2025. Maximum Number of Ways to Partition an Array

Problem Description

You are given an array nums with n integers. We want to determine the maximum number of ways we can partition this array into two non-empty parts by choosing a pivot index so that the sum of elements to the left of the pivot is equal to the sum of elements to the right of the pivot. You can also change one element in the array to another integer k to increase the number of such partitions.

Partitioning an array means picking an index pivot such that 1 <= pivot < n and ensuring the following condition is met: the sum of elements from nums[0] to nums[pivot - 1] should be equal to the sum of elements from nums[pivot] to nums[n - 1].

You are permitted to change at most one element in nums to k or you can choose to keep the array as it is. The goal is to find the maximum number of partitions that satisfy the condition described above after making at most one such change.

Intuition

To arrive at the solution, we need to think about how changing one element to k could affect the number of valid partitions. When we don't change any element, any pivot that splits the array into two equal sums is a valid partition. We can calculate the total sum of the array first and then iterate through the array while keeping track of the cumulative sum. This helps us to determine if a pivot is a valid partition for the unchanged array.

However, when considering the change of one element to k, we need to think about where this operation would be most beneficial. Changing an element affects the sums of the elements to the left and to the right of that element. It's helpful to keep track of cumulative sums from the left and the right, and update these sums as if the change to k has been made.

We use two dictionaries: left to count the frequency of the left sums and right for the right sums. For each element x in the array, if we replace x with k, we calculate the difference d = k - x. We then update our counters based on this difference to reflect the new sums if the element were replaced.

Finally, we iterate over each element, consider replacing it with k, and count how many times the left and right sums across the pivot will then be equal. We update the answer with this count whenever it results in more valid partitions. The idea is to maximize this count, which gives us the maximum possible number of ways to partition the array after changing at most one element.

Learn more about Prefix Sum patterns.

Solution Approach

The solution leverages the defaultdict from the collections module, a subclass of the built-in dict class. It's used to store key-value pairs where the keys represent sums of elements and the values count how many times each sum occurs.

Here's a step-by-step explanation of the implementation:

1. Initialize the cumulative sum list s containing the same value nums[0]. This list will help us keep track of the sum from the beginning of the array up to each index.

2. Create two dictionaries, right and left, using the defaultdict with integer default types. The right dictionary will keep track of sums from the right side of the array, and left will do so for the left side.

3. Iterate through the array, starting from index 1, to create the cumulative sum array s and populate the right dictionary where each sum prior to the current index will be used as a key, and its frequency as the value.

4. Set ans to 0. This variable will hold the count of the maximum number of ways to partition the array.

5. Check if the total sum s[-1] is even. If yes, set ans to the number of ways to partition the array without changing an element. This can be found directly from the right dictionary for half of the total sum, since that would mean the array can be bisected into two equal halves.

6. Iterate over each element and the values from the cumulative sum list together. In each iteration, calculate the difference d = k - x, which represents how the sum would change if we replaced an element with k.

7. For each value v in the cumulative sum list and each x in the nums, check if the total sum of the array after the change, which is s[-1] + d, is even. If it is, compute the count t as the sum of the values from the left and right dictionaries at the respective adjusted sums, which would represent a valid midpoint after the change. Compare this with our current ans and update ans if t is larger.

8. Increment the count of the current sum in the left dictionary by 1, and decrement the count of the current sum in the right dictionary by 1, reflecting that we are moving from right to left in terms of possible pivots.

9. Finally, return the value stored in ans, which represents the maximum ways we can partition the array after making at most one change.

The algorithm efficiently counts the number of ways to partition the array by maintaining cumulative sums and dynamically adjusting the counts when considering the impact of one element change.

Example Walkthrough

Let's consider an example array nums as [1, 2, 3, 4, 6] and suppose we want to change one element to k = 5. We want to calculate the maximum number of ways we can partition the array such that the sum of elements on either side of the pivot is equal.

1. Initialize the cumulative sum list s. After iterating over nums, s will be [1, 3, 6, 10, 16].

2. Create two dictionaries, right and left, using the defaultdict(int).

3. Populate the right dictionary, recording the frequence of sums to the right side of the array so that it looks like this: {16: 1, 15: 1, 13: 1, 10: 1, 6: 1} before any changes, signifying that the sum 16 occurs once, 15 occurs once, and so on.

4. Set ans to 0, ready to hold the maximum number of partitions.

5. The total sum s[-1] is 16, which is not even, so we cannot initially partition the array without changing an element.

6. Now, iterate over nums and s:

   • At index 0, x = 1, the difference d = k - x = 5 - 1 = 4. The total sum after the change would be 16 + 4 = 20, which is even.
   • At index 1, x = 2, d = 5 - 2 = 3. The total sum would be 16 + 3 = 19, which is not even, so no partition is possible here.
   • Continuing this process, we find that for index 2, x = 3, d = 5 - 3 = 2, and the total sum is even at 18, suggesting a potential partition.

7. When x = 3 and v = 6 at index 2, check for potential partitions:

   • The left sum would be 6 + 2 = 8, and the right sum (excluding the current element) would also need to be 8 for a valid partition.
   • We add 1 to left[8] as this sum has now occurred once.
   • We decrease right[10] to 0 since we have used that partition possibility.
   • We find that the partition could occur at index 3, since the sums on either side would then be equal.

8. Increment the left dictionary and decrement the right dictionary as we move along the array and continue checking for partitions after each element.

9. Return the value stored in ans, which is the maximum ways we can partition the array. In this case, ans would be updated to 1 as we found a valid partition when we considered changing the third element to 5, which would make the sums on either side of the pivot equal.

This walkthrough demonstrates how the algorithm works in a practical example, changing one element to k and finding the maximum number of partitions where the sums on either side of the pivot are equal.

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python function waysToPartition performs a series of operations on an array to determine the number of ways it can be partitioned such that the sums of the elements on either side of the partition are equal, with the option of changing one element to k.

First, the function creates an accumulated sum array s of the input array nums. This operation traverses the array once and has a time complexity of O(n), where n is the length of the input array.

Next, the function initializes a right dictionary (from the collections.defaultdict class) to count the occurrences of each prefix sum as we iterate from the end to the beginning of the accumulated sum array. Filling this dictionary also takes O(n).

The function then checks a special case where the total sum of the array is even, and a partition is possible without any change. Checking and updating the count takes constant time O(1).

Following this, the function iterates through nums and its accumulated sum s simultaneously and updates two dictionaries: left and right. Each iteration involves a constant amount of work - updating dictionaries and a few arithmetic operations. Since the iteration happens n times, we get O(n).

Within the same loop, the partition count is updated based on whether it's possible to create an equal partition by changing the current number to k. The checks and updates are performed in constant time for each iteration.

Considering all these operations together, since they are sequential, the total time complexity is O(n) + O(n) + O(1) + O(n) = O(n).

Space Complexity

The space complexity of the function involves the space taken by the input nums, the accumulated sum array s, and the left and right dictionaries.

The accumulated sum array s has the same length as nums, contributing O(n) space complexity.

Both left and right dictionaries could, in the worst case, store a distinct sum for every prefix and suffix, which also contributes up to O(n) space complexity.

Therefore, the total space complexity is the sum of the space complexities of these data structures, O(n) + O(n) + O(n) = O(n).

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