1392. Longest Happy Prefix

Problem Description

The problem presents a string processing task where we need to find the longest possible substring that serves two roles simultaneously: it is a prefix and a suffix of the given string s. A prefix is a start part of the string, and a suffix is an ending part of the string. It's important to note however that the 'happy prefix' we are looking for cannot be the entire string itself, it should be strictly non-empty and shorter than the full string.

Intuition

The intuition behind the solution is fairly straightforward. We want to find the longest sequence at the beginning of the string (s) that is repeated at the end. The naive approach would be to consider each possible prefix of the string and check if it's also a suffix.

To implement this, we can start checking the longest possible prefix/suffix first and narrow down to the shorter ones. This is done by slicing the string: for a prefix starting from the beginning of the string s and ending at s[:-i], check if there's an equivalent suffix starting at s[i:] and going to the end of the string. If such a match is found, then s[i:] is our longest 'happy prefix'. If no match is found all the way up to the shortest prefix (a single character), return an empty string.

The solution starts the loop from 1 (as opposed to starting from 0) because we want a non-empty prefix/suffix. Additionally, we loop until len(s) because s[:-len(s)] would give us an empty string which is not a valid 'happy prefix', and we're looking for a prefix/suffix pair that does not include the entire string itself.

Solution Approach

The provided solution uses a straightforward brute-force approach. There's no need for complex algorithms or additional data structures. The approach is based on string comparison and slicing only.

Here's a step-by-step explanation of the implemented solution:

1. The solution defines a method longestPrefix within the Solution class that takes a string s as its argument.

2. The method initiates a loop that starts at 1 and ends just before the length of the string. The loop iterates with i indicating the current size of the suffix and prefix that are being compared. By using s[:-i] and s[i:], we effectively slice the string to create both the prefix and the suffix to compare.

3. At each iteration, the algorithm checks whether the current prefix (s[:-i]) is equal to the current suffix (s[i:]). If these two substrings are the same, it means we have found a 'happy prefix'.

4. The loop starts checking from the longest possible happy prefix and goes down to the shortest. The reasoning behind this order is that we are interested in the longest happy prefix. If a match is found, it will be the longest one due to the order of the loop, and the method can return immediately without checking the rest.

5. If no 'happy prefix' is found throughout the loop, the method returns an empty string '', indicating that no such prefix exists for the given string.

To visualize, let's consider s = "abab" as an example:

• In the first iteration, i = 1, the prefix s[:-1] is "aba" and the suffix s[1:] is "bab". They do not match.
• In the second iteration, i = 2, the prefix s[:-2] is "ab" and the suffix s[2:] is "ab". They match, so "ab" is returned as the longest happy prefix.

This solution approach has a time complexity of O(n^2), where n is the length of the string, due to the fact that string comparison is done in each iteration over slices of the string that decrease in size by one character each time.

Example Walkthrough

Let's walk through a small example using the string s = "level" to illustrate the solution approach. We are looking for the longest substring that is both a prefix and a suffix of s but is not s itself.

1. Iteration 1: i = 1

   • Prefix: s[:-1] is "leve"
   • Suffix: s[1:] is "evel"
   • Comparison: "leve" is not equal to "evel", so we continue to the next iteration.

2. Iteration 2: i = 2

   • Prefix: s[:-2] is "lev"
   • Suffix: s[2:] is "vel"
   • Comparison: "lev" is not equal to "vel", so we continue to the next iteration.

3. Iteration 3: i = 3

   • Prefix: s[:-3] is "le"
   • Suffix: s[3:] is "el"
   • Comparison: "le" is not equal to "el", so we continue to the next iteration.

4. Iteration 4: i = 4

   • Prefix: s[:-4] is "l"
   • Suffix: s[4:] is "l"
   • Comparison: "l" is equal to "l"
   • Since we have found that the prefix is equal to the suffix, "l" is the longest "happy prefix".

In this case, since "l" is the longest substring that meets the criteria, the method returns "l". If no matching substrings had been found, the method would return an empty string ''.

The solution is efficient because it uses a loop to progressively shorten the prefixes and suffixes from the longest to the shortest, stopping early when a match is found, which is more optimal compared to checking in the opposite direction. Additionally, by not using additional data structures, the approach keeps memory usage minimal.

Solution Implementation

Time and Space Complexity

Time Complexity

The provided code snippet calculates the longest prefix of a given string s that is also a suffix by checking each substring. The outer for loop runs (len(s) - 1) times. For each iteration of the loop, the slicing operation [i:] and [:-i] has a time complexity of O(n), where n is the length of the string. Since the slicing happens twice for each iteration and the comparison also takes O(n) for each iteration, the total time complexity is:

O((n - 1) * 2n) = O(2n^2 - 2n) which simplifies to O(n^2).

Therefore, the time complexity of the code is O(n^2).

Space Complexity

Since the algorithm only uses a constant amount of extra space for variables like i and the space needed for storing the prefixes/suffixes which are being compared is also not additional as they are references to existing substrings in s, the space complexity is O(1). There is no additional space being used that grows with the input size.

Learn more about how to find time and space complexity quickly using problem constraints.