Today we learn how to count how many even and odd numbers there are in a sequence of integers. Like any problem, many solutions exist and here we analyse just one among many possible ones.

This is the complete code:

/* Write a function to count how many even and odd numbers there are in a sequence of integers. Examples: Input Returns [1, 6, 3, 2, 5, 12, 7] 3 even, 4 [-2, -12, 5, 8, 9, 10] 4 even, 2 odd */ #include <iostream> #include <tuple> using namespace std; // Foward declarations int countEven(int *seq); tuple<int , int> countEvenOdd(int *seq); main() { // Initialise sequences int seq1[] = {1, 6, 3, 2, 5, 12, 7} ; int seq2[] = {-2, -12, 5, 8, 9, 10}; // Run the function int nEven1 = countEven(seq1); int nEven2 = countEven(seq2); // Print outputs cout << "Number of even number" << endl; cout << " in seq1: " << nEven1 << endl; cout << " in seq2: " << nEven2 << endl; // Print both even and odd numbers with a different function int nEven, nOdd; tie(nEven , nOdd) = countEvenOdd(seq1); printf("seq1 has %d even numbers and %d odd numbers. \n", nEven, nOdd); tie(nEven , nOdd) = countEvenOdd(seq2); printf("seq2 has %d even numbers and %d odd numbers. \n", nEven, nOdd); return 0; } int countEven(int *seq){ int counter = 0; // Set up a counter for (int i=0; i < sizeof(seq)-1; i++) { // loop if (seq[i]%2 == 0) // divide by two: the number is even if remainder is zero counter += 1; // } return counter; } tuple<int , int> countEvenOdd(int *seq) { int even = 0; int odd =0; for (int i=0; i < sizeof(seq)-1; i++) // loop (seq[i]%2 == 0) ? even += 1 : odd += 1 ; return make_tuple(even, odd); }