How do you solve Maximum Number of Ways to Partition an Array on LeetCode?

Maximum Number of Ways to Partition an Array (LeetCode #2025) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Changing one element can significantly affect the partitioning possibilities.

What is the brute force solution for Maximum Number of Ways to Partition an Array?

The brute force approach for Maximum Number of Ways to Partition an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible pivot index and calculates the prefix and suffix sums for each pivot. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Ways to Partition an Array?

Maximum Number of Ways to Partition an Array uses the following concepts: Array, Hash Table, Counting, Enumeration, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Number of Ways to Partition an Array?

Always think about how you can optimize a brute-force solution. Use clear variable names and keep your code organized. Practice explaining your thought process as you code.

What are common mistakes when solving Maximum Number of Ways to Partition an Array?

Not considering the effect of changing each element on the overall sums. Overlooking edge cases where the array has negative numbers.

#2025

Maximum Number of Ways to Partition an Array

Hard

ArrayHash TableCountingEnumerationPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses prefix sums and a HashMap to efficiently count valid partitions without recalculating sums repeatedly. This reduces the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Calculate the total sum of the array and initialize a prefix sum and a HashMap to count occurrences of prefix sums.
2Step 2: Iterate through the array to fill the prefix sum and check for valid pivots.
3Step 3: For each index, calculate the required suffix sum and check if it exists in the HashMap.
4Step 4: For each element, consider changing it to k and adjust the prefix sum accordingly, checking for new valid partitions.
5Step 5: Return the maximum count of valid partitions found.

solution.py30 lines

1def maxWays(nums, k):
2    n = len(nums)
3    total_sum = sum(nums)
4    prefix_sum = 0
5    count = 0
6    prefix_count = {}
7
8    for i in range(n - 1):
9        prefix_sum += nums[i]
10        suffix_sum = total_sum - prefix_sum
11        if prefix_sum == suffix_sum:
12            count += 1
13        prefix_count[prefix_sum] = prefix_count.get(prefix_sum, 0) + 1
14
15    max_count = count
16
17    for i in range(n):
18        original = nums[i]
19        nums[i] = k
20        prefix_sum = 0
21        count = 0
22        for j in range(n - 1):
23            prefix_sum += nums[j]
24            suffix_sum = total_sum - prefix_sum
25            if prefix_sum == suffix_sum:
26                count += 1
27        max_count = max(max_count, count)
28        nums[i] = original
29
30    return max_count

ℹ

Complexity note: The time complexity is O(n) due to a single pass through the array to calculate prefix sums and check for valid partitions. The space complexity is O(n) for storing prefix sums in the HashMap.

1Changing one element can significantly affect the partitioning possibilities.
2Using prefix sums allows for efficient calculation of partition conditions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.