Confused with Two different implementation of Heap - c++

Question

Function 1
void min_heapify(int arr[],int n, int i){
int j, temp;
temp = arr[i];
j = 2 * i;
while (j <= n)
{
if (j < n && arr[j+1] < arr[j])
j = j + 1;
if (temp < arr[j])
break;
else if (temp >= arr[j])
{
arr[j/2] = arr[j];
j = 2 * j;
}
}
arr[j/2] = temp;
}
Function 2
void max_heapify(int arr[], int n, int i)
{
int largest = i; // Initialize largest as root
int l = 2*i + 1; // left = 2*i + 1
int r = 2*i + 2; // right = 2*i + 2
// If left child is larger than root
if (l < n && arr[l] < arr[largest])
largest = l;
// If right child is larger than largest so far
if (r < n && arr[r] < arr[largest])
largest = r;
// If largest is not root
if (largest != i)
{
swap(arr[i], arr[largest]);
// Recursively heapify the affected sub-tree
heapify(arr, n, largest);
}
}
Problem Details
Here the heapification work the same way to make a min_heap but the problem is, I used heap in this below problem to solve it but unfortunately function 2 which I implemented by watching MIT lecture didn't work for this problem, after looking some time in the web I found the 1st function which worked seamlessly for this problem. I'm just confused are not they the same function? ------
Problem
Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescended, to write a C/C++ program just to add a set of numbers. Such a problem will simply question your erudition. So, let’s add some flavor of ingenuity to it.
Addition operation requires cost now, and the cost is the summation of those two to be added. So, to add 1 and 10, you need a cost of 11. If you want to add 1, 2 and 3. There are several ways –
1 + 2 = 3, cost = 3
1 + 3 = 4, cost = 4
2 + 3 = 5, cost = 5
3 + 3 = 6, cost = 6
2 + 4 = 6, cost = 6
1 + 5 = 6, cost = 6
Total = 9
Total = 10
Total = 11
I hope you have understood already your mission, to add a set of integers so that the cost is minimal.
Input
Each test case will start with a positive number, N (2 ≤ N ≤ 5000) followed by N positive integers (all are less than 100000). Input is terminated by a case where the value of N is zero. This case should not be processed.
Output
For each case print the minimum total cost of addition in a single line.
SampleInput
3
1 2 3
4
1 2 3 4
0
SampleOutput
9
19

Answer 1

There is a problem with the swap function in function2.
C is call by value, so
swap(arr[i], arr[largest]);
cannot swap values in the array.
A swap function needs the addresses of the values to swap:
swap(int *v1, int *v2) {
int tmp = *v1;
*v1 = *v2;
*v2 = tmp;
}
And the call would be:
swap(&arr[i], &arr[largest]);

Answer 2

Ok I find out the solution there was a mistake in the condition check, in the if condition where we checking that if (left <= n) this was previously (left < n) this why it was not working for that problem. ok thank you.
void min_heapify(int arr[],int n, int i){
int lowest = i; // Initialize lowest as root
int left = 2*i ;
int right = 2*i + 1;
// If child is lower than root
if(left <= n && arr[left] < arr[lowest]){
lowest = left;
}
// If right child is lower than lowest
if(right <= n && arr[right] < arr[lowest]){
lowest = right;
}
// If lowest is not root
if(lowest != i){ // also break condition
swap(arr[i], arr[lowest]);
//Recursively heapify
min_heapify(arr, n, lowest);
}