How do you solve Count Partitions with Even Sum Difference on LeetCode?

Count Partitions with Even Sum Difference (LeetCode #3432) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The difference between two sums is even if both sums are either even or odd.

What is the brute force solution for Count Partitions with Even Sum Difference?

The brute force approach for Count Partitions with Even Sum Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check every possible partition of the array and calculate the sums of the left and right subarrays. If the difference between these sums is even, we count that partition. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Partitions with Even Sum Difference?

Count Partitions with Even Sum Difference uses the following concepts: Array, Math, Prefix Sum. The recommended patterns to study are: Prefix Sum, Array.

What are the interview tips for Count Partitions with Even Sum Difference?

Always think about how you can optimize a brute-force solution. Be clear about the conditions that make a partition valid. Practice explaining your thought process as you code.

What are common mistakes when solving Count Partitions with Even Sum Difference?

Not considering the case where the sums are both odd or both even. Failing to correctly update the left and right sums during iteration.

#3432

Count Partitions with Even Sum Difference

Easy

ArrayMathPrefix SumPrefix SumArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of calculating sums for each partition, we can use a prefix sum to efficiently determine the sums of the left and right subarrays. We also note that the difference is even if both sums are either even or odd.

⚙️

Algorithm

6 steps

1Step 1: Calculate the total sum of the array.
2Step 2: Initialize a counter for valid partitions and a variable for the left sum.
3Step 3: Loop through the array from index 0 to n-2 (inclusive) to consider each partition point.
4Step 4: Update the left sum and calculate the right sum as total_sum - left_sum.
5Step 5: Check the parity of left_sum and right_sum. If they are both even or both odd, increment the counter.
6Step 6: Return the counter after checking all partitions.

solution.py15 lines

1# Full working Python code
2
3def count_partitions(nums):
4    total_sum = sum(nums)
5    count = 0
6    left_sum = 0
7    for i in range(len(nums) - 1):
8        left_sum += nums[i]
9        right_sum = total_sum - left_sum
10        if (left_sum % 2 == right_sum % 2):
11            count += 1
12    return count
13
14# Example usage
15print(count_partitions([10, 10, 3, 7, 6]))  # Output: 4

ℹ

Complexity note: This complexity is linear because we only pass through the array a couple of times: once to calculate the total sum and once to check each partition.

1The difference between two sums is even if both sums are either even or odd.
2Using prefix sums allows us to avoid recalculating sums for each partition.

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