Longest Common Subsequence in DSA Go - Time & Space Complexity

Time Complexity: Longest Common Subsequence

O(m x n)

Understanding Time Complexity

We want to understand how the time needed to find the longest common subsequence grows as the input strings get longer.

How does the work increase when the strings become bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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 {
        dp[i][j] = max(dp[i-1][j], dp[i][j-1])
      }
    }
  }
  return dp[m][n]
}

func max(a, b int) int {
  if a > b {
    return a
  }
  return b
}
    

This code finds the length of the longest common subsequence between two strings using a table to store results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops over both strings to fill the dp table.
• How many times: Outer loop runs m times, inner loop runs n times, so total m x n times.

How Execution Grows With Input

As the strings get longer, the number of operations grows by multiplying their lengths.

Input Size (n)	Approx. Operations
10	100
100	10,000
1000	1,000,000

Pattern observation: Doubling the length roughly squares the work needed.

Final Time Complexity

Time Complexity: O(m x n)

This means the time needed grows proportionally to the product of the lengths of the two strings.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just loops through the strings once."

[OK] Correct: The algorithm uses nested loops, so it checks pairs of characters, making the work multiply, not just add.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves over simple recursion and why this approach is efficient for comparing sequences.

Self-Check

"What if we used recursion without memoization? How would the time complexity change?"

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