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Maximize the Number of Partitions After Operations

Number: 3233

Difficulty: Hard

Paid? No

Companies: HiLabs

Problem Description

Given a string s and an integer k, you are allowed to change at most one character in s to any other lowercase English letter. Then, repeatedly perform a partitioning procedure on s where in each step you remove the longest possible prefix that contains at most k distinct characters. The goal is to maximize the number of partitions (i.e. the number of times you remove a prefix) by choosing the optimal character modification (or none).

Key Insights

  • The greedy partitioning procedure always takes the longest possible prefix such that its distinct character count does not exceed k.
  • Without any modifications, if s has at most k distinct characters then the whole string becomes one partition.
  • By judiciously changing one character to a letter that is new relative to the current prefix, it is possible to force an earlier “violation” (i.e. the distinct count in the prefix would exceed k sooner) thereby creating shorter segments and increasing the total partition count.
  • The optimal solution may involve either not changing any character or replacing exactly one character – thus we must try both possibilities.
  • A simulation of the partitioning process (using a two‑pointer or sliding window method) is used to count the number of partitions given a string.
  • Though a brute-force approach of trying every index (and every candidate letter) leads to O(26 * n^2) worst-case time, it is acceptable for explanation purposes. In practice one would try to narrow the search to only indices where the change affects the partition boundaries, or use dynamic programming with precomputed “next partition” boundaries.

Space and Time Complexity

Time Complexity: O(26 * n^2) in the worst-case brute-force simulation (n = length of s).
Space Complexity: O(n) for storing temporary variables and the sliding window frequency counts.

Solution

We solve the problem in two main steps. First, we implement a helper function to simulate the greedy partitioning procedure: starting from the beginning of the string, extend the current partition (using a sliding window to track distinct letters) until the next letter would cause the count of distinct letters to exceed k, then “cut” the partition. The simulation returns the total number of partitions.

Second, we try two cases:

  1. No modification: simply count the partitions.
  2. For each index in s and for every possible letter different from the current one, construct a “modified” string with that one change and simulate the partitioning. Track the maximum partitions found.

We use simple data structures – such as a dictionary (or hash table) for frequency counting in the sliding window – and the two‑pointer technique. Comments in the code highlight the role of each variable and step.

Code Solutions

# Python solution with detailed comments

def simulate_partitions(s, k):
    """
    Simulate the greedy partitioning procedure.
    From the start of the string, expand the window until adding another
    character would make the distinct letters count exceed k.
    Return the number of partitions.
    """
    n = len(s)
    partitions = 0
    i = 0
    while i < n:
        freq = {}
        j = i
        # Extend the window while distinct char count is at most k
        while j < n:
            char = s[j]
            freq[char] = freq.get(char, 0) + 1
            # If distinct count exceeds k then break
            if len(freq) > k:
                # Removing s[j] so break from this partition
                break
            j += 1
        partitions += 1
        # Set starting index for next partition
        i = j
    return partitions

def maximize_partitions(s, k):
    # Baseline with no modification
    max_parts = simulate_partitions(s, k)
    n = len(s)
    letters = "abcdefghijklmnopqrstuvwxyz"
    
    # Try modifying each index to a different letter
    for i in range(n):
        original = s[i]
        for ch in letters:
            if ch == original:
                continue
            # Build the new string with a single modification at index i
            new_s = s[:i] + ch + s[i+1:]
            parts = simulate_partitions(new_s, k)
            if parts > max_parts:
                max_parts = parts
    return max_parts

# Example Test Cases
print(maximize_partitions("accca", 2))   # Expected output: 3 
print(maximize_partitions("aabaab", 3))  # Expected output: 1
print(maximize_partitions("xxyz", 1))    # Expected output: 4
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