## Smallest and largest elements in a C++ array :

In this program, we will find the smallest and largest elements in an array using C++. We will take the array elements as inputs from the user and find out the smallest and largest.

### Algorithm :

The following algorithm we will use to solve this problem :

- Ask the user to enter the total number of elements.
- Create one array with size as the total number of elements that user has entered. This array is to hold the user input elements.
- Create two variables to hold the
*largest*and*smallest*values of the array. Assign both the first array elementâ€™s value. - Iterate through the array elements one by one.
- Check for each element if it is smaller than the current
*smallest*value or larger than the current*largest*value. - If the above step is true, update the smallest and the largest value variable.

### C++ program :

```
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int count;
cout << "Enter total number of elements :" << endl;
cin >> count;
int values[count];
for (int i = 0; i < count; i++)
{
cout << "Enter the number for position " << i + 1 << endl;
cin >> values[i];
}
int smallest = values[0];
int largest = values[0];
for (int i = 1; i < count; i++) {
if (values[i] > largest)
{
largest = values[i];
}
if (values[i] < smallest)
{
smallest = values[i];
}
}
cout << "Smallest number is : " << smallest << endl;
cout << "Largest number is : " << largest << endl;
}
```

### Explanation :

- Ask the user to enter the total number of numbers. Store it on variable
*count*. *values*is an array of size*count*. This array will hold all user input numbers.- The first
*for loop*takes the user input numbers one by one and enters them in*values*array. - Assign the first number of
*values*as the smallest and largest of all. - The second loop runs from the second element of the array since we have assigned the first element as the current largest and smallest element.
- Check each element of the array if it is greater than the current largest or smaller than the current smallest element.
- Update the smallest and the largest element if the above step is
*true*. - Finally, print both smallest and largest numbers.

### Sample output :

```
Enter total number of elements :
4
Enter the number for position 1
1
Enter the number for position 2
2
Enter the number for position 3
3
Enter the number for position 4
4
Smallest number is : 1
Largest number is : 4
Enter total number of elements :
5
Enter the number for position 1
12
Enter the number for position 2
334
Enter the number for position 3
102
Enter the number for position 4
2220
Enter the number for position 5
1
Smallest number is : 1
Largest number is : 2220
```

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