Problem Description

Given a string s, we are tasked to find the maximum number of non-empty, unique substrings into which the string can be split. The goal is to make sure that after splitting, each substring is distinct, and their concatenation would result in the original string s. It's important to note that a substring in this context is defined as a contiguous block of characters within the string s.

Flowchart Walkthrough

Let's analyze the problem using the Flowchart. Here's how we would step through the flowchart to reach the conclusion:

1. Is it a graph?

   • No: The problem does not deal with nodes and edges typical in graph-related problems.

2. Need to solve for kth smallest/largest?

   • No: The goal is not to find the kth smallest or largest element.

3. Involves Linked Lists?

   • No: The problem is about strings and substrings, not linked structures.

4. Does the problem have small constraints?

   • Yes: The problem typically involves manipulating strings, which are manageable in size, allowing for exploration of multiple combinations or partitions.

5. Brute force / Backtracking?

   • Yes: The task requires exploring all possible ways to split the string into unique substrings, which is suitably handled using backtracking to generate and check each unique configuration.

Conclusion: The flowchart and analysis suggest using a backtracking approach for splitting the string into the maximum number of unique substrings. This involves generating all possible partitions and checking each one for uniqueness.

Intuition

To solve this problem, we can use a Depth First Search (DFS) approach. The intuition behind a DFS is to explore each possible substring starting from the beginning of the string and recursively split the rest of the string in a similar manner. For each recursive call, we track the index at which we're currently splitting the string and maintain a count of the unique substrings created so far. In addition, we keep a record of visited substrings in a set to ensure uniqueness.

Here's the step-by-step thought process:

1. Begin at the start of the string and consider every possible substring that can be formed starting from this position.
2. At each step, choose a substring and check if it is unique (i.e., not already in the set of visited substrings). If it's unique, add it to the set to track its inclusion.
3. Recurse with the next part of the string, starting immediately after the selected substring, while incrementing the count of unique substrings.
4. If you reach the end of the string, compare the count of unique substrings with a global variable that maintains the maximum count seen thus far, updating it if necessary.
5. After the recursion call, backtrack by removing the last chosen substring from the visited set, allowing it to be considered in different combinations.
6. By the end of the DFS exploration, the global variable, ans, will contain the maximum number of unique substrings we can obtain by splitting the string.

The use of DFS ensures we explore all possible combinations of substrings, and the use of a set (vis) makes sure that all chosen substrings are unique throughout the recursion.

Learn more about Backtracking patterns.

Solution Approach

The solution utilizes a recursive Depth First Search (DFS) algorithm to traverse through all possibilities, and uses a set as a data structure to ensure the uniqueness of substrings.

Here's a detailed walkthrough of the implementation:

• Define a recursive function dfs that takes two parameters: i, which represents the current starting index of the substring, and t, the count of unique substrings formed so far.
• If i reaches the length of string s, it means we have reached the end. We compare and update the global ans variable with t if it represents a higher count of unique substrings.
• Otherwise, iterate over the string s starting from index i + 1 to the end of the string. This loop represents all possible ends of the substring that starts at index i.
• In each iteration, we first check if the current substring s[i:j] hasn't been visited (is not in the set vis). If it is unique:
  • Add the current substring to the vis set.
  • Recurse with the new index j, which is the end of the current substring, and increment the count of unique substrings t + 1.
  • After recursion, remove the current substring s[i:j] from the set vis to allow for other potential unique substrings that could use the same segment of s.
• Initialize the vis set that holds the unique substrings and declare a variable ans which initially holds the value 1, since the string itself can be considered as a unique substring of length one.
• Finally, start the DFS with the initial call dfs(0, 0), and after recursive exploration of all possible splits, return the maximum count stored in ans.

The key patterns used here are recursion for exhaustive search and backtracking for exploring different paths without repeating the visited combinations. The set data structure is essential for assuring the uniqueness constraint for substrings.

Here's a snippet of the key function from the code:

def dfs(i, t):
    if i >= len(s):
        nonlocal ans
        ans = max(ans, t)
        return
    for j in range(i + 1, len(s) + 1):
        if s[i:j] not in vis:
            vis.add(s[i:j])
            dfs(j, t + 1)
            vis.remove(s[i:j])

This method ensures that all potential substring splits are considered, and the maximum number of unique substrings that can be obtained from s is found.

Example Walkthrough

Let's take a small example to illustrate the solution approach using the provided problem content. Consider the string s = "abab".

Following the steps of the depth-first search (DFS) and backtracking algorithm:

1. We begin with an empty set vis for maintaining unique substrings and a variable ans initialized to 1.
2. We start the DFS by calling dfs(0, 0), signifying that we start from the beginning of the string and currently have 0 unique substrings.

First, consider i = 0:

• We take the substring s[0:1] which is "a" and since it's not in vis, we add it to vis and call dfs(1, 1).

  For the recursive call dfs(1, 1):

  • Now i = 1; we take s[1:2] which is "b", add it to vis, and call dfs(2, 2).

    For the recursive call dfs(2, 2):

    • i = 2; we can't choose "a" again (as s[2:3]) because it is already in vis. Hence, we move to s[2:4], which is "ab", add it to vis, and call dfs(4, 3).

      For the recursive call dfs(4, 3):

      • i = 4; we have reached the end of the string. We compare and update ans with 3 which is greater than the initial value 1.
      • We backtrack and remove "ab" from vis.
      • The calls dfs(3, 3) and previous layers would not produce any further unique substrings, so we backtrack completely to the first call after "b" and s[1:2] is removed from vis.

  • Back in dfs(1, 1), we continue looking for substrings and try s[1:3] which is "ab". It's unique at this point, so we add it to vis and call dfs(3, 2).

    For the recursive call dfs(3, 2):

    • i = 3; we choose s[3:4] which is "b", add it to vis, and call dfs(4, 3).

      For the recursive call dfs(4, 3):

      • i = 4; we are at the end of the string. ans is again compared and updated if necessary, but it remains 3.

In each recursive call, after we reach the end, we backtrack by removing the last inserted substring from vis. This allows us to explore different combinations.

At the end of the DFS, ans will contain the value 3, which is the maximum number of non-empty, unique substrings that "abab" can be split into - these are "a", "b", and "ab".

Solution Implementation

Time and Space Complexity

Time Complexity

The provided code implements a depth-first search (DFS) approach to solve the problem of finding the maximum number of unique substrings that can be split from a given string.

To determine the time complexity, consider each step in the code:

• The dfs function is called recursively, considering all possible splits starting at different positions in the string until the end of the string is reached.
• In each function call, it performs a loop from i + 1 to len(s) + 1, exploring all the possibilities for the next starting point of a substring.
• A substring is added to the vis set only if it's not already seen. This operation takes O(k) where k is the length of the substring, as strings are immutable in Python and need to be hashed for insertion in a set.

Since we consider every possible split for a string of length n, and in each step we could potentially generate all possible substrings, the worst-case is exponential in nature with respect to the input size.

The worst-case time complexity is therefore O(n * 2^n), because at each of the n positions, we could make a decision to split or not to split the string, and we are doing this for all n positions except the last one which is determined by the choices made for the previous substrings.

Space Complexity

Looking at the space complexity:

• The set vis stores the unique substrings we have encountered. In the worst case, it can store all possible substrings of s, which amounts to O(2^n) substrings if we consider all possible splits.
• The recursion depth can go up to n in the case of splitting each character as a separate substring. Hence, the space complexity due to recursive calls is O(n).

Therefore, combining the storage for the set and the recursive call stack space, the overall space complexity is O(2^n + n), where O(2^n) is the dominant term, simplifying the space complexity to O(2^n).

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