Repeated String Match
Problem Description
Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b to be a substring of a after repeating it, return -1.
Notice: string "abc" repeated 0 times is "", repeated 1 time is "abc" and repeated 2 times is "abcabc".
Example 1:
Input: a = "abcd", b = "cdabcdab"
Output: 3
Explanation: We return 3 because by repeating a three times "abcdabcdabcd", b is a substring of it.
Example 2:
Input: a = "a", b = "aa"
Output: 2
Constraints:
- 1 <= a.length, b.length <= 104
aandbconsist of lowercase English letters.
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
The problem requires us to determine the minimum number of times string A has to be repeated such that B is a substring of it. A brute force solution involves repeatedly appending A to itself until B becomes a substring or we surpass a certain reasonable limit.
Steps:
- Start by observing that if the answer exists, the number of times we need to repeat
Ais at least|B|/|A|(integer division) and at most|B|/|A| + 2. - The reasoning behind the upper bound is that
Bcan start appearing mid-way within one of the copies ofAwhich might require one or two more copies ofAto cover whole ofB. - Create a repeated string starting from a single
A, and keep appendingAto this string. - For each iteration, check if
Bis a substring of the current repeated string. - If
Bis found, return the count of repetitions. If after|B|/|A| + 2repetitionsBis not found, return -1.
Code:
- Time Complexity: O(N * M), where
Nis the length ofAandMis the length ofB. This is due to the substring search operation. - Space Complexity: O(N * M), due to repeated string storage.
2. Optimal String Concatenation
Intuition:
The brute force solution already works fairly efficiently for most reasonable input sizes. However, we can optimize the repeated string construction and search process slightly using intuitive bounds.
Steps:
- Start by computing the minimum repetitions required using
ceil(|B|/|A|), which can be done using(B.length() + A.length() - 1) / A.length(). - Instead of constructing the repeated string each time from scratch, directly compute this base repeated string.
- Append one more
Ato handle boundary conditions. - Check if
Bis a substring afterminRepeatsandminRepeats + 1repetitions and return the respective count if true. - Return -1 if
Bis not a substring in both checks.
Code:
- Time Complexity: O(N + M), where
Nis the length ofAandMis the length ofB. This comes from building the string and subsequent substring check. - Space Complexity: O(N + M), due to storage of the repeated string.