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 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:
![$dp[i][j]=$$\sum_{j=1}^{M} f(N-j, K-1)](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20356%2037'%3E%3C/svg%3E "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 resultint dp[1000][1000];// 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[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; 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 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 = 3print(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
<script>// Javascript program to find the count of // Binary Strings of length N having// atmost M consecutive 1s or 0s// alternatively exactly K times // Array to contain the final result dp = Array(1000); for(i =0;i<1000;i++) dp[i] = Array(1000).fill(0); // Function to get the number // of desirable binary strings function solve(n , k , 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 var ans = 0; // Loop through every // possible m for (var j = 1; j <= m; j++) { ans += solve(n - j, k - 1, m); // document.write(ans); } return dp[n][k] = ans; } // Driver code var N = 7, K = 4, M = 3; document.write(solve(N, K, M));// This code contributed by umadevi9616</script> |
Output
16
Time complexity: O(N*K*M)
Auxiliary Space: O(1000*1000)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a DP to store the solution of the subproblems and initialize it with 0.
- Initialize the DP with base cases dp[0][0] = 1.
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in dp[n][k].
Implementation :
C++
// C++ code for above approach #include <bits/stdc++.h>using namespace std;// Function to get the number// of desirable binary stringsint countBinaryStrings(int n, int k, int m) { vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0)); // initialize the base cases dp[0][0] = 1; for (int i = 1; i <= m && i <= n; i++) { dp[i][1] = 1; } // compute the remaining values for (int i = 1; i <= n; i++) { for (int j = 2; j <= k; j++) { for (int l = 1; l <= m && l <= i; l++) { // update DP from previus stored values dp[i][j] += dp[i - l][j - 1]; } } } // return final answer return dp[n][k];}int main() { int n = 7, k = 4, m = 3; cout << countBinaryStrings(n, k, m) << endl; return 0;}// --- by bhardwajji |
Java
import java.util.*;public class Main { // Function to get the number // of desirable binary strings static int countBinaryStrings(int n, int k, int m) { int[][] dp = new int[n + 1][k + 1]; // initialize the base cases dp[0][0] = 1; for (int i = 1; i <= m && i <= n; i++) { dp[i][1] = 1; } // compute the remaining values for (int i = 1; i <= n; i++) { for (int j = 2; j <= k; j++) { for (int l = 1; l <= m && l <= i; l++) { // update DP from previous stored values dp[i][j] += dp[i - l][j - 1]; } } } // return final answer return dp[n][k]; } public static void main(String[] args) { int n = 7, k = 4, m = 3; System.out.println(countBinaryStrings(n, k, m)); }} |
Python3
def countBinaryStrings(n, k, m): # initialize dp matrix dp = [[0] * (k + 1) for i in range(n + 1)] # initialize base cases dp[0][0] = 1 for i in range(1, min(m+1, n+1)): dp[i][1] = 1 # compute remaining values for i in range(1, n+1): for j in range(2, k+1): for l in range(1, min(m+1, i+1)): dp[i][j] += dp[i-l][j-1] # return final answer return dp[n][k]# example usagen = 7k = 4m = 3print(countBinaryStrings(n, k, m)) |
C#
using System;class Program{ // Function to count desirable binary strings static int CountBinaryStrings(int n, int k, int m) { // Initialize a 2D array for dynamic programming int[,] dp = new int[n + 1, k + 1]; // Initialize base cases dp[0, 0] = 1; for (int i = 1; i <= m && i <= n; i++) { dp[i, 1] = 1; } // Compute the remaining values for (int i = 1; i <= n; i++) { for (int j = 2; j <= k; j++) { for (int l = 1; l <= m && l <= i; l++) { // Update dp from previously stored values dp[i, j] += dp[i - l, j - 1]; } } } // Return the final answer return dp[n, k]; } static void Main() { int n = 7, k = 4, m = 3; Console.WriteLine(CountBinaryStrings(n, k, m)); }} |
Output
16
Time complexity: O(N*K*M)
Auxiliary Space: O(N*K)
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