NOTE:PASS GOOGLE TEST CASES .NO NEED OF MAIN .USE RECURSION.

WRITE A C++ RECURSIVE FUNCTION TO Print all numbers in range [100,999] such that sum of digits at even position is equal to sum of digits at odd position

FUNCTION PROTOTYPE:

int* sum_of_digits(int start, int end, int size_of_resultant_array)

OUTPUT:

110 121 132 143 154 165 176 187 198 220 231 242 253 264 275 286 297 330 341 352 363 374 385 396 440 451 462 473 484 495 550 561 572 583 594 660 671 682 693 770 781 792 880 891 990

GOOGLE TEST CASES:

TEST(SumOFDigits, case1) { int* result = sum_of_digits(100, 999, 45); int* correct_answer = new int[45] {110, 121, 132, 143, 154, 165, 176, 187, 198, 220, 231, 242, 253, 264, 275, 286, 297, 330, 341, 352, 363, 374, 385, 396, 440, 451, 462, 473, 484, 495, 550, 561, 572, 583, 594, 660, 671, 682, 693, 770, 781, 792, 880, 891, 990}; for (int i = 0; i < 45; i++) { ASSERT_EQ(result[i], correct_answer[i]); }

CODE WITH LOOP

int sum_of_digits(int num, int start, int end) { int sum = 0; bool even = true; while (num > 0) { int digit = num % 10; if (even) { sum += digit; } else { sum -= digit; } num /= 10; even = !even; } return sum == 0; }

int main() {

const int start = 100; const int end = 999;

const int size_of_resultant_array = end - start + 1;

int arr[size_of_resultant_array];

int count = 0;

for (int i = start; i <= end; i++) {

if (sum_of_digits(i)) {

arr[count++] = i; } }

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

std::cout << arr[i] << " "; }

std::cout << std::endl; return 0; }