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Lecture 1 (Strings hard problems)

Z function

Given two strings, one is a text string and the other is a pattern string. Print the indexes of all occurrences of the pattern string in the text string using the Z function algorithm. Use 0-based indexing while returning the indices.

Input: text = "xyzabxyzabxyz", pattern = "xyz"
Output: [0, 5, 10]

Input: text = "cabcdab", pattern = "abc"
Output: [1]

Approach 1 (Brute Force)

• Build match function (that takes start idx)
• Call match function every idx

def z_function(text,pattern):
    T = len(text)
    P = len(pattern)


    def match(start):
        i =0
        while start < T and i < P and text[start]== pattern[i]:
            i+=1
            start +=1

        return i == P

    result = []

    for i in range(T):
        if match(i):
            result.append(i)

    return result

print(z_function("xyzabxyzabxyz","xyz"))
print(z_function("cabcdab","abc"))

[0, 5, 10]
[1]

Complexity

• T = len(text) and P = len(pattern)
• O(T * P)
• O(1)

Approach 2 (Z function)

• Build the z array
• Z[i] = len(substring) starting from index i that match prefix of string
• zbox : left ,right
• STEP 1: Use previous computations if inside the Z-box
• STEP 2: Extend the match as far as possible
• STEP 3: Update Z-box if we've extended beyond it Refer https://www.youtube.com/watch?v=CpZh4eF8QBw

def compute_z(s):
    n = len(s)
    z = [0] *n

    left ,right = 0,0

    for i in range(1,n):
        # STEP 1: Use previous computations if inside the Z-box
        if i <= right:
            z[i] = min(z[i - left], right - i + 1)
            
        # STEP 2: Extend the match as far as possible
        while i + z[i] <n and s[z[i]] == s[i + z[i]]:
            z[i] +=1

        # STEP 3: Update Z-box if we've extended beyond it
        if i + z[i] -1 > right:
            left = i
            right = i+ z[i] -1
    return z
        

def z_function(text,pattern):
    combined = pattern + "#" + text
    z = compute_z(combined)

    result = []
    for i in range(len(combined)):
        if z[i] == len(pattern):
            result.append(i - len(pattern)-1)

    return result

print(z_function("xyzabxyzabxyz","xyz"))
print(z_function("cabcdab","abc"))

[0, 5, 10]
[1]

Complexity

• T = len(text) and P = len(pattern)
• O(T + P)
• O (T +P)