LeetCode Entry

3432. Count Partitions with Even Sum Difference

05.12.2025 easy 2025 kotlin rust

Count Left sum - Right sum % 2

3432. Count Partitions with Even Sum Difference easy blog post substack youtube

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Problem TLDR

Count Left sum - Right sum % 2 #easy

Intuition

Just brute-force.

Approach

• a more interesting solution: A-B %2 == 0 only if both odd or both even.
• step i+0, […….even_sum][……….even_sum]
• step i+1, […….even_sum,odd][……….even_sum-odd]
• step i+1, […….even_sum,even][……….even_sum-even]
• basically, the even-ness will be the same for every partition
• same proof for odd-ness

Complexity

• Time complexity: \(O(n^2)\), or O(n) for clever solution

• Space complexity: \(O(1)\)

Code

// 11ms
    fun countPartitions(n: IntArray) =
        (n.size-1)* (1-n.sum()%2)

// 0ms
    pub fn count_partitions(n: Vec<i32>) -> i32 {
       (n.len()as i32-1)*(1-n.iter().sum::<i32>()%2)
    }