🧠 Find Minimum and Maximum in an Array

✅ Problem Statement

Given an array arr, find the minimum and maximum elements.

This is a fundamental problem to understand array traversal, comparisons, and optimization techniques.

🧩 Example

Input: arr = [3, 7, 1, 9, 2]
Output: Min = 1, Max = 9

🔍 Approach 1: Brute Force (O(n) time, O(1) space)

What: Traverse the array once, updating min and max values sequentially. Why: Simple and intuitive. How: Initialize min_val and max_val with the first element, then check each element in the array.

🐍 Python Code

def find_min_max(arr):
    if not arr:
        return None, None

    min_val = max_val = arr[0]
    for num in arr:
        if num < min_val:
            min_val = num
        if num > max_val:
            max_val = num
    return min_val, max_val

# Example
arr = [3, 7, 1, 9, 2]
print(find_min_max(arr))  # Output: (1, 9)

🧪 JavaScript Code

function findMinMax(arr) {
  let min = arr[0],
    max = arr[0];
  for (let num of arr) {
    if (num < min) min = num;
    if (num > max) max = num;
  }
  return { min, max };
}

// Example
const arr = [3, 7, 1, 9, 2];
console.log(findMinMax(arr)); // Output: { min: 1, max: 9 }

Execution Trace / Step-by-Step

arr = [3, 7, 1, 9, 2]
Step 1: min=3, max=3
Step 2: num=7 → max=7
Step 3: num=1 → min=1
Step 4: num=9 → max=9
Step 5: num=2 → no change
Final → min=1, max=9

🔍 Breaking It Down for Understanding

1. Single Pass Traversal: Check each element once → O(n) time.
2. Space Efficient: Only two variables used → O(1) space.
3. Mental Model: Imagine holding a “current min” and “current max” while walking through the array — update when a new extreme is found.
4. Pattern Recognition: Many array problems involve tracking extremes, counts, or sums, and this pattern is reusable.

📝 Summary

Approach	Time Complexity	Space Complexity	Notes
Single Pass	O(n)	O(1)	Simple, clear, efficient, optimal

Tip: Whenever you need minimum, maximum, or running stats in an array, a single-pass comparison is usually the most elegant solution.