Find Maximum and Minimum in Array using Loop - Algorithm and Examples

Problem Statement

Given an array of integers, your task is to find the maximum and minimum values present in the array using a simple loop (without any built-in functions).

  • The maximum value is the largest number in the array.
  • The minimum value is the smallest number in the array.

If the array is empty, there is no maximum or minimum, and the result should indicate that appropriately (e.g., null or a message).

Examples

Input Array Maximum Minimum Description
[10, 20, 5, 8, 30] 30 5 Normal case with unordered elements Visualization
[5, 5, 5, 5] 5 5 All elements are equalVisualization
[-10, -5, -3, -20] -3 -20 All elements are negativeVisualization
[100] 100 100 Only one element in the arrayVisualization
[] null null Empty array — no max or min can be determinedVisualization

Visualization Player

Solution

To find the maximum and minimum values in an array, we can simply scan through the array one element at a time and keep track of the largest and smallest values we encounter.

Case 1: When the Array Has Elements

We start by assuming the first element is both the maximum and the minimum. Then, we go through the rest of the array and compare each value:

  • If the current value is larger than our current maximum, we update the maximum.
  • If the current value is smaller than our current minimum, we update the minimum.

This way, by the time we reach the end of the array, we will have found the true maximum and minimum values.

Case 2: When the Array Has One Element

If the array has just a single element, then that element is both the maximum and the minimum, since there's nothing else to compare it to.

Case 3: When All Elements Are Equal

If every number in the array is the same, then that number is both the maximum and minimum. For example, in [7, 7, 7], the answer is max = 7, min = 7.

Case 4: When the Array is Empty

If the array is empty, we can't find either maximum or minimum values. There's simply no data to evaluate. In such cases, we should either return null, an empty result, or show a message like "No elements to compare"—depending on how we want to handle such edge cases.

Algorithm Steps

  1. Given an array of numbers arr.
  2. If array is empty, return null or -1.
  3. Initialize max_val and min_val with the first element of the array.
  4. Iterate through each element in the array starting from the second element.
  5. If the current element is greater than max_val, update max_val.
  6. If the current element is less than min_val, update min_val.
  7. After the loop, return max_val and min_val.

Code

Python
Java
JavaScript
C
C++
C#
Go
def find_max_min_loop(arr):
    if not arr:
        return None
    max_val = arr[0]
    min_val = arr[0]
    for num in arr[1:]:
        if num > max_val:
            max_val = num
        if num < min_val:
            min_val = num
    return max_val, min_val

# Sample Input
arr2 = [-5, -1, -9, 0, 3, 7]
result = find_max_min_loop(arr2)
if result:
    print("Maximum:", result[0])
    print("Minimum:", result[1])

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)Even in the best case, every element must be checked to ensure no smaller or larger values exist.
Average CaseO(n)Each element is visited once to compare and possibly update the maximum or minimum.
Worst CaseO(n)In the worst case, the loop still traverses the entire array to determine the max and min values.

Space Complexity

O(1)

Explanation: Only a fixed number of variables (e.g., max_val, min_val) are used, regardless of input size.

Detailed Step by Step Example

Let us take the following array and apply the logic to find the maximum and minimum elements.

{ "array": [6,3,8,2,7,4], "showIndices": true, "highlightIndices": [0] }
{ "array": [0,6,6], "showIndices": false, "highlightIndicesGreen": [1, 2], "emptyCompIndices": [0], "labels": { "1": "max", "2": "min" } }

Initialize max = 6 and min = 6 with the first element of the array.

Check index 1

Compare 3 at index=1 with current max = 6 and min = 6.

3 is smaller than current min. Update min = 3.

{ "array": [6,3,8,2,7,4], "showIndices": true, "highlightIndices": [1], "specialIndices": [], "labels": {"1":"i"} }
{ "array": [0,6,3], "showIndices": false, "highlightIndicesGreen": [1, 2], "emptyCompIndices": [0], "labels": { "1": "max", "2": "min" } }

Check index 2

Compare 8 at index=2 with current max = 6 and min = 3.

8 is greater than current max. Update max = 8.

{ "array": [6,3,8,2,7,4], "showIndices": true, "highlightIndices": [2], "specialIndices": [], "labels": {"2":"i"} }
{ "array": [0,8,3], "showIndices": false, "highlightIndicesGreen": [1, 2], "emptyCompIndices": [0], "labels": { "1": "max", "2": "min" } }

Check index 3

Compare 2 at index=3 with current max = 8 and min = 3.

2 is smaller than current min. Update min = 2.

{ "array": [6,3,8,2,7,4], "showIndices": true, "highlightIndices": [3], "specialIndices": [], "labels": {"3":"i"} }
{ "array": [0,8,2], "showIndices": false, "highlightIndicesGreen": [1, 2], "emptyCompIndices": [0], "labels": { "1": "max", "2": "min" } }

Check index 4

Compare 7 at index=4 with current max = 8 and min = 2.

No update required.

{ "array": [6,3,8,2,7,4], "showIndices": true, "highlightIndices": [4], "specialIndices": [], "labels": {"4":"i"} }
{ "array": [0,8,2], "showIndices": false, "highlightIndicesGreen": [1, 2], "emptyCompIndices": [0], "labels": { "1": "max", "2": "min" } }

Check index 5

Compare 4 at index=5 with current max = 8 and min = 2.

No update required.

{ "array": [6,3,8,2,7,4], "showIndices": true, "highlightIndices": [5], "specialIndices": [], "labels": {"5":"i"} }
{ "array": [0,8,2], "showIndices": false, "highlightIndicesGreen": [1, 2], "emptyCompIndices": [0], "labels": { "1": "max", "2": "min" } }

Final Result:

Maximum = 8, Minimum = 2

{ "array": [6,3,8,2,7,4], "showIndices": true, "labels": { "2": "max", "3": "min" }, "specialIndices": [] }
{ "array": [0,8,2], "showIndices": false, "highlightIndicesGreen": [1, 2], "emptyCompIndices": [0], "labels": { "1": "max", "2": "min" } }