KMP Pattern Matching Algorithm in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the prefix array with zeros.

DSA Java

int[] prefix = new int[pattern.length()];
for (int i = 0; i < prefix.length; i++) {
    prefix[i] = [1];
}

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

A0

Bpattern.length()

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing prefix array with ones or negative values.

Using pattern length as initial value.

The prefix array should be initialized with zeros because it stores the length of the longest prefix which is also a suffix for each substring.

2fill in blank

medium

Complete the code to update the prefix array when characters match.

DSA Java

if (pattern.charAt(i) == pattern.charAt(j)) {
    j++;
    prefix[i] = [1];
    i++;
}

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

Ai

Bj

C0

Di + j

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning prefix[i] = i instead of j.

Assigning prefix[i] = 0 when characters match.

When characters match, prefix[i] is set to j which is the length of the current longest prefix suffix.

3fill in blank

hard

Fix the error in the while loop condition to avoid infinite loop.

DSA Java

while (i < pattern.length()) {
    if (pattern.charAt(i) == pattern.charAt(j)) {
        j++;
        prefix[i] = j;
        i++;
    } else if (j > 0) {
        j = prefix[1];
    } else {
        prefix[i] = 0;
        i++;
    }
}

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

A(j - 1)

B(j + 1)

C{j - 1}

D[j - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or braces instead of square brackets for array access.

Using j+1 instead of j-1.

When characters don't match and j > 0, we update j to prefix[j-1] to try shorter prefix suffix.

4fill in blank

hard

Fill both blanks to complete the KMP search loop that finds pattern in text.

DSA Java

while (i < text.length()) {
    if (text.charAt(i) == pattern.charAt(j)) {
        i++;
        j++;
        if (j == [1]) {
            System.out.println("Pattern found at index " + (i - j));
            j = prefix[2];
        }
    } else if (j > 0) {
        j = prefix[j - 1];
    } else {
        i++;
    }
}

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

Apattern.length()

Btext.length()

C[j - 1]

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using text.length() instead of pattern.length() for comparison.

Resetting j to prefix[i] instead of prefix[j-1].

When j equals pattern length, a full match is found. Then j is reset to prefix[j-1] to continue searching.

5fill in blank

hard

Fill all three blanks to build the prefix array using the KMP preprocessing function.

DSA Java

public int[] buildPrefixArray(String pattern) {
    int[] prefix = new int[pattern.length()];
    int i = 1, j = 0;
    while (i < [1]) {
        if (pattern.charAt(i) == pattern.charAt(j)) {
            j++;
            prefix[i] = [2];
            i++;
        } else if (j > 0) {
            j = prefix[3];
        } else {
            prefix[i] = 0;
            i++;
        }
    }
    return prefix;
}

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

Apattern.length()

Bj

C[j - 1]

D[i - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop condition.

Assigning prefix[i] wrong value.

Incorrect array index for updating j.

The loop runs while i < pattern length. When characters match, prefix[i] = j. When mismatch and j > 0, j is updated to prefix[j-1].

Practice

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

easy

A. To count the number of vowels in a string

B. To sort characters in a string alphabetically

C. To reverse a string in linear time

D. To efficiently find occurrences of a pattern within a text

Solution

1. Step 1: Understand KMP's goal

   KMP algorithm is designed to search for a pattern inside a larger text efficiently.

2. Step 2: Compare with other options

   Sorting, reversing, or counting vowels are unrelated to pattern searching.

3. Final Answer:

   To efficiently find occurrences of a pattern within a text -> Option D

4. Quick Check:

   KMP = Pattern Searching [OK]

Hint: KMP is about fast pattern search, not string modification [OK]

Common Mistakes:

• Confusing KMP with sorting algorithms
• Thinking KMP reverses strings
• Assuming KMP counts characters

2. Which of the following is the correct way to initialize the prefix array (lps) in the KMP algorithm in Java?

easy

A. int[] lps = new int[pattern.length()];

B. int lps[] = new int[pattern.length() + 1];

C. int lps = new int[pattern.length()];

D. int[] lps = new int[pattern.length() - 1];

Solution

1. Step 1: Check array declaration syntax

   In Java, arrays are declared with type[], e.g., int[] lps.

2. Step 2: Check array size correctness

   Prefix array size equals pattern length, so new int[pattern.length()] is correct.

3. Final Answer:

   int[] lps = new int[pattern.length()]; -> Option A

4. Quick Check:

   Correct Java array init = int[] lps = new int[pattern.length()]; [OK]

Hint: Use int[] lps = new int[pattern.length()] for prefix array [OK]

Common Mistakes:

• Using incorrect array size (too big or too small)
• Wrong syntax like int lps instead of int[] lps
• Adding unnecessary +1 or -1 to length

3. What will be the output of the following Java code snippet using KMP to search pattern "ABABC" in text "ABABABCABABC"?

String text = "ABABABCABABC";
String pattern = "ABABC";
List<Integer> result = KMPSearch(pattern, text);
System.out.println(result);

Assume KMPSearch returns a list of starting indices where pattern is found.

medium

A. [1, 6]

B. [0, 5]

C. [2, 7]

D. [3, 8]

Solution

1. Step 1: Identify pattern occurrences in text

   Check text "ABABABCABABC" for "ABABC" starting indices: - At index 0: substring "ABABA" does not match fully - At index 2: substring "ABABC" matches fully - At index 5: substring "CABAB" does not match - At index 6: substring "ABABC" matches fully Actually, the pattern "ABABC" occurs at indices 2 and 7, but index 7 is out of range for the second match. Re-examining carefully: - At index 0: substring "ABABA" partial - At index 5: substring "CABAB" no - At index 6: substring "ABABC" matches - At index 7: substring "BABC" incomplete So matches at indices 2 and 6.

2. Step 2: Confirm no other matches

   Other indices do not match the full pattern exactly.

3. Final Answer:

   [2, 6] -> Option B

4. Quick Check:

   Pattern found at indices 2 and 6

Hint: Scan text for exact pattern start indices [OK]

Common Mistakes:

• Off-by-one errors in indexing
• Confusing partial matches as full matches
• Assuming matches start at 0 or 1 incorrectly

4. Consider this snippet of KMP prefix array (lps) computation in Java:

int len = 0;
int i = 1;
while (i < pattern.length()) {
  if (pattern.charAt(i) == pattern.charAt(len)) {
    len++;
    lps[i] = len;
    i++;
  } else {
    if (len != 0) {
      len = lps[len - 1];
    } else {
      lps[i] = 0;
      i++;
    }
  }
}

What is the bug in this code?

medium

A. Should increment i before assigning lps[i]

B. len should be reset to 0 instead of lps[len - 1]

C. No bug; code correctly computes lps array

D. Missing initialization of lps array before loop

Solution

1. Step 1: Analyze lps computation logic

   This code follows the standard KMP prefix array logic: - Increment len and assign lps[i] when chars match - Reset len using lps[len-1] when mismatch and len != 0 - Assign 0 and increment i when len == 0

2. Step 2: Verify no missing steps

   Initialization of lps array is assumed done before; loop logic is correct.

3. Final Answer:

   No bug; code correctly computes lps array -> Option C

4. Quick Check:

   Standard KMP lps logic = No bug; code correctly computes lps array [OK]

Hint: Standard KMP lps code is correct as given [OK]

Common Mistakes:

• Thinking i must increment before lps assignment
• Resetting len incorrectly to 0 instead of lps[len-1]
• Forgetting to initialize lps array before loop

5. Given the pattern "AABAACAABAA" and text "AABAACAABAABAABA", how many times will the KMP algorithm find the pattern in the text? Assume zero-based indexing and full matches only.

hard

A. 2

B. 1

C. 3

D. 4

Solution

1. Step 1: Search for pattern occurrences in text

   Check text "AABAACAABAABAABA" for pattern "AABAACAABAA": - At index 0: substring matches pattern fully - At index 7: substring "AABAACAABAA" matches fully - Other indices do not match fully

2. Step 2: Confirm matches

   Two full matches found at indices 0 and 7.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Pattern appears twice fully [OK]

Hint: Count only full pattern matches in text [OK]

Common Mistakes:

• Counting partial matches as full matches
• Missing the first match at index 0
• Assuming overlapping partial matches count