array a of size n, and wants to select exactly x elements from it, such that their sum is odd. These elements do not have to be consecutive. The elements of the array are not guaranteed to be distinct. code example

Example: n/3 number appears in array elements

// CPP program to find if any element appears 
// more than n/3. 
#include <bits/stdc++.h> 
using namespace std; 

int appearsNBy3(int arr[], int n) 
{ 
	int count1 = 0, count2 = 0; 
	int first=INT_MAX , second=INT_MAX ; 

	for (int i = 0; i < n; i++) { 

		// if this element is previously seen, 
		// increment count1. 
		if (first == arr[i]) 
			count1++; 

		// if this element is previously seen, 
		// increment count2. 
		else if (second == arr[i]) 
			count2++; 
	
		else if (count1 == 0) { 
			count1++; 
			first = arr[i]; 
		} 

		else if (count2 == 0) { 
			count2++; 
			second = arr[i]; 
		} 

		// if current element is different from 
		// both the previously seen variables, 
		// decrement both the counts. 
		else { 
			count1--; 
			count2--; 
		} 
	} 

	count1 = 0; 
	count2 = 0; 

	// Again traverse the array and find the 
	// actual counts. 
	for (int i = 0; i < n; i++) { 
		if (arr[i] == first) 
			count1++; 

		else if (arr[i] == second) 
			count2++; 
	} 

	if (count1 > n / 3) 
		return first; 

	if (count2 > n / 3) 
		return second; 

	return -1; 
} 

int main() 
{ 
	int arr[] = { 1, 2, 3, 1, 1 }; 
	int n = sizeof(arr) / sizeof(arr[0]); 
	cout << appearsNBy3(arr, n) << endl; 
	return 0; 
}