Count The Repetitions

shitgpt

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class Solution:
    def getMaxRepetitions(self, s1: str, n1: int, s2: str, n2: int) -> int:
        position_map = {}
        l1, l2 = len(s1), len(s2)
        total1 = l1 * n1
        i1 = i2 = 0

        while i1 < total1:
            if s1[i1 % l1] == s2[i2 % l2]:
                if (i1 % l1, i2 % l2) in position_map:
                    prevI1, prevI2 = position_map[(i1 % l1, i2 % l2)]
                    circle1 = i1 - prevI1
                    circle2 = i2 - prevI2
                    circle1Cnt = (total1 - i1) // circle1
                    i1 += circle1Cnt * circle1
                    i2 += circle1Cnt * circle2
                    if i1 >= total1:
                        break
                else:
                    position_map[(i1 % l1, i2 % l2)] = [i1, i2]
                i2 += 1
            i1 += 1

        return i2 // l2 // n2

Identifying Problem Isomorphism

“Count The Repetitions” can be mapped to “Repeated Substring Pattern”.

Reasoning:

Both involve the identification of repeated patterns within strings. In “Count The Repetitions”, you have to find out how many times a certain string can be repeated in another one. Similarly, in “Repeated Substring Pattern”, you need to determine if a given string can be constructed by taking a substring of it and appending multiple copies of the substring together. The difference lies in the detail, but the main concept, looking for repetitions in a string, is the same.

“Count The Repetitions” is more complex, as it not only requires identifying the repetitions but also counting them accurately.

The problem “Count The Repetitions” involves string manipulation and a deeper understanding of patterns. Here are 10 problems as preparation for tackling this problem:

  1. Repeated Substring Pattern: This problem can help you understand the basic idea of identifying repetition in strings.

  2. Repeated DNA Sequences: This problem is another case of looking for repeating patterns within a string.

  3. Longest Substring Without Repeating Characters: This problem introduces you to working with substrings.

  4. Implement strStr(): A classic problem for finding a needle in a haystack which can help you understand string matching.

  5. Minimum Window Substring: This problem requires a deeper understanding of string manipulation and sliding windows.

  6. ZigZag Conversion: This problem requires identifying and implementing a repetitive pattern in a string.

  7. Longest Palindromic Substring: This problem is about finding a specific pattern in a string.

  8. Permutation in String: This problem is about finding all permutations of a short string s1 in the longer one s2.

  9. Find All Anagrams in a String: This problem also deals with finding a pattern (an anagram of p) in a string s.

  10. Longest Common Subsequence: Involves finding common subsequences between two strings, which is a slightly simpler variant of the problem at hand.

These cover repetition, string manipulation, pattern identification, and sliding window technique.

We define str = [s, n] as the string str which consists of the string s concatenated n times.

For example, str == [“abc”, 3] ==“abcabcabc”. We define that string s1 can be obtained from string s2 if we can remove some characters from s2 such that it becomes s1.

For example, s1 = “abc” can be obtained from s2 = “abdbec” based on our definition by removing the bolded underlined characters. You are given two strings s1 and s2 and two integers n1 and n2. You have the two strings str1 = [s1, n1] and str2 = [s2, n2].

Return the maximum integer m such that str = [str2, m] can be obtained from str1.

Example 1:

Input: s1 = “acb”, n1 = 4, s2 = “ab”, n2 = 2 Output: 2

Example 2:

Input: s1 = “acb”, n1 = 1, s2 = “acb”, n2 = 1 Output: 1

Constraints:

1 <= s1.length, s2.length <= 100 s1 and s2 consist of lowercase English letters. 1 <= n1, n2 <= 106

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class Solution(object):
    def getMaxRepetitions(self, s1, n1, s2, n2):
        start = {}
        s1_round, s2_round, s2_idx = 0, 0, 0
        while s1_round < n1:
            s1_round += 1
            for ch in s1:
                if ch == s2[s2_idx]:
                    s2_idx += 1
                    if s2_idx == len(s2):
                        s2_round += 1
                        s2_idx = 0
            if s2_idx in start:
                prev_s1_round, prev_s2_round = start[s2_idx]
                circle_s1_round, circle_s2_round = s1_round - prev_s1_round, s2_round - prev_s2_round
                res = (n1 - prev_s1_round)
                left_s1_round = (n1 - prev_s1_round) % circle_s1_round + prev_s1_round
                for key in start:
                    val = start[key]
                    if val[0] == left_s1_round:
                        res += val[1]
                        break
                return res 
            else:
                start[s2_idx] = (s1_round, s2_round)
        return s2_round

Count The Repetitions - This problem requires counting the number of repetitions of a string within another string. This can be seen as a String Repetition Counting Problem.

Language Agnostic Coding Drills

Let’s break down the problem into units of learning:

  1. Understanding Problem: The key is to find how many times string s2 repeats in s1 after repeating s1 n1 times and s2 n2 times.

  2. String Traversal: The solution iterates over string characters. Given a string s1 and a number n1, you need to repeat the string s1 n1 times. Then, you have to find out how many times string s2 repeats in this longer string.

  3. Working with Dictionaries: The code uses a dictionary to keep track of instances when a specific character in s2 is found in s1. The key is the index of the character in s2, and the value is a tuple that records the rounds of s1 and s2.

  4. Conditional Checks: The code contains conditional checks to determine whether a specific character in s1 is equal to the character in s2 we are currently examining. If they are equal, it moves on to the next character in s2. If we reach the end of s2, it indicates that we’ve found a repetition of s2, so we reset the index for s2 and increment the count of s2 repetitions.

  5. Handling Repeating Patterns: The code checks whether the current character in s2 (indicated by s2_idx) has been seen before. If it has been seen, it means that a repeating pattern has been found. It calculates the lengths of s1 and s2 before the repeating pattern and within the repeating pattern (circle). It then calculates how many such repeating patterns fit into the remaining s1 repetitions, and whether there are any leftover s1 repetitions after fitting in the repeating patterns.

  6. Calculating Final Result: The code calculates the remaining s1 repetitions and uses it to find the corresponding s2 repetitions. It sums this with the s2 repetitions found within the repeating patterns and returns the total s2 repetitions divided by n2, as the question asks for the maximum number of s2 repetitions that we can get for each instance of s2.

  7. String Matching and Indexing: This problem requires matching the characters of one string to another and also understanding the concept of indexing in string to check each character.

  8. Looping and Iteration: Understanding how to iterate over the string characters and when to break the loop based on certain conditions is another crucial aspect of this problem.

To solve the problem, the code initially starts with no repetitions of s1 or s2 and goes through each character in s1. If a character matches the current character in s2, it moves on to the next character in s2 and checks whether it has gone through all characters in s2. When it finds a repeating pattern, it uses this pattern to quickly calculate the total s2 repetitions in the remaining s1 repetitions. This is an efficient solution to the problem because it avoids unnecessary repetitions by identifying repeating patterns.

Targeted Drills in Python

Here are the coding drills for each concept involved in the problem.

  1. String Traversal: Given a string, iterate over its characters and print each one.
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s = "HelloWorld"
for ch in s:
    print(ch)
  1. Working with Dictionaries: Create a dictionary, add items to it, access its values by keys.
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d = {}  # empty dictionary
d['apple'] = 1  # add a key-value pair
print(d['apple'])  # access value by key
  1. Conditional Checks: Write an if-else block that checks if a variable num is even. If it’s even, print “even”, else print “odd”.
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num = 10
if num % 2 == 0:
    print("even")
else:
    print("odd")
  1. Handling Repeating Patterns: Given a list of integers, detect if there is any repeating sequence in it.
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lst = [1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4]
seen = set()
for i in lst:
    if i in seen:
        print(f"Repeating sequence found at {i}")
    else:
        seen.add(i)
  1. Calculating Final Result: Given two integers a and b, calculate their sum, difference, product, and quotient.
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a = 10
b = 5
print(a + b)  # sum
print(a - b)  # difference
print(a * b)  # product
print(a / b)  # quotient
  1. String Matching and Indexing: Given two strings, iterate over the characters and print “Match” if characters at the same index in both strings are the same.
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s1 = "Hello"
s2 = "Hemlo"
for i in range(len(s1)):
    if s1[i] == s2[i]:
        print("Match")
  1. Looping and Iteration: Given a number n, print all numbers from 1 to n.
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n = 10
for i in range(1, n + 1):
    print(i)

These drills can help you strengthen the fundamentals required for solving the problem. Each of these drills focuses on a specific concept involved in the problem. By practicing these drills, you can get a better understanding of these concepts which can help in solving the problem more efficiently.