grid47 Exploring patterns and algorithms
Aug 30, 2024
5 min read
Solution to LeetCode 686: Repeated String Match Problem
You are given two strings, a and b. You need to find the minimum number of times you should repeat string a so that string b becomes a substring of the repeated a. If it is impossible, return -1.
Problem
Approach
Steps
Complexity
Input: You are given two strings `a` and `b` consisting of lowercase English letters.
Example: a = 'xyz', b = 'yzxyz'
Constraints:
• 1 <= a.length, b.length <= 10^4
• a and b consist of lowercase English letters.
Output: Return the minimum number of times string `a` should be repeated such that string `b` becomes a substring of it. If not possible, return -1.
Example: For a = 'xyz' and b = 'yzxyz', the output is 2.
Constraints:
• The output should be an integer representing the minimum number of times string `a` should be repeated.
Goal: To find the minimum number of repetitions of string `a` that contains string `b` as a substring.
Steps:
• 1. Start by iterating over string `a` and try matching its substring with string `b`.
• 2. If the substring of repeated `a` matches `b`, return the number of repetitions.
• 3. If no such match is found, return -1.
Goal: The problem has constraints ensuring the strings are manageable in size and consist of lowercase letters.
Steps:
• The length of both strings `a` and `b` are between 1 and 10^4.
• Both strings `a` and `b` consist of lowercase English letters.
Assumptions:
• Both strings are not empty and contain only lowercase letters.
• Input: a = 'xyz', b = 'yzxyz'
• Explanation: Repeating string `a` two times gives 'xyzxyz', which contains 'yzxyz' as a substring.
• Input: a = 'ab', b = 'ababab'
• Explanation: Repeating string `a` three times gives 'ababab', which is exactly string `b`.
Approach: The approach is to iterate through possible repetitions of string `a` and check if string `b` is a substring of the repeated string.
Observations:
• We need to repeat string `a` multiple times to check if string `b` can fit within it as a substring.
• We can start from the smallest repetition and gradually increase until we find a match or exhaust all possibilities.
Steps:
• 1. Iterate through each repetition of string `a`.
• 2. For each repetition, check if string `b` is a substring of the repeated string.
• 3. If a match is found, return the number of repetitions.
• 4. If no match is found after sufficient repetitions, return -1.
Empty Inputs:
• There will always be non-empty strings, so no need to handle empty strings.
Large Inputs:
• The solution should handle large strings efficiently (up to length 10^4).
Special Values:
• When `a` is exactly equal to `b`, the answer is 1.
Constraints:
• The time complexity should be efficient enough to handle the largest inputs within the given constraints.
intrepeatedStringMatch(string a, string b) {
for(int i =0, j =0; i < a.size(); i++) {
for(j =0; j < b.size() && a[(i + j) % a.size()] == b[j]; j++);
if(j == b.size())
return (j + i -1)/ a.size() +1;
}
return-1;
}
1 : Function Definition
intrepeatedStringMatch(string a, string b) {
Define the method 'repeatedStringMatch' that takes two strings, 'a' and 'b', and returns the minimum number of times 'a' must be repeated to contain 'b' as a substring.
2 : Loop Initialization
for(int i =0, j =0; i < a.size(); i++) {
Start an outer loop to iterate through string 'a'. Initialize two indices, 'i' for 'a' and 'j' for 'b'.
Start an inner loop to check if the substring of 'a' starting from index 'i' matches the string 'b'. The modulo operator ensures that the loop wraps around to the start of 'a' if necessary.
4 : Substring Match Check
if(j == b.size())
If the entire string 'b' has been matched within 'a', the condition is true.
5 : Return Calculation
return (j + i -1)/ a.size() +1;
If a match is found, calculate the number of repetitions of 'a' needed to include 'b'. This is done by dividing the total length by the length of 'a' and adding 1.
6 : Return -1
return-1;
If no valid match is found after all iterations, return -1 to indicate that it's not possible to form 'b' by repeating 'a'.
Best Case: O(n), where n is the length of string b
Average Case: O(n * m), where n is the length of string a and m is the length of string b
Worst Case: O(n * m), where n is the length of string a and m is the length of string b
Description: The time complexity depends on how many times we need to repeat string `a` and check if `b` is a substring.
Best Case: O(n), if we use efficient substring matching techniques.
Worst Case: O(n * m), where n is the length of string a and m is the length of string b
Description: The space complexity is dependent on the string storage and substring checking process.
