Given **N** positive integers **A _{1}, A_{2}, …, A_{N}**, the task is to determine the parity of the expression S.

For the given

**N**numbers, the expression S is given as:

**Examples:**

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 A_{i} 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 + (A

_{1}+ A_{2}+ A_{3}) + ((A_{1}* A_{2}) + (A_{2}* A_{3}) + (A_{3}* A_{1}) + (A_{1}* A_{2}* A_{3}) - Now, the same expression is restructured as follows:

(1 + A

_{1}) + (A_{2}+ A_{1}* A_{2}) + (A_{3}+ A_{3}* A_{1}) + (A_{2}* A_{3}+ A_{1}* A_{2}* A_{3})=> (1 + A

_{1}) + A_{2}* (1 + A_{1}) + A_{3}* (1 + A_{1}) + A_{2}* A_{3}* (1 + A_{1}) - On taking (1 + A
_{1}) common from the above equation,

(1 + A

_{1}) * (1 + A_{2}+ A_{2}+ (A_{2}* A_{3}))=> (1 + A

_{1}) * (1 + A_{2}+ A_{3}* (1 + A_{2}) - Finally, on taking (1 + A
_{2}) common, the final expression becomes:

(1 + A

_{1}) * (1 + A_{2}) * (1 + A_{3}) - By symmetry, for N elements, the expression S becomes:

(1 + A

_{1}) * (1 + A_{2}) * (1 + A_{3}) … * (1 + A_{N}) - 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 <bits/stdc++.h> ` `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 code ` `int` `main() ` `{ ` ` ` ` ` `int` `N = 3; ` ` ` `vector<` `int` `> A = { 2, 3, 1 }; ` ` ` `getParity(N, A); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java program to determine the ` `// parity of the given mathematical ` `// expression ` `class` `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 code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `N = ` `3` `; ` ` ` `int` `[] A = { ` `2` `, ` `3` `, ` `1` `}; ` ` ` `getParity(N, A); ` `} ` `} ` ` ` `// This code is contributed by AnkitRai01 ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 program to determine the ` `# parity of the given mathematical ` `# expression ` `def` `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 code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `N ` `=` `3` ` ` `A ` `=` `[ ` `2` `, ` `3` `, ` `1` `] ` ` ` ` ` `getParity(N, A) ` ` ` `# This code is contributed by mohit kumar 29 ` |

*chevron_right*

*filter_none*

## C#

`// C# program to determine the ` `// parity of the given mathematical ` `// expression ` `using` `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 ` |

*chevron_right*

*filter_none*

**Output:**

Even

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

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.