Longest Common Subsequence in DSA Go - Build from Scratch

Longest Common Subsequence

📖 Scenario: Imagine you are comparing two versions of a document to find the longest sequence of words that appear in both versions in the same order. This helps in understanding the common parts between the two documents.

🎯 Goal: You will build a program in Go that finds the length of the longest common subsequence (LCS) between two given strings. The LCS is the longest sequence of characters that appear in both strings in the same order, but not necessarily contiguously.

📋 What You'll Learn

Create two string variables with exact values

Create a 2D slice to store LCS lengths

Implement nested loops to fill the LCS table using dynamic programming

Print the final LCS length

💡 Why This Matters

🌍 Real World

Finding common sequences helps in text comparison, DNA sequence analysis, and version control systems.

💼 Career

Understanding LCS is useful for software developers working on diff tools, bioinformatics, and data comparison tasks.

1

Create the input strings

Create two string variables called str1 and str2 with the exact values "AGGTAB" and "GXTXAYB" respectively.

DSA Go

# Your code here

Hint

Use := to create and assign the strings str1 and str2.

2

Create the LCS table

Create a 2D slice of integers called lcs with dimensions len(str1)+1 by len(str2)+1. Initialize all values to zero.

DSA Go

package main

import "fmt"

func main() {
    str1 := "AGGTAB"
    str2 := "GXTXAYB"

    // Create 2D slice lcs with zeros
    // Your code here
}

Hint

Use make to create a slice of slices. Use a for loop to initialize each inner slice.

3

Fill the LCS table using nested loops

Use nested for loops with variables i from 1 to len(str1) and j from 1 to len(str2). Inside the loops, if str1[i-1] equals str2[j-1], set lcs[i][j] = lcs[i-1][j-1] + 1. Otherwise, set lcs[i][j] to the maximum of lcs[i-1][j] and lcs[i][j-1]. Use if statements and if-else blocks as needed.

DSA Go

package main

import "fmt"

func main() {
    str1 := "AGGTAB"
    str2 := "GXTXAYB"

    lcs := make([][]int, len(str1)+1)
    for i := range lcs {
        lcs[i] = make([]int, len(str2)+1)
    }

    // Use nested loops to fill lcs
    // Your code here
}

Hint

Use two for loops to compare characters and fill the table. Use if to check equality and assign values accordingly.

4

Print the length of the Longest Common Subsequence

Print the value of lcs[len(str1)][len(str2)] using fmt.Println to display the length of the longest common subsequence.

DSA Go

package main

import "fmt"

func main() {
    str1 := "AGGTAB"
    str2 := "GXTXAYB"

    lcs := make([][]int, len(str1)+1)
    for i := range lcs {
        lcs[i] = make([]int, len(str2)+1)
    }

    for i := 1; i  lcs[i][j-1] {
                    lcs[i][j] = lcs[i-1][j]
                } else {
                    lcs[i][j] = lcs[i][j-1]
                }
            }
        }
    }

    // Print the LCS length
    // Your code here
}

Hint

Use fmt.Println to print the value at lcs[len(str1)][len(str2)].

Practice

1. What does the Longest Common Subsequence (LCS) between two strings represent?

easy

A. The longest substring that appears exactly the same in both strings.

B. The longest sequence of characters that appear in both strings in the same order, but not necessarily consecutively.

C. The shortest sequence that can be formed by combining both strings.

D. The total number of characters in both strings combined.

Solution

1. Step 1: Understand the definition of LCS

   The LCS is the longest sequence that appears in both strings in the same order, but characters do not need to be next to each other.

2. Step 2: Differentiate from substring

   A substring requires characters to be consecutive, but LCS allows gaps.

3. Final Answer:

   The longest sequence of characters that appear in both strings in the same order, but not necessarily consecutively. -> Option B

4. Quick Check:

   LCS = longest ordered sequence [OK]

Hint: LCS is about order, not consecutive characters [OK]

Common Mistakes:

• Confusing LCS with longest substring
• Thinking LCS characters must be consecutive
• Mixing LCS with string concatenation

2. Which of the following Go function signatures correctly represents a function to compute the length of the LCS of two strings s1 and s2?

easy

A. func lcs(s1, s2 string) int

B. func lcs(s1, s2 int) string

C. func lcs(s1, s2 []int) int

D. func lcs(s1, s2 string) string

Solution

1. Step 1: Identify input types for LCS function

   The LCS function takes two strings as input, so parameters should be of type string.

2. Step 2: Identify output type for LCS function

   The function returns the length of the LCS, which is an integer.

3. Final Answer:

   func lcs(s1, s2 string) int -> Option A

4. Quick Check:

   Input strings, output int length [OK]

Hint: LCS inputs are strings, output is integer length [OK]

Common Mistakes:

• Using wrong input types like int or slices
• Returning string instead of length
• Mixing input and output types

3. What is the output of the following Go code snippet?

package main
import "fmt"
func lcs(s1, s2 string) int {
    m, n := len(s1), len(s2)
    dp := make([][]int, m+1)
    for i := range dp {
        dp[i] = make([]int, n+1)
    }
    for i := 1; i <= m; i++ {
        for j := 1; j <= n; j++ {
            if s1[i-1] == s2[j-1] {
                dp[i][j] = dp[i-1][j-1] + 1
            } else {
                if dp[i-1][j] > dp[i][j-1] {
                    dp[i][j] = dp[i-1][j]
                } else {
                    dp[i][j] = dp[i][j-1]
                }
            }
        }
    }
    return dp[m][n]
}
func main() {
    fmt.Println(lcs("AGGTAB", "GXTXAYB"))
}

medium

A. 4

B. 5

C. 6

D. 7

Solution

1. Step 1: Understand the input strings

   Strings are "AGGTAB" and "GXTXAYB". We find their LCS length using dynamic programming.

2. Step 2: Trace the LCS calculation

   The LCS is "GTAB" which has length 4, but checking carefully the code and strings, the LCS is actually "GTAB" (length 4) or "GTTAB"? The correct LCS is "GTAB" length 4 or "GTAB" length 4? Actually, the LCS is "GTAB" length 4. But the code returns 4 or 5? Let's check carefully: The LCS is "GTAB" (G T A B) length 4. But the code returns dp[m][n], which is 4. But options have 4 and 5. Let's verify carefully: The LCS is "GTAB" length 4. So output is 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   LCS length of "AGGTAB" and "GXTXAYB" = 4 [OK]

Hint: Trace DP table or recall classic LCS example [OK]

Common Mistakes:

• Confusing LCS length with longest substring length
• Off-by-one errors in indexing
• Misreading input strings

4. The following Go code attempts to compute the LCS length but has a bug. What is the bug?

func lcs(s1, s2 string) int {
    m, n := len(s1), len(s2)
    dp := make([][]int, m)
    for i := range dp {
        dp[i] = make([]int, n)
    }
    for i := 1; i <= m; i++ {
        for j := 1; j <= n; j++ {
            if s1[i-1] == s2[j-1] {
                dp[i][j] = dp[i-1][j-1] + 1
            } else {
                if dp[i-1][j] > dp[i][j-1] {
                    dp[i][j] = dp[i-1][j]
                } else {
                    dp[i][j] = dp[i][j-1]
                }
            }
        }
    }
    return dp[m-1][n-1]
}

medium

A. The return statement should be dp[m][n] instead of dp[m-1][n-1].

B. The loops should start from 0 instead of 1.

C. The dp slice is not initialized with size m+1 and n+1, causing index out of range errors.

D. The comparison s1[i-1] == s2[j-1] should be s1[i] == s2[j].

Solution

1. Step 1: Check dp slice initialization

   The dp slice is created with size m and n, but the loops access dp[i][j] for i and j up to m and n inclusive, causing index out of range errors.

2. Step 2: Correct dp size for safe indexing

   dp should be initialized with size m+1 and n+1 to safely access dp[i][j] where i and j go up to m and n.

3. Final Answer:

   The dp slice is not initialized with size m+1 and n+1, causing index out of range errors. -> Option C

4. Quick Check:

   dp size must be m+1 x n+1 [OK]

Hint: dp size must cover all indices used in loops [OK]

Common Mistakes:

• Initializing dp with size m and n instead of m+1 and n+1
• Starting loops at 1 but dp size too small
• Incorrect return index

5. Given two strings s1 = "ABCBDAB" and s2 = "BDCABA", write a Go function that returns the actual Longest Common Subsequence string, not just its length. Which approach correctly reconstructs the LCS from the DP table?

hard

A. Sort both strings and find common characters in sorted order.

B. Build the DP table and return dp[m][n] as the LCS string directly.

C. Use recursion without memoization to find the LCS string.

D. Build the DP table for LCS length, then backtrack from dp[m][n] moving diagonally when characters match, else move in direction of larger dp value.

Solution

1. Step 1: Build DP table for LCS length

   Use dynamic programming to fill dp table with lengths of LCS for substrings of s1 and s2.

2. Step 2: Backtrack to reconstruct LCS string

   Start from dp[m][n], move diagonally up-left if characters match (add character to result), else move in direction of larger dp value until reaching dp[0][0].

3. Final Answer:

   Build the DP table for LCS length, then backtrack from dp[m][n] moving diagonally when characters match, else move in direction of larger dp value. -> Option D

4. Quick Check:

   DP + backtracking reconstructs LCS string [OK]

Hint: Backtrack DP table diagonally on matches to get LCS string [OK]

Common Mistakes:

• Returning dp[m][n] as string instead of length
• Not backtracking to reconstruct sequence
• Sorting strings loses original order