Day 5 - Search Pattern (KMP-Algorithm).md

Difficulty	Source	Tags
Medium	160 Days of Problem Solving	Strings Pattern Searching

🚀 Day 5. Search Pattern (KMP Algorithm) 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given two strings:

• txt: The text string in which the pattern is to be searched.
• pat: The pattern string to search for.

The task is to print all indices in txt where pat starts, using 0-based indexing. Return an empty list if no occurrences are found.

🔍 Example Walkthrough:

Input:
txt = "abcab", pat = "ab"
Output:
[0, 3]

Explanation:
The pattern ab appears at indices 0 and 3 in the text string.

Input:
txt = "aaaaa", pat = "aa"
Output:
[0, 1, 2, 3]

Input:
txt = "hello", pat = "world"
Output:
[]

Constraints:

• 1 <= txt.length, pat.length <= 10^6
• Both strings consist of lowercase English alphabets.

🎯 My Approach:

The Knuth-Morris-Pratt (KMP) algorithm is an efficient pattern matching algorithm that avoids unnecessary comparisons, making it well-suited for this task.

1. Compute the Longest Prefix Suffix (LPS) Array:

   • Preprocess the pattern string pat to build the LPS array.
   • The LPS array stores the length of the longest prefix which is also a suffix for substrings of pat.
   • This preprocessing helps in skipping characters during comparisons.

2. Search Using the LPS Array:

   • Traverse the txt and use the pat LPS array to efficiently find matches.
   • If a mismatch occurs, use the LPS array to skip unnecessary comparisons.

3. Output the Indices:

   • Store the starting indices of matches found in txt.

🕒 Time and Auxiliary Space Complexity

• Preprocessing (LPS Array): O(m), where m is the length of the pattern string.
• Searching: O(n), where n is the length of the text string.
• Overall Time Complexity: O(n + m).

Auxiliary Space Complexity:

• Space for LPS Array: O(m).

📝 Solution Code

Code (C)

void computeLPSArray(char* pat, int m, int* lps) {
    int len = 0;
    int i = 1;
    lps[0] = 0;

    while (i < m) {
        if (pat[i] == pat[len]) {
            len++;
            lps[i] = len;
            i++;
        } else {
            if (len != 0) {
                len = lps[len - 1];
            } else {
                lps[i] = 0;
                i++;
            }
        }
    }
}

int* search(char pat[], char txt[], int* resultSize) {
    int m = strlen(pat);
    int n = strlen(txt);

    int* lps = (int*)malloc(m * sizeof(int));
    if (!lps) {
        fprintf(stderr, "Memory allocation failed for lps\n");
        exit(1);
    }

    computeLPSArray(pat, m, lps);

    int* result = (int*)malloc(10 * sizeof(int)); 
    if (!result) {
        fprintf(stderr, "Memory allocation failed for result\n");
        free(lps);
        exit(1);
    }
    int resCapacity = 10;
    int resIndex = 0;

    int i = 0, j = 0;
    while (i < n) {
        if (txt[i] == pat[j]) {
            i++;
            j++;
        }

        if (j == m) {
            if (resIndex >= resCapacity) {
                resCapacity *= 2;
                result = (int*)realloc(result, resCapacity * sizeof(int));
                if (!result) {
                    fprintf(stderr, "Memory reallocation failed for result\n");
                    free(lps);
                    exit(1);
                }
            }
            result[resIndex++] = i - j;
            j = lps[j - 1];
        } else if (i < n && txt[i] != pat[j]) {
            if (j != 0) {
                j = lps[j - 1];
            } else {
                i++;
            }
        }
    }
    *resultSize = resIndex;
    free(lps);

    return result;
}

Code (C++)

class Solution {
  public:
    vector<int> search(string& pat, string& txt) {
        int m = pat.size();
        int n = txt.size();
        vector<int> lps(m, 0);  
        vector<int> result;
        int len = 0; 
        for (int i = 1; i < m; ) {
            if (pat[i] == pat[len]) {
                len++;
                lps[i] = len;
                i++;
            } else {
                if (len != 0) {
                    len = lps[len - 1];
                } else {
                    lps[i] = 0;
                    i++;
                }
            }
        }
        int i = 0; 
        int j = 0; 
        while (i < n) {
            if (txt[i] == pat[j]) {
                i++;
                j++;
            }
            if (j == m) {
                result.push_back(i - j); 
                j = lps[j - 1];
            } else if (i < n && txt[i] != pat[j]) {
                if (j != 0) {
                    j = lps[j - 1];
                } else {
                    i++;
                }
            }
        }
        return result;
    }
};

Code (Java)

class Solution {
    private void computeLPSArray(String pat, int m, int[] lps) {
        int len = 0;
        int i = 1;
        while (i < m) {
            if (pat.charAt(i) == pat.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            } else {
                if (len != 0) {
                    len = lps[len - 1];
                } else {
                    lps[i] = 0;
                    i++;
                }
            }
        }
    }

    ArrayList<Integer> search(String pat, String txt) {
        ArrayList<Integer> result = new ArrayList<>();
        int m = pat.length();
        int n = txt.length();
        int[] lps = new int[m];

        computeLPSArray(pat, m, lps);

        int i = 0, j = 0;
        while (i < n) {
            if (txt.charAt(i) == pat.charAt(j)) {
                i++;
                j++;
            }

            if (j == m) {
                result.add(i - j);
                j = lps[j - 1];
            } else if (i < n && txt.charAt(i) != pat.charAt(j)) {
                if (j != 0) {
                    j = lps[j - 1];
                } else {
                    i++;
                }
            }
        }
        return result;
    }
}

Code (Python)

class Solution:
    def computeLPSArray(self, pat, m, lps):
        len = 0
        i = 1
        while i < m:
            if pat[i] == pat[len]:
                len += 1
                lps[i] = len
                i += 1
            else:
                if len != 0:
                    len = lps[len - 1]
                else:
                    lps[i] = 0
                    i += 1

    def search(self, pat, txt):
        result = []
        m = len(pat)
        n = len(txt)
        lps = [0] * m

        self.computeLPSArray(pat, m, lps)

        i = 0
        j = 0
        while i < n:
            if txt[i] == pat[j]:
                i += 1
                j += 1

            if j == m:
                result.append(i - j)
                j = lps[j - 1]
            elif i < n and txt[i] != pat[j]:
                if j != 0:
                    j = lps[j - 1]
                else:
                    i += 1
        return result

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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