KMP Pattern Matching Algorithm in DSA Go - Time & Space Complexity

Time Complexity: KMP Pattern Matching Algorithm

O(n + m)

Understanding Time Complexity

We want to understand how fast the KMP algorithm finds a pattern inside a text.

Specifically, how the work grows as the text and pattern get longer.

Scenario Under Consideration

Analyze the time complexity of the following KMP pattern matching code.

func KMPSearch(text, pattern string) []int {
    lps := computeLPSArray(pattern)
    var result []int
    i, j := 0, 0
    for i < len(text) {
        if text[i] == pattern[j] {
            i++
            j++
            if j == len(pattern) {
                result = append(result, i-j)
                j = lps[j-1]
            }
        } else if j != 0 {
            j = lps[j-1]
        } else {
            i++
        }
    }
    return result
}

func computeLPSArray(pattern string) []int {
    lps := make([]int, len(pattern))
    length := 0
    i := 1
    for i < len(pattern) {
        if pattern[i] == pattern[length] {
            length++
            lps[i] = length
            i++
        } else if length != 0 {
            length = lps[length-1]
        } else {
            lps[i] = 0
            i++
        }
    }
    return lps
}
    

This code finds all occurrences of a pattern inside a text efficiently using the KMP algorithm.

Identify Repeating Operations

Look at the loops and repeated steps in the code.

• Primary operation: The main loop scans the text once, and the LPS computation scans the pattern once.
• How many times: Each character in text and pattern is visited at most twice during the process.

How Execution Grows With Input

As the text length (n) and pattern length (m) grow, the work grows roughly in a straight line.

Input Size (n)	Approx. Operations
10 (text), 3 (pattern)	About 13 steps
100 (text), 10 (pattern)	About 110 steps
1000 (text), 50 (pattern)	About 1050 steps

Pattern observation: The total steps grow roughly as the sum of text and pattern lengths, not their product.

Final Time Complexity

Time Complexity: O(n + m)

This means the algorithm checks each character in the text and pattern only a few times, making it very efficient.

Common Mistake

[X] Wrong: "KMP takes O(n * m) time because it compares every pattern character with every text character."

[OK] Correct: KMP uses the LPS array to skip unnecessary comparisons, so it never rechecks characters unnecessarily, keeping time linear.

Interview Connect

Knowing KMP's time complexity shows you understand efficient string searching, a useful skill for many coding challenges and real-world text processing.

Self-Check

"What if we used a naive search instead of KMP? How would the time complexity change?"

Practice

1. What is the main purpose of the KMP (Knuth-Morris-Pratt) algorithm in pattern matching?

easy

A. To efficiently find occurrences of a pattern in a text without rechecking characters

B. To sort the characters of the pattern alphabetically

C. To reverse the text before searching

D. To count the number of vowels in the text

Solution

1. Step 1: Understand KMP's goal

   KMP is designed to find pattern matches in text efficiently by avoiding repeated comparisons.

2. Step 2: Compare options with KMP's function

   Only To efficiently find occurrences of a pattern in a text without rechecking characters describes this efficient search without rechecking characters.

3. Final Answer:

   To efficiently find occurrences of a pattern in a text without rechecking characters -> Option A

4. Quick Check:

   KMP purpose = efficient pattern search [OK]

Hint: KMP avoids rechecking characters during pattern search [OK]

Common Mistakes:

• Thinking KMP sorts or modifies the text
• Confusing KMP with simple substring search
• Assuming KMP counts characters instead of searching

2. Which of the following is the correct way to declare the prefix table (lps array) in Go for KMP algorithm?

easy

A. lps := []string{len(pattern)}

B. lps := make([]int, len(pattern))

C. lps := make([]bool, len(pattern))

D. lps := new(int)

Solution

1. Step 1: Identify correct Go syntax for integer slice

   In Go, make([]int, size) creates an integer slice of given size.

2. Step 2: Check options for correct slice creation

   lps := make([]int, len(pattern)) correctly creates an int slice for lps. Others use wrong types or syntax.

3. Final Answer:

   lps := make([]int, len(pattern)) -> Option B

4. Quick Check:

   Go slice for lps = make([]int, len(pattern)) [OK]

Hint: Use make([]int, size) to create integer slices in Go [OK]

Common Mistakes:

• Using wrong slice type like []string or []bool
• Using new(int) which creates a pointer, not slice
• Incorrect slice initialization syntax

3. What will be the output of the following Go code snippet that uses KMP to find pattern occurrences?

package main
import "fmt"

func computeLPS(pattern string) []int {
    lps := make([]int, len(pattern))
    length := 0
    i := 1
    for i < len(pattern) {
        if pattern[i] == pattern[length] {
            length++
            lps[i] = length
            i++
        } else {
            if length != 0 {
                length = lps[length-1]
            } else {
                lps[i] = 0
                i++
            }
        }
    }
    return lps
}

func main() {
    pattern := "ABABC"
    lps := computeLPS(pattern)
    fmt.Println(lps)
}

medium

A. [0 0 0 0 0]

B. [0 1 2 3 4]

C. [1 2 3 4 5]

D. [0 0 1 2 0]

Solution

1. Step 1: Compute LPS array for pattern "ABABC"

   Check characters stepwise: - i=1: 'B' != 'A' => lps[1]=0 - i=2: 'A' == 'A' => length=1, lps[2]=1 - i=3: 'B' == 'B' => length=2, lps[3]=2 - i=4: 'C' != 'A' => length=lps[1]=0, lps[4]=0

2. Step 2: Final LPS array is [0, 0, 1, 2, 0]

   This matches [0 0 1 2 0] exactly.

3. Final Answer:

   [0 0 1 2 0] -> Option D

4. Quick Check:

   LPS for "ABABC" = [0 0 1 2 0] [OK]

Hint: Trace pattern chars to build LPS stepwise [OK]

Common Mistakes:

• Confusing LPS values with pattern indices
• Not resetting length correctly on mismatch
• Assuming all zeros or increasing sequence

4. Identify the bug in this KMP prefix table computation code snippet in Go:

func computeLPS(pattern string) []int {
    lps := make([]int, len(pattern))
    length := 0
    i := 0
    for i < len(pattern) {
        if pattern[i] == pattern[length] {
            length++
            lps[i] = length
            i++
        } else {
            if length != 0 {
                length = lps[length-1]
            } else {
                lps[i] = 0
                i++
            }
        }
    }
    return lps
}

medium

A. The function should return a string, not []int

B. The length variable should be initialized to 1

C. The loop should start with i = 1, not 0

D. The lps array should be of type []bool

Solution

1. Step 1: Check loop initialization for LPS computation

   In KMP, i starts from 1 because lps[0] is always 0 and no comparison needed at index 0.

2. Step 2: Identify incorrect i initialization

   Here i starts at 0, causing incorrect comparisons and possible infinite loop.

3. Final Answer:

   The loop should start with i = 1, not 0 -> Option C

4. Quick Check:

   Start i=1 for LPS loop [OK]

Hint: LPS loop always starts at index 1 [OK]

Common Mistakes:

• Starting loop at 0 causing wrong LPS values
• Initializing length incorrectly
• Wrong return type or array type

5. Given the text "ABABDABACDABABCABAB" and pattern "ABABCABAB", how many times does the pattern occur in the text using KMP algorithm?

hard

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Understand the pattern and text

   The pattern "ABABCABAB" appears in the text at indices 10 and 12 (0-based indexing), overlapping occurrences.

2. Step 2: Count occurrences using KMP logic

   KMP finds two matches without rechecking characters, so total occurrences = 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Pattern occurs twice in text [OK]

Hint: Use KMP to find all pattern matches efficiently [OK]

Common Mistakes:

• Counting overlapping matches incorrectly
• Overcounting occurrences
• Assuming zero matches due to complexity