#3432

Easy

LeetCode Problem Workspace

Count Partitions with Even Sum Difference

Count the number of partitions with an even sum difference from an array of integers.

Problem Statement

You are given an integer array nums of length n. A partition is defined as an index i where 0 <= i < n - 1, splitting the array into two non-empty subarrays. The task is to return the number of partitions where the difference between the sum of the left and right subarrays is even.

To solve this, focus on understanding the parity (even or odd) of the sums of the subarrays formed by each partition. If the difference between the sums is even, that partition is valid. Efficiently calculate the sum differences for various partitions to count the valid ones.

Examples

Example 1

Input: nums = [10,10,3,7,6]

Output: 4

The 4 partitions are:

Example 2

Input: nums = [1,2,2]

Output: 0

No partition results in an even sum difference.

Example 3

Input: nums = [2,4,6,8]

Output: 3

All partitions result in an even sum difference.

Constraints

2 <= n == nums.length <= 100
1 <= nums[i] <= 100

Solution Approach

Prefix Sum Calculation

Compute the prefix sum of the array, which helps quickly calculate the sum of elements in any subarray. This enables an efficient way to determine the sum of elements on both sides of each partition and check the parity of the difference.

Parity Tracking

Track the parity of the prefix sum. Instead of computing the sum for each partition from scratch, track whether the sum up to each index is even or odd. Use this information to determine if the sum difference between the left and right subarrays is even.

Iterate with Efficient Checks

Loop through the array and check the parity of the sums as you go. For each partition, determine if the difference between the two subarrays has even parity. This approach allows for an optimal solution without redundant calculations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on how you compute the prefix sums and track the parity. A prefix sum approach with parity checks can solve the problem in O(n) time. The space complexity is also O(n) due to the storage of prefix sums and parity information.

What Interviewers Usually Probe

Focus on efficient sum calculation and parity checking to avoid unnecessary recomputations.
Be cautious of using brute-force methods that lead to high time complexity.
Look for optimized ways to track the parity of subarray sums to minimize redundant work.

Common Pitfalls or Variants

Common pitfalls

Ignoring the parity of prefix sums and recalculating sums for each partition.
Not leveraging prefix sum and parity, resulting in a brute-force O(n^2) solution.
Overlooking edge cases where partitions result in no valid even sum differences.

Follow-up variants

Optimize further for larger arrays by focusing on parity without storing full sums.
Explore dynamic programming alternatives for this problem if required by the interviewer.
Consider modifying the problem to track odd sum differences or another mathematical property.

FAQ

How do I efficiently count partitions with even sum differences?

You can use prefix sums combined with parity tracking to efficiently determine whether the sum difference of subarrays at each partition is even.

What is the best time complexity for solving 'Count Partitions with Even Sum Difference'?

The optimal solution runs in O(n) time, where n is the length of the array, by utilizing prefix sums and tracking parity during iteration.

Can I solve this problem using brute force?

Brute force is possible but inefficient, with a time complexity of O(n^2). It's best to leverage prefix sums and parity checks for a more optimal approach.

What pattern is involved in 'Count Partitions with Even Sum Difference'?

The problem uses the Array plus Math pattern, specifically applying prefix sums and parity logic to determine valid partitions.

How does GhostInterview help with array and math problems?

GhostInterview guides you through efficient strategies for solving array and math problems, including optimal solutions and key insights like prefix sums and parity.

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Solution

Solution 1: Prefix Sum

We use two variables $l$ and $r$ to represent the sum of the left subarray and the right subarray, respectively. Initially, $l = 0$ and $r = \sum_{i=0}^{n-1} \textit{nums}[i]$.

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class Solution:
    def countPartitions(self, nums: List[int]) -> int:
        l, r = 0, sum(nums)
        ans = 0
        for x in nums[:-1]:
            l += x
            r -= x
            ans += (l - r) % 2 == 0
        return ans

Continue Practicing

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