LeetCode #3432 — EASY

Count Partitions with Even Sum Difference

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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 such that:

Left subarray contains indices [0, i].
Right subarray contains indices [i + 1, n - 1].

Return the number of partitions where the difference between the sum of the left and right subarrays is even.

Example 1:

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

Output: 4

Explanation:

The 4 partitions are:

[10], [10, 3, 7, 6] with a sum difference of 10 - 26 = -16, which is even.
[10, 10], [3, 7, 6] with a sum difference of 20 - 16 = 4, which is even.
[10, 10, 3], [7, 6] with a sum difference of 23 - 13 = 10, which is even.
[10, 10, 3, 7], [6] with a sum difference of 30 - 6 = 24, which is even.

Example 2:

Input: nums = [1,2,2]

Output: 0

Explanation:

No partition results in an even sum difference.

Example 3:

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

Output: 3

Explanation:

All partitions result in an even sum difference.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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 such that: Left subarray contains indices [0, i]. Right subarray contains indices [i + 1, n - 1]. Return the number of partitions where the difference between the sum of the left and right subarrays is even.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[10,10,3,7,6]

Example 2

[1,2,2]

Example 3

[2,4,6,8]

Step 02

Core Insight

What unlocks the optimal approach

If the parity of the sum is even, the partition is valid; otherwise, there is no partition.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3432: Count Partitions with Even Sum Difference
class Solution {
    public int countPartitions(int[] nums) {
        int l = 0, r = 0;
        for (int x : nums) {
            r += x;
        }
        int ans = 0;
        for (int i = 0; i < nums.length - 1; ++i) {
            l += nums[i];
            r -= nums[i];
            if ((l - r) % 2 == 0) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3432: Count Partitions with Even Sum Difference
func countPartitions(nums []int) (ans int) {
	l, r := 0, 0
	for _, x := range nums {
		r += x
	}
	for _, x := range nums[:len(nums)-1] {
		l += x
		r -= x
		if (l-r)%2 == 0 {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3432: Count Partitions with Even Sum Difference
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

// Accepted solution for LeetCode #3432: Count Partitions with Even Sum Difference
impl Solution {
    pub fn count_partitions(nums: Vec<i32>) -> i32 {
        let mut l: i64 = 0;
        let mut r: i64 = nums.iter().map(|&x| x as i64).sum();
        let mut ans: i32 = 0;

        for &x in nums[..nums.len() - 1].iter() {
            l += x as i64;
            r -= x as i64;
            if (l - r) % 2 == 0 {
                ans += 1;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3432: Count Partitions with Even Sum Difference
function countPartitions(nums: number[]): number {
    let l = 0;
    let r = nums.reduce((a, b) => a + b, 0);
    let ans = 0;
    for (const x of nums.slice(0, -1)) {
        l += x;
        r -= x;
        ans += (l - r) % 2 === 0 ? 1 : 0;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.