# Count sequences of given length having non-negative prefix sums that can be generated by given values

Given two integers M and X, the task is to find the number of sequences of length M that can be generated comprising of X and -X such that their respective counts are equal and the prefix sum upto each index of the resulting sequence is non-negative.

Examples:

Input: M = 4, X = 5
Output: 2
Explanation:
There are only 2 possible sequences that have all possible prefix sums non-negative:

1. {+5, +5, -5, -5}
2. {+5, -5, +5, -5}

Input: M = 6, X = 2
Output: 5
Explanation:
There are only 5 possible sequences that have all possible prefix sums non-negative:

1. {+2, +2, +2, -2, -2, -2}
2. {+2, +2, -2, -2, +2, -2}
3. {+2, -2, +2, -2, +2, -2}
4. {+2, +2, -2, +2, -2, -2}
5. {+2, -2, +2, +2, -2, -2}

Naive Approach: The simplest approach is to generate all possible arrangements of size M with the given integers +X and -X and find the prefix sum of each arrangement formed and count those sequence whose prefix sum array has only non-negative elements. Print the count of such sequence after the above steps.

Time Complexity: O((M*(M!))/((M/2)!)2
Auxiliary Space: O(M)

Efficient Approach: The idea is to observe the pattern that for any sequence formed the number of positive X that has occurred at each index is always greater than or equal to the number of negative X occurred. This is similar to the pattern of Catalan Numbers. In this case, check that at any point the number of positive X occurred is always greater than or equal to the number of negative X occurred which is the pattern of Catalan numbers. So the task is to find Nth Catalan number where N = M/2. where, KN is Nth the Catalan Number and is the binomial coefficient.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach     #include  using namespace std;     // Function to find the Binomial  // Coefficient C(n, r)  unsigned long int binCoff(unsigned int n,                            unsigned int r)  {      // Stores the value C(n, r)      unsigned long int val = 1;      int i;         // Update C(n, r) = C(n, n - r)      if (r > (n - r))          r = (n - r);         // Find C(n, r) iteratively      for (i = 0; i < r; i++) {          val *= (n - i);          val /= (i + 1);      }         // Return the final value      return val;  }     // Function to find number of sequence  // whose prefix sum at each index is  // always non-negative  void findWays(int M)  {      // Find n      int n = M / 2;         unsigned long int a, b, ans;         // Value of C(2n, n)      a = binCoff(2 * n, n);         // Catalan number      b = a / (n + 1);         // Print the answer      cout << b;  }     // Driver Code  int main()  {      // Given M and X      int M = 4, X = 5;         // Function Call      findWays(M);         return 0;  }

## Java

 // Java program for the above approach  import java.io.*;     class GFG{     // Function to find the Binomial  // Coefficient C(n, r)  static long binCoff(long n, long r)  {             // Stores the value C(n, r)      long val = 1;      int i;         // Update C(n, r) = C(n, n - r)      if (r > (n - r))          r = (n - r);         // Find C(n, r) iteratively      for(i = 0; i < r; i++)      {          val *= (n - i);          val /= (i + 1);      }         // Return the final value      return val;  }     // Function to find number of sequence  // whose prefix sum at each index is  // always non-negative  static void findWays(int M)  {             // Find n      int n = M / 2;         long a, b, ans;         // Value of C(2n, n)      a = binCoff(2 * n, n);         // Catalan number      b = a / (n + 1);         // Print the answer      System.out.print(b);  }     // Driver Code  public static void main(String[] args)  {         // Given M and X      int M = 4, X = 5;         // Function Call      findWays(M);  }  }     // This code is contributed by akhilsaini

## Python3

 # Python3 program for the above approach     # Function to find the Binomial  # Coefficient C(n, r)  def binCoff(n, r):         # Stores the value C(n, r)      val = 1        # Update C(n, r) = C(n, n - r)      if (r > (n - r)):          r = (n - r)         # Find C(n, r) iteratively      for i in range(0, r):          val *= (n - i)          val //= (i + 1)         # Return the final value      return val     # Function to find number of sequence  # whose prefix sum at each index is  # always non-negative  def findWays(M):         # Find n      n = M // 2        # Value of C(2n, n)      a = binCoff(2 * n, n)         # Catalan number      b = a // (n + 1)         # Print the answer      print(b)     # Driver Code  if __name__ == '__main__':         # Given M and X      M = 4     X = 5        # Function Call      findWays(M)     # This code is contributed by akhilsaini

## C#

 // C# program for the above approach  using System;     class GFG{     // Function to find the Binomial  // Coefficient C(n, r)  static long binCoff(long n, long r)  {             // Stores the value C(n, r)      long val = 1;      int i;         // Update C(n, r) = C(n, n - r)      if (r > (n - r))          r = (n - r);         // Find C(n, r) iteratively      for(i = 0; i < r; i++)      {          val *= (n - i);          val /= (i + 1);      }         // Return the final value      return val;  }     // Function to find number of sequence  // whose prefix sum at each index is  // always non-negative  static void findWays(int M, int X)  {             // Find n      int n = M / 2;         long a, b;         // Value of C(2n, n)      a = binCoff(2 * n, n);         // Catalan number      b = a / (n + 1);         // Print the answer      Console.WriteLine(b);  }     // Driver Code  public static void Main()  {             // Given M and X      int M = 4;      int X = 5;         // Function Call      findWays(M, X);  }  }     // This code is contributed by akhilsaini

Output:

2



Time Complexity: O(M)
Auxiliary Space: O(1)

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.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : akhilsaini