Minimum and Maximum in Array - CA06

My Thinking and Approach

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Introduction

In this problem, I was given an array of integers and asked to find the minimum and maximum elements in the array.

This is a basic problem but important for understanding array traversal.

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Problem Statement

• Given an array of integers arr

• Find:

  • Minimum element
  • Maximum element

• Return both values

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My Initial Thought

At first, I considered:

• Sorting the array
• Taking the first element as minimum
• Taking the last element as maximum

But sorting takes more time and is not necessary.

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Key Observation

Instead of sorting:

• I can traverse the array once
• Keep track of minimum and maximum values

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Optimized Approach

I decided to:

• Initialize min and max with first element
• Traverse the array
• Update values when needed

Logic:

• If current element < min → update min
• If current element > max → update max

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My Approach (Step-by-Step)

1. Initialize:

• min_val = arr[0]

• max_val = arr[0]

  1. Traverse the array from index 1
  2. For each element:

• If element < min_val → update min_val

• If element > max_val → update max_val

  1. Return both values

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Code (Python)

Below is the implementation clearly separated inside a code block:

```python id="p5j0ps"
def findMinMax(arr):
min_val = arr[0]
max_val = arr[0]

for i in range(1, len(arr)):
    if arr[i] < min_val:
        min_val = arr[i]
    if arr[i] > max_val:
        max_val = arr[i]

return [min_val, max_val]

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## Example Walkthrough

### Input:



```text id="htc0zt"
arr = [1, 4, 3, 5, 8, 6]

Steps:

• Start with min = 1, max = 1
• Compare with 4 → max = 4
• Compare with 3 → no change
• Compare with 5 → max = 5
• Compare with 8 → max = 8
• Compare with 6 → no change

Output:

```text id="x4jsm3"
[1, 8]

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## Complexity Analysis

| Type             | Complexity |
| ---------------- | ---------- |
| Time Complexity  | O(n)       |
| Space Complexity | O(1)       |

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## Key Takeaways

* No need to sort the array
* Single traversal is enough
* Efficient use of variables improves performance

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## Conclusion

This problem helped me understand how to find minimum and maximum values efficiently using a single traversal.

It is a basic but important concept in array manipulation.

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