Longest Common Subsequence in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the 2D slice for dynamic programming.

DSA Go

dp := make([][]int, len(s1)+1)
for i := range dp {
    dp[i] = make([]int, [1])
}

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

Alen(s2)+1

Blen(s1)+1

Clen(s2)

Dlen(s1)

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(s1)+1 for columns instead of len(s2)+1

Forgetting the +1 for base case initialization

We create a 2D slice with rows for s1 and columns for s2, so each row has length len(s2)+1.

2fill in blank

medium

Complete the code to check if characters match in the LCS logic.

DSA Go

if s1[i-1] == [1][j-1] {
    dp[i][j] = dp[i-1][j-1] + 1
}

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

As1

Bs2

Cdp

Dlcs

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing s1[i-1] with s1[j-1]

Using dp or lcs instead of s2

We compare characters from s1 and s2 at positions i-1 and j-1 respectively.

3fill in blank

hard

Fix the error in updating dp when characters do not match.

DSA Go

else {
    dp[i][j] = [1](dp[i-1][j], dp[i][j-1])
}

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

Asum

Bmin

Cmax

Dabs

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max

Using sum which is incorrect for LCS

When characters don't match, we take the maximum of the two possible subsequences.

4fill in blank

hard

Fill both blanks to correctly backtrack the LCS string.

DSA Go

if s1[i-1] == s2[j-1] {
    lcs = string(s1[i-1]) + lcs
    i = i [1] 1
} else if dp[i-1][j] [2] dp[i][j-1] {
    i = i - 1
} else {
    j = j - 1
}

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

A-

B>

C<

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - for i decrement

Using < instead of > in comparison

We move i back by 1 when characters match, and compare dp[i-1][j] > dp[i][j-1] to decide direction.

5fill in blank

hard

Fill all three blanks to complete the LCS function return statement and helper function call.

DSA Go

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

func [2](s1, s2 string, dp [][]int, i, j int) string {
    if i == 0 || j == 0 {
        return ""
    }
    if s1[i-1] == s2[j-1] {
        return [3](s1, s2, dp, i-1, j-1) + string(s1[i-1])
    }
    if dp[i-1][j] > dp[i][j-1] {
        return lcsHelper(s1, s2, dp, i-1, j)
    }
    return lcsHelper(s1, s2, dp, i, j-1)
}

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

AlcsHelper

BlongestCommonSubsequence

Dmax

Attempts:

3 left

💡 Hint

Common Mistakes

Using longestCommonSubsequence instead of helper function

Using max instead of recursive call

The main function calls lcsHelper to build the LCS string recursively.

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