C++ Code to Determine if an Array Can Be Colored With Equal Parity Sums

Here is a C++ code that solves the problem:
\

#include <iostream>\
#include <vector>\
\
using namespace std;\
\
bool canColorArray(vector<int>& nums) {\
    int n = nums.size();\
    int sum = 0;\
    int oddCount = 0;\
    int evenCount = 0;\
\
    // Calculate the sum of all elements and count the number of odd and even elements\
    for (int i = 0; i < n; i++) {\
        sum += nums[i];\
        if (nums[i] % 2 == 0) {\
            evenCount++;\
        } else {\
            oddCount++;\
        } \
    }\
\
    // If the sum is odd, it is not possible to color the array in the required way\
    if (sum % 2 != 0) {\
        return false;\
    }\
\
    // If there are only odd or even elements, it is not possible to color the array in the required way\
    if (oddCount == 0 || evenCount == 0) {\
        return false;\
    }\
\
    return true;\
}\
\
int main() {\
    // Example usage\
    vector<int> nums = {1, 2, 4, 3, 2, 3, 5, 4};\
    if (canColorArray(nums)) {\
        cout << "It is possible to color the array in the required way." << endl;\
    } else {\
        cout << "It is not possible to color the array in the required way." << endl;\
    }\
\
    return 0;\
}\
```\
\
The code starts by calculating the sum of all elements in the given array and counting the number of odd and even elements. If the sum is odd or there are only odd or even elements, it is not possible to color the array in the required way. Otherwise, it is possible to color the array.