Programming

In this programming tutorial we will find the minimum and maximum element of an array using Tournament Method.

Related: Find the minimum and maximum number in an array

For the given array of integer values find the minimum and maximum value. Note! There will always be at least 1 number in the array. The numbers can be negative, zero, positive and can repeat.

Lets say we are given the following array of integer values.

```
[1, 2, 4, 6, 5, 3]
```

And we have to find the min and max values.

Tournament method finds the min and max value by dividing the array into sub-arrays till min-max is found for the sub-arrays. Then from the min-max of those sub-arrays we find the min-max of the full array. Divide and Conquer.

To find the minimum and maximum values using Tournament method we perform the following steps.

Steps:

- Set the
`start`

index to the first element of the array. - Set the
`end`

index to the last element of the array. - If
`start`

is equal to`end`

then there is only one item in the array and it is the both the minimum and the maximum value. - If
`end - start = 1`

then there are two items in the array. Compare and find the min and max values. - If there are more items in the array then find the
`middle`

index.

Where,`middle = floor((start + end) / 2)`

This divides the array into two parts.

Left half from index`start`

to`middle`

.

Right half from index`middle+1`

to`end`

. - We recursively go through step 3 to 5 till we end up with a pair of min and max values.

Python

```
def min_max(array, start, end):
# only 1 element in the array
if start == end:
return (array[start], array[end])
# 2 elments in the array
elif end - start == 1:
min_number = min(array[start], array[end])
max_number = max(array[start], array[end])
return (min_number, max_number)
# more than 2 elements in the array
else:
middle = int(start + end) / 2
min_num1, max_num1 = min_max(array, start, middle)
min_num2, max_num2 = min_max(array, middle + 1, end)
return (min(min_num1, min_num2), max(max_num1, max_num2))
numbers = [1, 2, 4, 5, 6, 3]
start = 0
end = len(numbers) - 1
min_number, max_number = min_max(numbers, start, end)
assert min_number == 1
assert max_number == 6
print('min', min_number, 'max', max_number)
```

```
'min', 1, 'max', 6
```

Time complexity of this is `O(n)`

.

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