I am trying to find the total number of odd values in integers which are elements of an array. For example, the entered array is {113, 220, 235} and the output is:

```
First element has 3 odd numbers.
Second element has 0 odd numbers.
Third element has 2 odd numbers.

```

Till now I have come across the following code.

```
 int optionKS;
 int* arrPtrKS;
 int sizeKS;
 int oddDigitcount;

 cout << "\nWhat is the size of the array? ";
         cin >> sizeKS;
         cout << "\n";
         arrPtrKS = new int[sizeKS];

         for (int i = 0; i < sizeKS; i++) {
             cout << "  Value #" << i + 1 << " ";
             cin >> arrPtrKS[i];
         }

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

     while (arrPtrKS[i] > 0) {
         if (arrPtrKS[i] % 2) {
             oddDigitcount++;
         }

         arrPtrKS[i] /= 10;
     }

 }

```

The above code gives the total number of odd numbers in the whole array but I'm looking for the number of odd values in each element of the array. I have tried to think of every possible way but still couldn't figure out the solution. What would be the best possible to tackle this issue?