Leetcode 3432. Count Partitions with Even Sum Difference

Let's understand what we're being asked to do. We have an array like [2, 4, 6, 8] and need to count how many split indices exist where the sum difference between left and right subarrays is even.

For example, at split index i=1, the left subarray is [2] (sum=2) and the right subarray is [4,6,8] (sum=18). The difference is 18-2=16, which is even.

Key insight: Let's trace through all possible splits. At i=0: left=[] (sum=0), right=[2,4,6,8] (sum=20), difference=20 (even). At i=1: left=[2] (sum=2), right=[4,6,8] (sum=18), difference=16 (even). At i=2: left=[2,4] (sum=6), right=[6,8] (sum=14), difference=8 (even).

Important observation: The difference is rightSum - leftSum. But notice that rightSum + leftSum = totalSum (always). So rightSum - leftSum = totalSum - 2*leftSum. This means the difference is even if and only if the total sum is even!

Edge cases to consider: What if the array has only one element? (no valid splits, return 0). What if all elements are odd? What if the total sum is odd? (then ALL differences are odd, return 0).

The straightforward approach is to iterate through each possible split index from 0 to n-2. For each split, calculate the left subarray sum (from index 0 to i) and the right subarray sum (from index i+1 to n-1). Compute the absolute difference between these sums and check if it's even. Count how many splits satisfy this condition. This approach works but requires recalculating sums for each split, leading to O(n²) time complexity in the worst case, or O(n) with prefix sums.

def count_even_split_indices(nums):
    """
    Count split indices where left-right sum difference is even.
    Key insight: difference is even iff total sum is even.
    """
    n = len(nums)
    
    # Edge case: need at least 2 elements to split
    if n < 2:
        return 0
    
    # Calculate total sum to check if it's even
    total_sum = 0
    for num in nums:
        total_sum += num
    
    # If total sum is even, all splits have even difference
    if total_sum % 2 == 0:
        return n - 1
    
    # Brute force: check each split index
    count = 0
    
    for i in range(n - 1):
        # Calculate left subarray sum (0 to i)
        left_sum = 0
        for j in range(i + 1):
            left_sum += nums[j]
        
        # Calculate right subarray sum (i+1 to n-1)
        right_sum = 0
        for k in range(i + 1, n):
            right_sum += nums[k]
        
        # Check if difference is even
        difference = abs(left_sum - right_sum)
        if difference % 2 == 0:
            count += 1
    
    return count

The difference between right and left sums can be expressed as rightSum - leftSum. But we know that rightSum + leftSum = totalSum (the entire array sum).

Solving these two equations: rightSum = (totalSum + difference)/2 and leftSum = (totalSum - difference)/2. For the difference to be even, we need totalSum - 2*leftSum to be even.

This happens if and only if totalSum itself is even! The value of leftSum doesn't matter - only the parity of totalSum determines whether the difference is even.

This observation transforms a seemingly complex problem (checking each split's sum difference) into a trivial one: just check if the total sum is even!

If totalSum is even, then every single split (all n-1 of them) will have an even difference. If totalSum is odd, then no split will have an even difference. The answer is either n-1 or 0.

Think about splitting a pile of coins into two groups. The difference in coin counts between the groups depends on the total number of coins, not on where you split.

If you have an even total (say 10 coins), any split like 3-7, 4-6, or 5-5 will have an even difference (4, 2, 0 respectively). If you have an odd total (say 11 coins), any split like 3-8, 4-7, or 5-6 will have an odd difference (5, 3, 1 respectively).

The split position doesn't change the parity of the difference - only the total determines it!

The optimal approach leverages a mathematical insight: the difference between right and left sums is rightSum - leftSum = totalSum - 2*leftSum. For this difference to be even, totalSum must be even (since 2*leftSum is always even). Therefore, we only need to compute the total sum of the array once. If the total sum is even, return n-1 (all splits have even differences). If the total sum is odd, return 0 (no splits have even differences). This eliminates the need to check each split individually.

def count_even_split_differences(nums):
    """
    Count split indices where sum difference is even.
    Key insight: rightSum - leftSum = totalSum - 2*leftSum.
    Since 2*leftSum is always even, difference is even iff totalSum is even.
    """
    # Calculate total sum of array
    total_sum = 0
    for num in nums:
        total_sum += num
    
    # Get array length
    n = len(nums)
    
    # Check if total sum is even
    if total_sum % 2 == 0:
        # All splits have even difference
        result = n - 1
    else:
        # No splits have even difference
        result = 0
    
    return result

• Mathematical invariant recognition: When a problem asks about relationships between subarrays created by splits, look for global invariants that hold regardless of split position - here, the parity of (left_sum - right_sum) depends only on total_sum parity, not the split index itself.
• Parity algebra insight: The difference (left_sum - right_sum) is even if and only if total_sum is even, because left_sum + right_sum = total_sum, so left_sum - right_sum = 2*left_sum - total_sum, which has the same parity as total_sum. This eliminates the need to compute any actual sums.
• O(1) optimization through mathematical proof: Instead of iterating through all n-1 possible splits and computing sums (O(n) time), prove that all splits behave identically - if the condition holds for one split, it holds for all, reducing the solution to a single parity check.
• Common pitfall - unnecessary iteration: Many developers instinctively write loops to check each split index, wasting O(n) time when the answer is deterministic based on total sum parity alone. Always ask if iteration is truly necessary or if a mathematical property makes all iterations equivalent.
• Pattern extension to subarray problems: When problems involve all possible ways to partition an array, check if the answer depends on partition-independent properties (sum, parity, count) rather than specific partition positions - this often reduces complexity from O(n) or O(n²) to O(1).
• Real-world load balancing application: This parity principle applies to resource distribution systems where you need to split workloads evenly - knowing that even distribution is only possible when total resources have certain properties helps quickly determine feasibility without testing all split configurations.