Zero Sum Subarray

Zero Sum Subarray

January 20, 2024

Problem #

Given a non-empty string s and an integer k, rearrange the string such that the same characters are at least distance k from each other. If it is not possible to rearrange the string, return an empty string.


  1. Input: s = “aabbcc”, k = 3 Output: “abcabc” or “bacbac” Explanation: The same characters are at least distance 3 from each other.

Solution Approach #

The solution involves using a max heap to ensure the most frequent characters are placed first. After placing a character, decrease its frequency and temporarily store it until k characters have been placed since its last placement.

Algorithm Steps #

  1. Count the frequency of each character in the string.
  2. Use a max heap to store characters by their frequencies.
  3. Pop the most frequent character from the heap, add it to the result string, decrease its frequency, and store it in a temporary queue.
  4. If the temporary queue size is k, push the front element back to the heap.
  5. Continue until the heap is empty.
  6. If the resulting string’s length is equal to the original string, return it; otherwise, return an empty string.

Code (Python) #

from collections import Counter, deque
import heapq

def rearrangeString(s, k):
    if k == 0:
        return s

    count = Counter(s)
    max_heap = [(-freq, char) for char, freq in count.items()]
    queue = deque()
    result = []

    while max_heap:
        freq, char = heapq.heappop(max_heap)

        queue.append((freq + 1, char))  # Increment frequency since it's used once

        if len(queue) == k:
            freq, char = queue.popleft()
            if freq < 0:
                heapq.heappush(max_heap, (freq, char))

    return ''.join(result) if len(result) == len(s) else ''

# Test the function
print(rearrangeString("aabbcc", 3))  # Output: "abcabc" or "bacbac"

Time Complexity #

O(n log n), where n is the length of the string. The heap operations contribute to the log n factor.

Space Complexity #

O(n), for storing the character frequencies and the heap.

Solution Analysis #

  1. Function Definition:

    • maxSumTwoNoOverlap(nums, k) takes an array nums and an integer k as input.
  2. Initialization:

    • n: Length of the nums array.
    • max_sum: A list initialized with zeros to store the maximum sum of non-overlapping subarrays up to each index.
    • window_sum: The sum of the first subarray of size k.
  3. First Sliding Window:

    • This loop calculates the maximum sum of a single subarray of size k ending at each index i.
    • window_sum += nums[i] - nums[i - k]: Updates the sum of the current window by adding the new element and subtracting the element that is no longer in the window.
    • max_sum[i] = max(max_sum[i - 1], window_sum): Updates max_sum[i] with the greater of the existing max sum or the current window sum.
  4. Dynamic Programming for Non-overlapping Subarrays:

    • The second loop uses dynamic programming to find the maximum sum of two non-overlapping subarrays.
    • It iterates through the array again, updating max_sum[i] to be the maximum of either the existing value or the sum of the maximum subarray ending k elements before i plus the sum of the current window.
  5. Returning the Result:

    • Finally, the function returns the maximum value from the max_sum array, which represents the maximum sum of two non-overlapping subarrays.

The thought process #

Coming up with a solution to an algorithmic problem, like the “Rearrange String k Distance Apart” question, typically involves a series of logical steps and considerations. Here’s an insight into the thought process:

Understanding the Problem #

First, I thoroughly analyze the problem statement to understand the requirements and constraints:

  • Objective: Rearrange a string so that the same characters are at least k distance apart.
  • Constraints: The input string can have any characters, and k is a non-negative integer.

Identifying Key Challenges #

Next, I identify the main challenges posed by the problem:

  • Frequency Management: Characters with higher frequencies pose a bigger challenge for spacing.
  • Distance Maintenance: Ensuring that the same characters are placed at least k distance apart.

Considering Potential Solutions #

With the challenges in mind, I think about potential solutions and data structures that could be useful:

  • Heap for Frequency Management: A max heap can help in managing characters by frequency, ensuring that the most frequent characters are dealt with first.
  • Queue for Distance Tracking: A queue can effectively keep track of recently placed characters to maintain the required distance k.

Solution Formulation #

With a potential approach in mind, I start formulating the solution:

  1. Count Frequencies: Determine the frequency of each character using a hashmap or Counter.
  2. Build a Max Heap: Use the heap to sort characters by their frequencies in descending order.
  3. Place Characters: Pop characters from the heap and add them to the result. Decrement the frequency and use a queue to keep track of the order and timing of placed characters.
  4. Manage Distance: Use the queue to ensure a character re-enters the heap only after k other characters have been placed, maintaining the distance requirement.

Writing the Code #

Translate the formulated solution into code, ensuring it handles edge cases and adheres to the problem constraints.

Analyzing Time and Space Complexity #

Finally, evaluate the time and space complexity to understand the efficiency of the solution.

Refinement #

After the initial solution is devised, I consider possible optimizations or alternative approaches, refining the solution for better efficiency or readability.

Testing and Validation #

Finally, test the solution with various inputs, including edge cases, to ensure correctness and robustness.

This systematic approach, combining understanding, planning, and iterative refinement, is essential for solving complex algorithmic problems effectively.