Python-Exercise

Python program to find maximum and minimum values in a list

4 min read

Given a list of numbers, find the maximum and minimum values present in the list.

Example

Input: [10, 3, 56, 23, 0, -5, 56]

Output:

Maximum = 56
Minimum = -5

Approach

We can solve this problem in several ways: use Python’s built-in max() and min() functions for simplicity and readability; iterate through the list once to compute both values in O(n) time without extra space; or sort the list and pick the first and last elements (simpler but slower for large lists).

Method 1 — Using built-in max() and min()

Idea: Directly call max() and min() for a concise solution.

Code:

# Method 1: simplest — direct built-ins
nums = [10, 3, 56, 23, 0, -5, 56]
minimum = min(nums)
maximum = max(nums)
print("Minimum =", minimum)
print("Maximum =", maximum)

Explanation (2–4 lines): Using min() and max() directly is the simplest and most readable approach. Both functions scan the list once. They raise a ValueError for an empty list, so ensure the list is not empty before calling them.

Method 2 — Using built-in max() and min()

Idea: Directly call max(list) and min(list). This is the most readable and Pythonic approach.

Code:

# Method 1: built-in functions
def find_min_max_builtin(arr):
    if not arr:
        raise ValueError("Empty list has no min or max")
    return min(arr), max(arr)

# Example usage
nums = [10, 3, 56, 23, 0, -5, 56]
minimum, maximum = find_min_max_builtin(nums)
print("Minimum =", minimum)
print("Maximum =", maximum)

Explanation (2–4 lines):
min() and max() scan the iterable once and return the smallest and largest elements. They are concise and efficient for most use cases. If the list is empty, these functions raise a ValueError, so handle that case explicitly.

Method 3 — Single pass loop

Idea: Iterate through the list once, maintaining current min_val and max_val. This avoids creating extra copies or relying on built-ins.

Code:

# Method 2: single pass loop
def find_min_max_loop(arr):
    if not arr:
        raise ValueError("Empty list has no min or max")
    min_val = max_val = arr[0]
    for x in arr[1:]:
        if x < min_val:
            min_val = x
        elif x > max_val:
            max_val = x
    return min_val, max_val

# Example usage
nums = [10, 3, 56, 23, 0, -5, 56]
minimum, maximum = find_min_max_loop(nums)
print("Minimum =", minimum)
print("Maximum =", maximum)

Explanation :
Initialize both min_val and max_val with the first element, then update them while scanning the rest. This guarantees a single pass (O(n)) and constant extra space.

Method 4 — Using sorted()

Idea: Sort the list and take the first (minimum) and last (maximum) elements. This is easy to write but slower for large lists.

Code:

# Method 4: sort the list
def find_min_max_sorted(arr):
    if not arr:
        raise ValueError("Empty list has no min or max")
    s = sorted(arr)
    return s[0], s[-1]

# Example usage
nums = [10, 3, 56, 23, 0, -5, 56]
minimum, maximum = find_min_max_sorted(nums)
print("Minimum =", minimum)
print("Maximum =", maximum)

Explanation :
Sorting places values in order; the smallest is at index 0 and the largest at index -1. Complexity is dominated by sorting (O(n log n)). Use this when you also need the list ordered.

Method 5 — Using functools.reduce()

Idea: Use reduce() to compute min and max by applying a comparison function cumulatively.

Code:

from functools import reduce

# reduce to get max
def find_max_reduce(arr):
    if not arr:
        raise ValueError("Empty list has no max")
    return reduce(lambda a, b: a if a > b else b, arr)

# reduce to get min
def find_min_reduce(arr):
    if not arr:
        raise ValueError("Empty list has no min")
    return reduce(lambda a, b: a if a < b else b, arr)

# Example usage
nums = [10, 3, 56, 23, 0, -5, 56]
print("Minimum =", find_min_reduce(nums))
print("Maximum =", find_max_reduce(nums))

Explanation :
reduce() applies a binary function across the list to accumulate a single result. This is more functional but less direct than min()/max() or a loop.

Edge cases to consider

  • Empty list: raise an error or decide on sentinel behavior.
  • All elements equal: min and max are the same number.
  • Lists with mixed numeric types (e.g., int and float) — Python supports comparison between them.
  • Lists with non-numeric comparable elements (e.g., strings) — min()/max() still work lexicographically.

Time and Space Complexity

  • Method 1 (built-ins simple): Time O(n), Space O(1) (ignoring iterator overhead).
  • Method 2 (built-ins with function): Time O(n), Space O(1).
  • Method 3 (single pass loop): Time O(n), Space O(1).
  • Method 4 (sorted): Time O(n log n), Space O(n) if sorted creates a new list.
  • Method 5 (reduce): Time O(n), Space O(1) (function call overhead).

Leave a Comment