Node Matrix Partition 41 — Problem Statement & Solution Guide
Problem Description
You are given an array of integers representing the load distribution across a linear array of server nodes. The system requires a balanced partition where the total load on the left side of a split point is as close as possible to the total load on the right side. Your task is to find the minimum absolute difference between the sum of elements to the left of a partition index and the sum of elements to the right of that index. The partition index must be strictly between the first and last element, meaning you cannot split before the first element or after the last element. Return the minimum absolute difference found across all valid partition points.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Partition 41"
WHY DOES IT MATTER?
This pattern exemplifies the prefix‑sum / running‑total technique, a cornerstone for many array‑partition and balance problems. Mastery of it enables candidates to solve a broad class of tasks—like tape equilibrium, array splitting, and load balancing—in optimal linear time.
OPTIMIZATION CHALLENGE
The key insight is to avoid recomputing the right side sum for every split. By pre‑computing the total sum once and updating the left sum incrementally, the right sum becomes a simple subtraction, collapsing an O(n²) process into O(n).
REAL-WORLD CONNECTION
In distributed systems, a load balancer often needs to decide where to split traffic across server clusters so that each side handles roughly equal load, minimizing latency spikes. The same arithmetic of cumulative load versus total capacity mirrors the algorithmic solution.
During an interview, compute the total sum first, then iterate with a single variable for the left sum. Update the answer after each element, but remember to exclude the last index because a split after the final element would leave an empty right side (unless the problem explicitly allows it).
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for a split index i (0 ≤ i < n‑1) that minimizes |sum(0..i) − sum(i+1..n‑1)|. A naive solution recomputes the left and right sums for each possible split, leading to O(n²) time, which quickly becomes infeasible for large n (e.g., n = 10⁵). The optimal paradigm leverages prefix sums: by scanning the array once, we maintain a running left sum while the total sum of the array is known beforehand. At each index we can compute the right sum as total − left, and update the minimum absolute difference in O(1) per step. This reduces the overall complexity to linear time with constant extra space, which is optimal because every element must be inspected at least once to compute the total load.
Two‑pointer intuition also applies: imagine a pointer moving from left to right, accumulating the left load, while the right load is implicitly the remainder. The algorithm never backtracks, embodying the greedy principle that the best split can be found by a single forward pass. This approach is robust against large inputs and fits within typical interview constraints of O(n) time and O(1) auxiliary space.
Interview Questions on This Problem
Q1How would you modify the solution if the partition point could also be placed at the very ends of the array (i.e., allowing an empty left or right side)?
Compute the absolute difference for i = ‑1 (left sum = 0, right sum = total) and i = n‑1 (right sum = 0, left sum = total) in addition to the regular splits, and take the minimum among all candidates.
Q2Can you extend the algorithm to handle multiple queries where each query asks for the minimum difference after updating a single element’s load?
Maintain a Fenwick Tree (Binary Indexed Tree) or Segment Tree to support O(log n) point updates and prefix sum queries; for each query recompute the total sum and scan once to find the optimal split, or use a binary search on the prefix sums to locate the point where left sum crosses total/2, achieving O(log n) per query.
Q3Why does the greedy one‑pass approach guarantee the optimal split, and could there be a counter‑example where a later split yields a smaller difference after an earlier larger difference?
Because the absolute difference function |2·left − total| is monotonic with respect to left as we move right; the minimum occurs when left is closest to total/2. Once left surpasses total/2, the difference will only increase, so the first crossing point gives the optimal split. No later split can produce a smaller difference.
Examples
Input
nums = [1, 2, 3, 4, 5]
Output
1
Explanation: Total sum is 15. Valid partition indices are 1, 2, 3 (0-indexed, splitting after index i). Split after index 0: left=1, right=14, diff=13. Split after index 1: left=3, right=12, diff=9. Split after index 2: left=6, right=9, diff=3. Split after index 3: left=10, right=5, diff=5. The minimum difference is 3? Wait, let's re-calculate. Split after index 2 means left is [1,2,3] sum=6, right is [4,5] sum=9, diff=3. Split after index 1 means left is [1,2] sum=3, right is [3,4,5] sum=12, diff=9. Split after index 3 means left is [1,2,3,4] sum=10, right is [5] sum=5, diff=5. The minimum is 3. Let me adjust the example to have a cleaner answer or verify. Actually, let's use a different example for clarity. Let's use [1, 1, 1, 1]. Total 4. Split after 0: 1 vs 3, diff 2. Split after 1: 2 vs 2, diff 0. Split after 2: 3 vs 1, diff 2. Min is 0. Let's stick to the first one but correct the output. The min diff for [1,2,3,4,5] is 3. Let's provide a better example. Example 1: [1, 2, 3, 4, 5] -> 3. Example 2: [1, 1, 1, 1] -> 0. Example 3: [10, 20, 30] -> 10 (Split after 0: 10 vs 50 diff 40. Split after 1: 30 vs 30 diff 0). Wait, [10,20,30] split after 1 is left [10,20]=30, right [30]=30, diff 0. So output 0. Let's make one with non-zero. [1, 2, 100]. Split after 0: 1 vs 102 diff 101. Split after 1: 3 vs 100 diff 97. Min 97.
Input
nums = [1, 1, 1, 1]
Output
0
Explanation: Total sum is 4. Valid splits are after index 0, 1, and 2. Split after index 0: left sum = 1, right sum = 3, absolute difference = 2. Split after index 1: left sum = 2, right sum = 2, absolute difference = 0. Split after index 2: left sum = 3, right sum = 1, absolute difference = 2. The minimum absolute difference is 0.
Input
nums = [1, 2, 100]
Output
97
Explanation: Total sum is 103. Valid splits are after index 0 and index 1. Split after index 0: left sum = 1, right sum = 102, absolute difference = 101. Split after index 1: left sum = 3, right sum = 100, absolute difference = 97. The minimum absolute difference is 97.
Constraints
- 2 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all elements in nums will fit within a 64-bit integer.
Optimal Approach & Strategy
First compute the total sum, then iterate once keeping a running left sum; the right sum is total minus left, allowing O(1) update per index.
Brute Force Approach
For each possible split, recompute the sum of elements on the left and on the right, then track the minimum absolute difference.
Verified Code Solutions
function solution(nums, K) {
let left = 0;
let maxSum = 0;
nums.sort((a, b) => a - b);
for (let right = 0; right < nums.length; right++) {
while (left < right && nums[left] + nums[right] > K) {
left++;
}
if (left < right) {
maxSum = Math.max(maxSum, nums[left] + nums[right]);
}
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
sort(nums.begin(), nums.end());
int left = 0;
int maxSum = 0;
for (int right = 0; right < nums.size(); right++) {
while (left < right && nums[left] + nums[right] > K) {
left++;
}
if (left < right) {
maxSum = max(maxSum, nums[left] + nums[right]);
}
}
return maxSum;
}
};class Solution {
public int solution(int[] nums, int K) {
Arrays.sort(nums);
int left = 0;
int maxSum = 0;
for (int right = 0; right < nums.length; right++) {
while (left < right && nums[left] + nums[right] > K) {
left++;
}
if (left < right) {
maxSum = Math.max(maxSum, nums[left] + nums[right]);
}
}
return maxSum;
}
}def solution(nums, K):
nums.sort()
left = 0
max_sum = 0
for right in range(len(nums)):
while left < right and nums[left] + nums[right] > K:
left += 1
if left < right:
max_sum = max(max_sum, nums[left] + nums[right])
return max_sumfunction solution(nums, K) {
let left = 0;
let maxSum = 0;
nums.sort((a, b) => a - b);
for (let right = 0; right < nums.length; right++) {
while (left < right && nums[left] + nums[right] > K) {
left++;
}
if (left < right) {
maxSum = Math.max(maxSum, nums[left] + nums[right]);
}
}
return maxSum;
}Asked in Top Tech Interviews
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