Substring Search Patterns in DSA Go - Time & Space Complexity

Time Complexity: Substring Search Patterns

O(n * m)

Understanding Time Complexity

When searching for a smaller string inside a bigger one, we want to know how long it takes as the strings grow.

We ask: How does the search time change when the text or pattern gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func substringSearch(text, pattern string) int {
    n := len(text)
    m := len(pattern)
    for i := 0; i <= n-m; i++ {
        j := 0
        for j < m && text[i+j] == pattern[j] {
            j++
        }
        if j == m {
            return i
        }
    }
    return -1
}
    

This code checks each possible position in the text to see if the pattern matches starting there.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Comparing characters of pattern with text at each position.
• How many times: Outer loop runs about (n - m + 1) times; inner loop runs up to m times per outer loop.

How Execution Grows With Input

As text length (n) and pattern length (m) grow, the number of comparisons grows roughly by n times m.

Input Size (n)	Approx. Operations
10 (m=3)	About 8 * 3 = 24 comparisons
100 (m=5)	About 96 * 5 = 480 comparisons
1000 (m=10)	About 991 * 10 = 9,910 comparisons

Pattern observation: Operations grow roughly by multiplying text length and pattern length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time to search grows proportionally to the product of text length and pattern length.

Common Mistake

[X] Wrong: "The search always takes time proportional to the text length only."

[OK] Correct: The pattern length matters too because each position may require checking multiple characters.

Interview Connect

Understanding this helps you explain why simple search can be slow and why smarter methods exist, showing your grasp of algorithm efficiency.

Self-Check

"What if we used a more advanced algorithm like KMP? How would the time complexity change?"

Practice

1. What does the strings.Contains(s, substr) function in Go do?

easy

A. Returns the index of substr in s

B. Checks if substr is present anywhere in s

C. Splits s by substr

D. Replaces substr with another string in s

Solution

1. Step 1: Understand the function purpose

   strings.Contains checks if one string is inside another.

2. Step 2: Check the return type

   It returns true if substr is found anywhere in s, otherwise false.

3. Final Answer:

   Checks if substr is present anywhere in s -> Option B

4. Quick Check:

   Contains means presence check = A [OK]

Hint: Contains means checking presence, not position or replacement [OK]

Common Mistakes:

• Confusing Contains with Index or Replace functions
• Thinking it returns position instead of boolean
• Assuming it modifies the original string

2. Which of the following is the correct syntax to find the index of substring "go" in string s using Go's standard library?

easy

A. index := strings.Index(s, "go")

B. index := s.IndexOf("go")

C. index := s.find("go")

D. index := strings.Find(s, "go")

Solution

1. Step 1: Recall Go's strings package functions

   The correct function to find substring index is strings.Index.

2. Step 2: Check syntax correctness

   The function is called as strings.Index(s, "go"), returning the first index or -1 if not found.

3. Final Answer:

   index := strings.Index(s, "go") -> Option A

4. Quick Check:

   Index function in strings package = B [OK]

Hint: Use strings.Index for substring position in Go [OK]

Common Mistakes:

• Using methods on string variable directly (strings are not objects)
• Using Python or JavaScript syntax like find or IndexOf
• Calling non-existent functions like strings.Find

3. What will be the output of the following Go code?

package main
import (
  "fmt"
  "strings"
)
func main() {
  s := "hello golang"
  pos := strings.Index(s, "go")
  fmt.Println(pos)
}

medium

A. 6

B. -1

C. 0

D. Error

Solution

1. Step 1: Understand strings.Index behavior

   It returns the first index where substring "go" appears in "hello golang".

2. Step 2: Find substring position

   Counting from 0: h(0) e(1) l(2) l(3) o(4) ' '(5) g(6) o(7)... So "go" starts at index 6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Index of "go" in "hello golang" = 6 [OK]

Hint: Count characters from zero to find substring index [OK]

Common Mistakes:

• Counting from 1 instead of 0
• Assuming substring not found returns 0
• Confusing Index with Contains boolean

4. Identify the error in the following Go code snippet for substring search:

package main
import (
  "fmt"
  "strings"
)
func main() {
  s := "golang"
  if s.Contains("go") {
    fmt.Println("Found")
  }
}

medium

A. Incorrect string literal syntax

B. Missing import for fmt package

C. No error, code runs fine

D. strings package has no Contains method on string variable

Solution

1. Step 1: Check method call on string variable

   In Go, strings are not objects and do not have methods like Contains directly.

2. Step 2: Correct usage of Contains function

   The correct call is strings.Contains(s, "go") from the strings package, not s.Contains("go").

3. Final Answer:

   strings package has no Contains method on string variable -> Option D

4. Quick Check:

   Use strings.Contains(s, substr), not s.Contains(substr) [OK]

Hint: Call Contains from strings package, not on string variable [OK]

Common Mistakes:

• Calling methods on string variables as if they were objects
• Forgetting to import strings package
• Assuming Contains returns index

5. Given a list of strings words := []string{"go", "golang", "gopher", "gone"}, which code snippet correctly finds all words containing substring "go" and stores them in result?

hard

A. result := []string{}
for _, w := range words {
if strings.Index(w, "go") > 5 {
result = append(result, w)
}
}

B. var result []string
for i := 0; i < len(words); i++ {
if words[i].Contains("go") {
result = append(result, words[i])
}
}

C. var result []string
for _, w := range words {
if strings.Contains(w, "go") {
result = append(result, w)
}
}

D. result := []string{}
for _, w := range words {
if strings.Index(w, "go") == -1 {
result = append(result, w)
}
}

Solution

1. Step 1: Identify correct substring check

   Use strings.Contains(w, "go") to check if substring exists.

2. Step 2: Append matching words to result slice

   For each word containing "go", append it to result slice.

3. Final Answer:

   var result []string for _, w := range words { if strings.Contains(w, "go") { result = append(result, w) } } -> Option C

4. Quick Check:

   Contains + append in loop = D [OK]

Hint: Use strings.Contains inside loop and append matches [OK]

Common Mistakes:

• Calling Contains as method on string variable
• Appending words that do NOT contain substring
• Checking Index == -1 instead of != -1