### Problem Statement :

```Variadic functions are functions which take a variable number of arguments. In C programming, a variadic function will contribute to the flexibility of the program that you are developing.

The declaration of a variadic function starts with the declaration of at least one named variable, and uses an ellipsis as the last parameter, e.g.

int printf(const char* format, ...);

In this problem, you will implement three variadic functions named sum(), min() and max() to calculate sums, minima, maxima of a variable number of arguments. The first argument passed to the variadic function is the count of the number of arguments, which is followed by the arguments themselves.

Input Format

The first line of the input consists of an integer  number_of_test_cases  .
Each test case tests the logic of your code by sending a test implementation of 3, 5 and 10 elements respectively.
You can test your code against sample/custom input.
The error log prints the parameters which are passed to the test implementation. It also prints the sum, minimum element and maximum element corresponding to your code.

Constraints

1  <=  number_of_test_cases  <= 50
1  <=  element  <= 1000000

Output Format

"Correct Answer" is printed corresponding to each correct execution of a test implementation."Wrong Answer" is printed otherwise.```

### Solution :

```                            ```Solution in C :

#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define MIN_ELEMENT 1
#define MAX_ELEMENT 1000000

int  sum (int count,...) {
va_list ap;
int i, n;
va_start(ap, count);
n = 0;
for (i=0;i<count;i++){
n += va_arg(ap, int);
}
va_end(ap);
return n;
}

int min(int count,...) {
va_list ap;
int i, current, minimum;
va_start(ap, count);
minimum = 1000001;
for (i=0;i<count;i++){
current = va_arg(ap, int);
if (current < minimum)
minimum = current;
}
va_end(ap);
return minimum;

}

int max(int count,...) {
va_list ap;
int i, current, maximum;
va_start(ap, count);
maximum = 0;
for (i=0;i<count;i++){
current = va_arg(ap, int);
if (current > maximum)
maximum = current;
}
va_end(ap);
return maximum;

}

int test_implementations_by_sending_three_elements() {
srand(time(NULL));

int elements[3];

elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

fprintf(stderr, "Sending following three elements:\n");
for (int i = 0; i < 3; i++) {
fprintf(stderr, "%d\n", elements[i]);
}

int elements_sum = sum(3, elements[0], elements[1], elements[2]);
int minimum_element = min(3, elements[0], elements[1], elements[2]);
int maximum_element = max(3, elements[0], elements[1], elements[2]);

fprintf(stderr, "Elements sum is %d\n", elements_sum);
fprintf(stderr, "Minimum element is %d\n", minimum_element);
fprintf(stderr, "Maximum element is %d\n\n", maximum_element);

int expected_elements_sum = 0;
for (int i = 0; i < 3; i++) {
if (elements[i] < minimum_element) {
return 0;
}

if (elements[i] > maximum_element) {
return 0;
}

expected_elements_sum += elements[i];
}

return elements_sum == expected_elements_sum;
}

int test_implementations_by_sending_five_elements() {
srand(time(NULL));

int elements[5];

elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[3] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[4] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

fprintf(stderr, "Sending following five elements:\n");
for (int i = 0; i < 5; i++) {
fprintf(stderr, "%d\n", elements[i]);
}

int elements_sum = sum(5, elements[0], elements[1], elements[2], elements[3], elements[4]);
int minimum_element = min(5, elements[0], elements[1], elements[2], elements[3], elements[4]);
int maximum_element = max(5, elements[0], elements[1], elements[2], elements[3], elements[4]);

fprintf(stderr, "Elements sum is %d\n", elements_sum);
fprintf(stderr, "Minimum element is %d\n", minimum_element);
fprintf(stderr, "Maximum element is %d\n\n", maximum_element);

int expected_elements_sum = 0;
for (int i = 0; i < 5; i++) {
if (elements[i] < minimum_element) {
return 0;
}

if (elements[i] > maximum_element) {
return 0;
}

expected_elements_sum += elements[i];
}

return elements_sum == expected_elements_sum;
}

int test_implementations_by_sending_ten_elements() {
srand(time(NULL));

int elements[10];

elements[0] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[1] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[2] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[3] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[4] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[5] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[6] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[7] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[8] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;
elements[9] = rand() % (MAX_ELEMENT - MIN_ELEMENT + 1) + MIN_ELEMENT;

fprintf(stderr, "Sending following ten elements:\n");
for (int i = 0; i < 10; i++) {
fprintf(stderr, "%d\n", elements[i]);
}

int elements_sum = sum(10, elements[0], elements[1], elements[2], elements[3], elements[4],
elements[5], elements[6], elements[7], elements[8], elements[9]);
int minimum_element = min(10, elements[0], elements[1], elements[2], elements[3], elements[4],
elements[5], elements[6], elements[7], elements[8], elements[9]);
int maximum_element = max(10, elements[0], elements[1], elements[2], elements[3], elements[4],
elements[5], elements[6], elements[7], elements[8], elements[9]);

fprintf(stderr, "Elements sum is %d\n", elements_sum);
fprintf(stderr, "Minimum element is %d\n", minimum_element);
fprintf(stderr, "Maximum element is %d\n\n", maximum_element);

int expected_elements_sum = 0;
for (int i = 0; i < 10; i++) {
if (elements[i] < minimum_element) {
return 0;
}

if (elements[i] > maximum_element) {
return 0;
}

expected_elements_sum += elements[i];
}

return elements_sum == expected_elements_sum;
}

int main ()
{
int number_of_test_cases;
scanf("%d", &number_of_test_cases);

while (number_of_test_cases--) {
if (test_implementations_by_sending_three_elements()) {
} else {
}

if (test_implementations_by_sending_five_elements()) {
} else {
}

if (test_implementations_by_sending_ten_elements()) {
} else {
}
}

return 0;
}```
```

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <