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:
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 evenInput: 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 <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 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 // 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 |
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 // 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 |
Javascript
<script>// Javascript program to determine the // parity of the given mathematical // expression function getParity(n, A) { // Iterating through the // given integers for (let x in A) { if ((x & 1) == 1) { // If any odd number // is present, then S // is even parity document.write("Even"); return; } } // Else, S is odd parity document.write("Odd"); } // Driver Code let N = 3; let A = [ 2, 3, 1 ]; getParity(N, A); </script> |
Even
Time Complexity: O(N), where N is the number of given numbers.
Space Complexity : O(1)
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