Network Protocol Partition 10 — Problem Statement & Solution Guide
Problem Description
Given an array nums of N integers that represent sequential network‑protocol metrics, you may choose any split position i (0 ≤ i ≤ N) and divide the array into a left segment nums[0..i‑1] and a right segment nums[i..N‑1]. Let S_left be the sum of the left segment (0 if the segment is empty) and S_right be the sum of the right segment. Your goal is to select a split that minimizes the absolute difference |S_left − S_right|. Output this minimum possible difference.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Partition 10"
WHY DOES IT MATTER?
Balancing prefix and suffix sums is a fundamental technique for equilibrium and partition problems.
OPTIMIZATION CHALLENGE
The key is reducing repeated sum calculations from O(N^2) to O(N) by using a rolling total.
REAL-WORLD CONNECTION
It mirrors load‑balancing traffic across two network paths to minimize latency disparity.
Initialize total sum once, then update left sum in‑place; avoid recomputing right sum inside the loop.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to finding an index i that balances the prefix sum and suffix sum of an array, a classic prefix‑suffix partition task. A naive O(N^2) scan recomputes sums for each split, which explodes for N up to 10^5, while the optimal O(N) solution leverages a single pass with running totals, updating the absolute difference in constant time per element.
By pre‑computing the total sum and iterating left‑to‑right, we maintain S_left incrementally and derive S_right as total‑S_left, allowing immediate evaluation of the split metric. This paradigm—prefix sum with rolling computation—is a staple for array partition, sliding window, and equilibrium index problems, delivering linear scalability and minimal memory overhead.
Interview Questions on This Problem
Q1How would you compute the optimal split index in a single pass?
Maintain a running left sum while iterating, compute right sum as total‑left, and track the minimum absolute difference.
Q2Why is a prefix‑sum array unnecessary for this problem?
Because the left sum can be updated incrementally, eliminating the need for extra O(N) storage.
Q3What edge case must you handle when the array is empty or has one element?
Both splits (i=0 and i=N) are valid; the algorithm should correctly treat empty segments as sum 0.
Examples
Input
4 4 1 2 3
Output
0
Explanation: Total sum = 10, half of total = 5. Prefix sums are 4, 5, 7, 10. The split after the second element gives S_left = 5 and S_right = 5, so the absolute difference is |5‑5| = 0, which is the smallest achievable.
Input
5 7 3 2 5 8
Output
1
Explanation: Total sum = 25, half = 12.5. Prefix sums are 7, 10, 12, 17, 25. The split after the third element yields S_left = 12 and S_right = 13, giving |12‑13| = 1. No other split produces a smaller difference.
Input
5 1 -2 3 -4 5
Output
1
Explanation: Total sum = 3, half = 1.5. Prefix sums are 1, -1, 2, -2, 3. The split after the third element gives S_left = 2 and S_right = 1, so the difference is |2‑1| = 1, which is minimal.
Constraints
- 1 <= nums.length <= 100000
- -10^9 <= nums[i] <= 10^9
- The answer fits in a 64‑bit signed integer
- Time limit: O(N) per test case
- Memory limit: O(1) extra space besides the input array
Optimal Approach & Strategy
Compute total sum once, then scan once updating a running left sum and deriving right sum on the fly.
Brute Force Approach
Iterate every possible split, recompute left and right sums from scratch, and track the minimum difference.
Verified Code Solutions
function solution(nums, K) {
let partitionValue = 0;
for (let num of nums) {
if (num > K) {
partitionValue++;
}
}
return partitionValue;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int partitionValue = 0;
for (int num : nums) {
if (num > K) {
partitionValue++;
}
}
return partitionValue;
}
};class Solution {
public int solution(int[] nums, int K) {
int partitionValue = 0;
for (int num : nums) {
if (num > K) {
partitionValue++;
}
}
return partitionValue;
}
}def solution(nums, K):
partition_value = 0
for num in nums:
if num > K:
partition_value += 1
return partition_valuefunction solution(nums, K) {
let partitionValue = 0;
for (let num of nums) {
if (num > K) {
partitionValue++;
}
}
return partitionValue;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.