Disk space monitoring script in Bash Scripting - Time & Space Complexity

Time Complexity: Disk space monitoring script

O(n^2)

Understanding Time Complexity

We want to understand how the time to run a disk space monitoring script changes as the number of disks or partitions grows.

Specifically, how does the script's work increase when checking more disk entries?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#!/bin/bash

# Check disk usage for all mounted filesystems
for disk in $(df -h | tail -n +2 | awk '{print $1}'); do
  usage=$(df -h | grep "^$disk" | awk '{print $5}' | sed 's/%//')
  if [ "$usage" -gt 80 ]; then
    echo "Warning: Disk $disk is $usage% full"
  fi
done

This script loops through each disk or partition, checks its usage percentage, and warns if usage is above 80%.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping over each disk entry from the disk list.
• How many times: Once for each disk or partition found by df.
• Inside the loop: Running df and grep commands again for each disk.

How Execution Grows With Input

As the number of disks (n) increases, the script runs the loop n times, with each iteration running df -h (O(n)) and grep (O(n)) over all n entries.

Input Size (n)	Approx. Operations
10	About 10 loops × ~10 ops each = ~100 total
100	About 100 loops × ~100 ops each = ~10,000 total
1000	About 1000 loops × ~1000 ops each = ~1,000,000 total

Pattern observation: The total work grows roughly in proportion to the square of the number of disks.

Final Time Complexity

Time Complexity: O(n^2)

This means the script's running time grows quadratically as the number of disks increases.

Common Mistake

[X] Wrong: "The script runs in constant time because it just checks disk usage once."

[OK] Correct: The script actually loops over each disk, so more disks mean more checks and longer run time.

Interview Connect

Understanding how scripts scale with input size helps you write efficient monitoring tools and explain your code clearly in real work or interviews.

Self-Check

"What if we stored the output of df -h once before the loop instead of calling it inside the loop? How would the time complexity change?"

Practice

1. What is the main purpose of a disk space monitoring script in a server environment?

easy

A. To check disk usage and alert when space is low

B. To delete files automatically when disk is full

C. To speed up the disk read/write operations

D. To backup all files to another disk

Solution

1. Step 1: Understand the script's goal

   The script's goal is to monitor disk usage to prevent running out of space.

2. Step 2: Identify the correct purpose

   Among the options, only checking disk usage and alerting matches the monitoring function.

3. Final Answer:

   To check disk usage and alert when space is low -> Option A

4. Quick Check:

   Disk monitoring = alert on low space [OK]

Hint: Monitoring means checking and alerting, not deleting or backing up [OK]

Common Mistakes:

• Confusing monitoring with automatic deletion
• Thinking it speeds up disk operations
• Assuming it performs backups

2. Which of the following is the correct syntax to get the disk usage percentage of the root partition using df and awk in bash?

easy

A. df -h / | awk '{print $5}'

B. df -h / | sed '2q;d'

C. df / | awk 'NR==1 {print $5}'

D. df -h / | awk 'NR==2 {print $5}'

Solution

1. Step 1: Understand df -h / output

   The command shows disk usage for root; the second line (NR==2) has the data.

2. Step 2: Use awk 'NR==2 {print $5}' to get usage percent

   This prints the 5th column (usage %) from the second line, which is correct.

3. Final Answer:

   df -h / | awk 'NR==2 {print $5}' -> Option D

4. Quick Check:

   Second line, fifth column = usage percent [OK]

Hint: Use NR==2 to skip header line in df output [OK]

Common Mistakes:

• Printing all lines instead of second line
• Using NR==1 which is header line
• Using sed incorrectly to get the line

3. Given this script snippet:

usage=$(df / | awk 'NR==2 {print $5}' | sed 's/%//')
if [ "$usage" -gt 80 ]; then
  echo "Warning: Disk usage is above 80%"
else
  echo "Disk usage is normal"
fi

What will be the output if the disk usage is 75%?

medium

A. Disk usage is normal

B. No output

C. Syntax error in script

D. Warning: Disk usage is above 80%

Solution

1. Step 1: Extract usage value

   The script removes % sign and stores 75 in usage variable.

2. Step 2: Compare usage with 80

   75 is not greater than 80, so the else branch runs.

3. Final Answer:

   Disk usage is normal -> Option A

4. Quick Check:

   75 <= 80 means normal usage [OK]

Hint: Compare numeric values after removing % sign [OK]

Common Mistakes:

• Forgetting to remove % sign before comparison
• Using string comparison instead of numeric
• Misreading the if condition

4. Identify the error in this disk space check script snippet:

usage=$(df / | awk 'NR==2 {print $5}')
if [ $usage -gt 90 ]; then
  echo "Disk almost full"
fi

What problem will occur when running this script?

medium

A. No error, script runs fine

B. Syntax error due to missing quotes around $usage

C. df command will fail

D. awk command syntax error

Solution

1. Step 1: Check variable usage in test

   The variable $usage contains a value like '85%'. Without quotes, test may fail if value has spaces or special chars.

2. Step 2: Identify missing % removal and quotes

   Also, % sign is not removed, causing numeric comparison error. Missing quotes cause syntax error.

3. Final Answer:

   Syntax error due to missing quotes around $usage -> Option B

4. Quick Check:

   Always quote variables in test to avoid syntax errors [OK]

Hint: Quote variables and remove % before numeric comparison [OK]

Common Mistakes:

• Not quoting variables in if test
• Not removing % sign before numeric test
• Assuming df or awk commands fail

5. You want to create a bash script that checks all mounted partitions and sends an alert if any partition exceeds 85% usage. Which approach correctly implements this?

hard

A. Use du command on root directory and alert if usage exceeds 85%

B. Run df -h once and alert if total disk usage exceeds 85%

C. Use a loop over df -h output lines, extract usage %, compare with 85, and echo alert if exceeded

D. Check only root partition usage and ignore others

Solution

1. Step 1: Understand requirement to check all partitions

   We must check each mounted partition's usage, not just total or root.

2. Step 2: Loop through df -h output lines and compare usage

   Extract usage percent per line, compare with 85, and alert if exceeded.

3. Final Answer:

   Use a loop over df -h output lines, extract usage %, compare with 85, and echo alert if exceeded -> Option C

4. Quick Check:

   Loop and compare each partition usage [OK]

Hint: Loop through df output lines to check each partition [OK]

Common Mistakes:

• Checking only total or root partition
• Using du instead of df for disk usage
• Not looping through all partitions