Why Longest Common Prefix in DSA Go?

The Big Idea

Discover how a simple trick can save you from endless letter comparisons!

The Scenario

Imagine you have a list of words like "flower", "flow", "flight". You want to find the part at the start that all these words share. Doing this by looking at each word one by one and comparing letters can be tiring and confusing, especially if the list is very long.

The Problem

Checking each word manually means reading every letter again and again. It is slow and easy to make mistakes, like missing a letter or stopping too early. If the list grows, this manual checking becomes almost impossible to do quickly or correctly.

The Solution

The Longest Common Prefix method quickly finds the shared beginning part of all words by comparing letters step-by-step in a smart way. It stops as soon as it finds a difference, saving time and avoiding errors.

Before vs After

✗ Before

words := []string{"flower", "flow", "flight"}
prefix := ""
for i := 0; i < len(words[0]); i++ {
  for _, word := range words {
    if i == len(word) || word[i] != words[0][i] {
      return prefix
    }
  }
  prefix += string(words[0][i])
}

✓ After

func longestCommonPrefix(words []string) string {
  if len(words) == 0 {
    return ""
  }
  prefix := words[0]
  for _, word := range words[1:] {
    for len(prefix) > 0 && !strings.HasPrefix(word, prefix) {
      prefix = prefix[:len(prefix)-1]
    }
  }
  return prefix
}

What It Enables

This method lets you quickly find the shared start of many words, which helps in searching, grouping, and organizing text data efficiently.

Real Life Example

When you type in a search box, the system suggests words that start with what you typed. Finding the longest common prefix helps the system know what letters to show next.

Key Takeaways

Manual letter-by-letter checking is slow and error-prone.

Longest Common Prefix finds shared beginnings quickly by smart comparison.

This helps in text searching, autocomplete, and data grouping tasks.

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