Open In App

# 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 X and -X such that their respective counts are equal and the prefix sum up to 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 sequences 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 that occurred. This is similar to the pattern of Catalan Numbers. In this case, check that at any point the number of positive X that occurred is always greater than or equal to the number of negative X that occurred which is the pattern of Catalan numbers. So the task is to find the 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-negativevoid 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 Codeint main(){    // Given M and X    int M = 4, X = 5;     // Function Call    findWays(M);     return 0;}

## Java

 // Java program for the above approachimport 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-negativestatic 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 Codepublic 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-negativedef 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 Codeif __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 approachusing 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-negativestatic 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 Codepublic static void Main(){         // Given M and X    int M = 4;    int X = 5;     // Function Call    findWays(M, X);}} // This code is contributed by akhilsaini

## Javascript

 

Output:

2

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