty Level : Medium
• Last Updated : 03 May, 2021

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:
The 3 configurations are:
11001
10011
11011
Explanation:
Notice that the groups of 1’s and 0’s alternate exactly K times
Input: 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: • 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:
1. When N is 0 and K is 0, then return 1
2. When N is 0 but K is not 0, then return 0
3. When N is not 0 but K is 0, then return 0
4. 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 using namespace std; // Array to contain the final resultint dp; // Function to get the number// of desirable binary stringsint 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 codeint main(){     int N = 7, K = 4, M = 3;    cout << solve(N, K, M);}

## Java

 // Java program to find the count of// Binary Strings of length N having// atmost M consecutive 1s or 0s// alternatively exactly K timesimport java.util.*; class GFG{ // Array to contain the final resultstatic int [][]dp = new int; // Function to get the number// of desirable binary stringsstatic 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 codepublic 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

## 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 resultrows, cols = (1000, 1000)dp = [[0 for i in range(cols)]         for j in range(rows)] # Function to get the number# of desirable binary stringsdef 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 codeN = 7K = 4M = 3 print(solve(N, K, M)) # This code is contributed by ishayadav181

## C#

 // C# program to find the count of// binary strings of length N having// atmost M consecutive 1s or 0s// alternatively exactly K timesusing System; class GFG{ // Array to contain the readonly resultstatic int [,]dp = new int[1000, 1000]; // Function to get the number// of desirable binary stringsstatic 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 codepublic static void Main(String[] args){    int N = 7, K = 4, M = 3;         Console.Write(solve(N, K, M));}} // This code is contributed by gauravrajput1

## Javascript

 
Output:
16

Time complexity: O(N*K*M)

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