# Parity of the given mathematical expression using given N numbers

• Difficulty Level : Expert
• Last Updated : 13 Sep, 2021

Given N positive integers A1, A2, …, AN, the task is to determine the parity of the expression S.
For the given N numbers, the expression S is given as: Examples:

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Input: N = 3, A1 = 2, A2 = 3, A3 = 1
Output: Even
Explanation:
S = 1 + (2 + 3 + 1) + (2*3 + 3*1 + 1*2) + (2*3*1) = 24, which is even

Input: N = 2, A1 = 2, A2 = 4
Output: Odd
Explanation:
S = 1 + (2 + 4) + (2 * 4) = 15, which is odd

Naive Approach: The naive approach for this problem is to plug in all the values of Ai in the given expression and find the parity of the given expression. This method doesn’t work for higher values of N as the multiplication is not a constant operation for higher-ordered numbers. And also, the value might become so large that it might cause integer overflow.

Efficient Approach: The idea is to perform some processing on the expression and reduce the expression into simpler terms so that the parity can be checked without computing the value. Let N = 3. Then:

• The expression S is:

1 + (A1 + A2 + A3) + ((A1 * A2) + (A2 * A3) + (A3 * A1) + (A1 * A2 * A3)

• Now, the same expression is restructured as follows:

(1 + A1) + (A2 + A1 * A2) + (A3 + A3 * A1) + (A2 * A3 + A1 * A2 * A3)
=> (1 + A1) + A2 * (1 + A1) + A3 * (1 + A1) + A2 * A3 * (1 + A1

• On taking (1 + A1) common from the above equation,

(1 + A1) * (1 + A2 + A2 + (A2 * A3))
=> (1 + A1) * (1 + A2 + A3 * (1 + A2

• Finally, on taking (1 + A2) common, the final expression becomes:

(1 + A1) * (1 + A2) * (1 + A3

• By symmetry, for N elements, the expression S becomes:

(1 + A1) * (1 + A2) * (1 + A3) … * (1 + AN

• Clearly, for a number to become even parity, the answer must be even. It is known that the answer is even if any of the numbers are even.
• Therefore, the idea is to check if any of the numbers in the given input is odd. If it is, then on adding one, it becomes even and the value is even parity.

Below is the implementation of the above approach:

## C++

 // C++ program to determine the// parity of the given mathematical// expression #include using namespace std; void getParity(    int n,    const vector<int>& A){     // Iterating through the    // given integers    for (auto x : A) {        if (x & 1) {             // If any odd number            // is present, then S            // is even parity            cout << "Even" << endl;            return;        }    }     // Else, S is odd parity    cout << "Odd" << endl;} // Driver codeint main(){     int N = 3;    vector<int> A = { 2, 3, 1 };    getParity(N, A);     return 0;}

## Java

 // Java program to determine the// parity of the given mathematical// expressionclass GFG{ static void getParity(int n, int []A){     // Iterating through the    // given integers    for (int x : A)    {        if ((x & 1) == 1)        {             // If any odd number            // is present, then S            // is even parity            System.out.println("Even");            return;        }    }     // Else, S is odd parity    System.out.println("Odd");} // Driver codepublic static void main(String[] args){    int N = 3;    int [] A = { 2, 3, 1 };    getParity(N, A);}} // This code is contributed by AnkitRai01

## Python3

 # Python3 program to determine the# parity of the given mathematical# expressiondef getParity(n, A):     # Iterating through    # the given integers    for x in A:        if (x & 1):             # If any odd number            # is present, then S            # is even parity            print("Even")            return     # Else, S is odd parity    print("Odd")     # Driver codeif __name__ == '__main__':     N = 3    A = [ 2, 3, 1 ]         getParity(N, A) # This code is contributed by mohit kumar 29

## C#

 // C# program to determine the// parity of the given mathematical// expressionusing System; public class GFG{     static void getParity(int n, int []A)    {             // Iterating through the        // given integers        foreach (int x in A)        {            if ((x & 1) == 1)            {                     // If any odd number                // is present, then S                // is even parity                Console.WriteLine("Even");                return;            }        }             // Else, S is odd parity        Console.WriteLine("Odd");    }         // Driver code    public static void Main(string[] args)    {        int N = 3;        int [] A = { 2, 3, 1 };        getParity(N, A);    }} // This code is contributed by AnkitRai01

## Javascript

 
Output:
Even

Time Complexity: O(N), where N is the number of given numbers.

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