Longest Common Prefix in DSA Go - Time & Space Complexity

Time Complexity: Longest Common Prefix

O(n * m)

Understanding Time Complexity

We want to know how the time to find the longest common prefix grows as we add more words or longer words.

How does the work increase when the input list or word length grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func longestCommonPrefix(strs []string) string {
    if len(strs) == 0 {
        return ""
    }
    prefix := strs[0]
    for i := 1; i < len(strs); i++ {
        for len(prefix) > 0 && !strings.HasPrefix(strs[i], prefix) {
            prefix = prefix[:len(prefix)-1]
        }
    }
    return prefix
}

This code finds the longest common prefix among a list of strings by comparing prefixes and shortening them until all strings match.

Identify Repeating Operations

• Primary operation: Nested loops where the outer loop goes through each string and the inner loop shortens the prefix until it matches.
• How many times: Outer loop runs n-1 times (n = number of strings), inner loop runs up to length of prefix each time.

How Execution Grows With Input

As the number of strings (n) or the length of the prefix (m) grows, the work increases roughly by multiplying these two.

Input Size (n)	Approx. Operations
10 strings, prefix length 5	About 50 checks
100 strings, prefix length 10	About 1000 checks
1000 strings, prefix length 20	About 20,000 checks

Pattern observation: The operations grow roughly by multiplying the number of strings by the prefix length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows proportionally with the number of strings and the length of the prefix we check.

Common Mistake

[X] Wrong: "The time depends only on the number of strings, not their length."

[OK] Correct: The prefix length matters because we compare characters one by one, so longer prefixes mean more work.

Interview Connect

Understanding this helps you explain how your solution scales with input size, a key skill interviewers look for.

Self-Check

"What if we sorted the strings first and only compared the first and last strings? How would the time complexity change?"

Practice

1. What does the Longest Common Prefix (LCP) of a list of strings represent?

easy

A. The longest starting substring common to all strings

B. The longest ending substring common to all strings

C. The longest substring found anywhere in all strings

D. The shortest substring unique to each string

Solution

1. Step 1: Understand the prefix concept

   A prefix is a substring starting at the beginning of a string.

2. Step 2: Define longest common prefix

   The longest common prefix is the longest substring starting at the beginning that all strings share.

3. Final Answer:

   The longest starting substring common to all strings -> Option A

4. Quick Check:

   Prefix means start of string = longest common start substring [OK]

Hint: Focus on the start of strings, not anywhere else [OK]

Common Mistakes:

• Confusing prefix with substring anywhere
• Thinking about suffix instead of prefix
• Assuming shortest substring is the answer

2. Which of the following Go code snippets correctly initializes a slice of strings named strs with values "flower", "flow", "flight"?

easy

A. var strs = {"flower", "flow", "flight"}

B. strs := [3]string{"flower", "flow", "flight"}

C. strs := []string{"flower", "flow", "flight"}

D. strs := string{"flower", "flow", "flight"}

Solution

1. Step 1: Identify correct slice initialization syntax

   In Go, slices are initialized with []type{values} syntax.

2. Step 2: Check each option

   strs := []string{"flower", "flow", "flight"} uses []string with curly braces correctly. strs := [3]string{"flower", "flow", "flight"} uses array syntax but is valid but fixed size, not slice. var strs = {"flower", "flow", "flight"} uses invalid braces. strs := string{"flower", "flow", "flight"} uses wrong type syntax.

3. Final Answer:

   strs := []string{"flower", "flow", "flight"} -> Option C

4. Quick Check:

   Slice syntax = []type{values} [OK]

Hint: Use []string{...} to create string slices in Go [OK]

Common Mistakes:

• Using array syntax when slice is needed
• Missing type brackets []
• Using curly braces without type

3. Given a list of strings ["dog", "racecar", "car"], what should the longest common prefix function return and why?

easy

A. "d" because all strings start with 'd'

B. "" because no common prefix exists

C. "c" because 'car' is the shortest string

D. "r" because 'racecar' starts with 'r'

Solution

1. Step 1: Compare prefixes of all strings

   "dog" starts with 'd', "racecar" with 'r', "car" with 'c'. No common starting character exists.

2. Step 2: Determine longest common prefix

   Since no common starting substring exists, the longest common prefix is empty string "".

3. Final Answer:

   "" because no common prefix exists -> Option B

4. Quick Check:

   No common start means empty prefix [OK]

Hint: If first chars differ, prefix is empty [OK]

Common Mistakes:

• Assuming shortest string is prefix
• Picking first string's first letter
• Ignoring all strings must share prefix

4. What is the output of the following Go code that finds the longest common prefix?

package main
import "fmt"
func longestCommonPrefix(strs []string) string {
    if len(strs) == 0 {
        return ""
    }
    prefix := strs[0]
    for _, s := range strs[1:] {
        for len(prefix) > 0 && len(s) >= len(prefix) && s[:len(prefix)] != prefix {
            prefix = prefix[:len(prefix)-1]
        }
    }
    return prefix
}
func main() {
    strs := []string{"inter", "internet", "internal"}
    fmt.Println(longestCommonPrefix(strs))
}

medium

A. int

B. in

C. inte

D. inter

Solution

1. Step 1: Initialize prefix and compare with each string

   Prefix starts as "inter". Compare with "internet": len(s)=8 >=5 and "internet"[:5] == "inter", so no reduction.

2. Step 2: Compare prefix "inter" with "internal"

   len(s)=8 >=5 and "internal"[:5] == "inter", so prefix remains "inter".

3. Final Answer:

   inter -> Option D

4. Quick Check:

   Common start of all strings = "inter" [OK]

Hint: Shrink prefix until all strings start with it [OK]

Common Mistakes:

• Not shrinking prefix correctly
• Comparing prefix with wrong substring
• Returning initial prefix without checks

5. Identify the bug in this Go function for longest common prefix and select the fix:

func longestCommonPrefix(strs []string) string {
    if len(strs) == 0 {
        return ""
    }
    prefix := strs[0]
    for _, s := range strs[1:] {
        for len(prefix) > 0 && s[:len(prefix)] != prefix {
            prefix = prefix[:len(prefix)-1]
        }
    }
    return prefix
}

medium

A. Add check that len(s) >= len(prefix) before s[:len(prefix)]

B. Initialize prefix as empty string instead of strs[0]

C. Change for loop to iterate over all strings including first

D. Return strs[0] directly without loop

Solution

1. Step 1: Analyze substring slicing in loop

   The code slices s[:len(prefix)] without checking if s is long enough, causing runtime panic if s is shorter than prefix.

2. Step 2: Fix by adding length check

   Adding condition len(s) >= len(prefix) before slicing prevents out-of-range error.

3. Final Answer:

   Add check that len(s) >= len(prefix) before s[:len(prefix)] -> Option A

4. Quick Check:

   Prevent slice out-of-range by length check [OK]

Hint: Always check slice length before slicing [OK]

Common Mistakes:

• Ignoring string length before slicing
• Changing prefix initialization unnecessarily
• Skipping loop over all strings