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Split a String Into the Max Number of Unique Substrings

Number: 1715

Difficulty: Medium

Paid? No

Companies: Amazon, Meta, Google

Problem Description

Given a string s, split it into a list of non-empty substrings such that the concatenation of these substrings forms the original string and all substrings are unique. The goal is to maximize the number of unique substrings that can be formed.

Key Insights

  • Use a backtracking approach to try all possible ways to split the string.
  • Maintain a set to track substrings that have already been used.
  • Since the maximum string length is 16, exploring all split combinations is feasible.
  • Update the maximum count found whenever a valid split is achieved.

Space and Time Complexity

Time Complexity: O(2^n * n) in the worst case, where n is the length of the string (each recursive call may iterate through remaining characters). Space Complexity: O(n) due to the recursion stack and the storage for the set of substrings.

Solution

We can solve this problem using a recursive backtracking approach:

  1. Define a recursive function that takes the current index in the string and a set of substrings used so far.
  2. At every recursive call, iterate over all possible end indexes for the next substring.
  3. If the current substring is not already in the set, add it and make a recursive call using the updated index.
  4. When the end of the string is reached, update the result if the number of substrings used is greater than the maximum found so far.
  5. Use backtracking to remove the substring from the set when going back up the recursive chain. This approach ensures that all valid partition combinations are explored, ultimately finding the maximum number of unique substrings.

Code Solutions

# Function to split a string into the maximum number of unique substrings.
def maxUniqueSplit(s: str) -> int:
    # This variable keeps track of the maximum count of unique substrings found.
    max_count = 0
    
    # Define the recursive backtracking function.
    def backtrack(start: int, seen: set):
        nonlocal max_count
        # If we've reached the end of the string,
        # update the maximum count if needed.
        if start == len(s):
            max_count = max(max_count, len(seen))
            return
        # Try every possible end index for the current substring.
        for end in range(start + 1, len(s) + 1):
            substring = s[start:end]
            # Only consider the substring if it's not been used before.
            if substring not in seen:
                seen.add(substring)
                backtrack(end, seen)
                # Backtrack: remove the substring from the set.
                seen.remove(substring)
    
    # Start the recursion with initial index 0 and an empty set.
    backtrack(0, set())
    return max_count

# Example usage:
print(maxUniqueSplit("ababccc"))  # Expected output: 5
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