Longest Common Prefix in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return an empty string if the input list is empty.

DSA Go

func longestCommonPrefix(strs []string) string {
    if len(strs) == [1] {
        return ""
    }
    return ""
}

Drag options to blanks, or click blank then click option'

A0

B-1

C1

Dlen(strs)

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if len(strs) == 1 instead of 0

Using a negative number for length check

If the list has zero elements, its length is 0, so we check if len(strs) == 0 to return an empty string.

2fill in blank

medium

Complete the code to set the prefix as the first string in the list.

DSA Go

func longestCommonPrefix(strs []string) string {
    if len(strs) == 0 {
        return ""
    }
    prefix := [1]
    return prefix
}

Drag options to blanks, or click blank then click option'

Astrs[len(strs)]

Bstrs[1]

Cstrs[0]

Dstrs[len(strs)-1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using strs[1] which is the second string

Using strs[len(strs)] which is out of range

The first string in the list is at index 0, so prefix should be strs[0].

3fill in blank

hard

Fix the error in the loop condition to iterate over all strings in the list.

DSA Go

for i := 1; i [1] len(strs); i++ {
    // loop body
}

Drag options to blanks, or click blank then click option'

A>

B<=

C>=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' which causes no iterations

Using '<=' which causes out of range error

To loop from 1 up to but not including len(strs), use i < len(strs).

4fill in blank

hard

Fill both blanks to check if the prefix is longer than the current string and to compare characters.

DSA Go

for i := 0; i < len(prefix); i++ {
    if i [1] len(strs[j]) || prefix[i] [2] strs[j][i] {
        prefix = prefix[:i]
        break
    }
}

Drag options to blanks, or click blank then click option'

A>=

B<

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>=' for length check

Using '==' instead of '!=' for character comparison

We check if i is greater or equal to the length of the current string to avoid index error, and if characters differ, we shorten the prefix.

5fill in blank

hard

Fill all three blanks to complete the nested loop that finds the longest common prefix.

DSA Go

for j := 1; j [1] len(strs); j++ {
    for i := 0; i [2] len(prefix); i++ {
        if i [3] len(strs[j]) || prefix[i] != strs[j][i] {
            prefix = prefix[:i]
            break
        }
    }
}

Drag options to blanks, or click blank then click option'

A<

B<=

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' in loops causing index out of range

Using '<' instead of '>=' in condition causing errors

The outer loop runs while j < len(strs), the inner loop runs while i < len(prefix), and the condition checks if i >= len(strs[j]) to avoid index errors.

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