Network Node Partition 21 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the partitioning of a linear network topology represented by an array of node weights. The goal is to divide the sequence into exactly K contiguous segments such that the maximum sum of any single segment is minimized. This metric represents the peak load on the most heavily utilized partition, and minimizing it ensures balanced traffic distribution across the network infrastructure.
Given an array of positive integers representing node weights and an integer K, determine the minimum possible value of the maximum subarray sum among all valid partitions into K contiguous parts. A valid partition must cover all elements without overlap or omission.
Input: An array 'weights' of length N and an integer 'K' representing the number of partitions.
Output: Return the minimum possible maximum sum of any partition.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Partition 21"
WHY DOES IT MATTER?
Binary search on the answer transforms a combinatorial partitioning problem into a series of linear feasibility checks, dramatically reducing complexity from exponential to logarithmic in the answer space.
OPTIMIZATION CHALLENGE
The key insight is that feasibility is monotonic: if a maximum sum works, any larger sum will also work. This allows binary search to prune the search space efficiently.
REAL-WORLD CONNECTION
In distributed file systems, data is split into shards; this algorithm models how to partition files to minimize the largest shard size, ensuring balanced storage and load.
When implementing, always use 64-bit integers for sums to avoid overflow, and carefully handle the edge case where K equals the array length—each element becomes its own segment.
COMPLEXITY AT A GLANCE
O(N log S)O(1)Core Theory — Why This Approach?
The problem asks for the minimal possible maximum segment sum when partitioning an array into exactly K contiguous parts. A naive approach would enumerate all ways to cut the array, which grows exponentially with N and is infeasible for large inputs. The optimal solution observes that the answer lies between the largest single element (a lower bound) and the total sum of the array (an upper bound). By performing a binary search over this range and using a greedy check to see if the array can be split into at most K parts without exceeding a candidate maximum sum, we can find the minimal feasible maximum in O(N log S) time, where S is the total sum. This paradigm—binary search on the answer combined with a linear feasibility test—is a classic pattern for partitioning and scheduling problems.
Interview Questions on This Problem
Q1How would you solve the 'Split Array Largest Sum' problem in O(N log S) time, and why is binary search on the answer appropriate?
I would binary search on the maximum allowed segment sum between max(A) and sum(A). For each mid, I would greedily count how many segments are needed if no segment exceeds mid. If the count <= K, mid is feasible and I search left; otherwise I search right. Binary search works because feasibility is monotonic: if a sum is feasible, any larger sum is also feasible.
Q2In a fintech platform, you need to balance transaction batches across servers. How does the partitioning algorithm help, and what edge cases must you guard against?
The algorithm ensures the heaviest batch is minimized, preventing any server from becoming a bottleneck. Edge cases include when K equals the number of transactions (each gets its own batch), when K is 1 (all transactions in one batch), and when transaction sizes vary widely, which can cause integer overflow if not using 64-bit types.
Q3During a startup interview, you’re asked to explain how you would extend this algorithm to handle dynamic updates to the array. What data structure would you use and why?
I would use a segment tree or binary indexed tree to maintain prefix sums and support point updates in O(log N). Then, for each query, I could recompute the binary search bounds quickly and run the greedy check in O(N) or use a more advanced structure to reduce it further.
Examples
Input
weights = [4, 2, 1, 7, 3], K = 2
Output
10
Explanation: We need to split the array into 2 parts. Possible splits: 1. [4,2,1] and [7,3] -> Sums: 7, 10 -> Max: 10 2. [4,2] and [1,7,3] -> Sums: 6, 11 -> Max: 11 3. [4] and [2,1,7,3] -> Sums: 4, 13 -> Max: 13 The minimum of the maximums is 10.
Input
weights = [1, 2, 3, 4, 5], K = 3
Output
6
Explanation: We need to split into 3 parts. 1. [1,2], [3], [4,5] -> Sums: 3, 3, 9 -> Max: 9 2. [1,2,3], [4], [5] -> Sums: 6, 4, 5 -> Max: 6 3. [1], [2,3], [4,5] -> Sums: 1, 5, 9 -> Max: 9 4. [1,2], [3,4], [5] -> Sums: 3, 7, 5 -> Max: 7 The minimum possible maximum sum is 6.
Input
weights = [10, 10, 10, 10], K = 4
Output
10
Explanation: Since K equals the length of the array, each element forms its own partition. The sums are [10, 10, 10, 10]. The maximum sum is 10.
Input
weights = [1, 1, 1, 1, 1], K = 1
Output
5
Explanation: Since K is 1, the entire array is a single partition. The sum is 1+1+1+1+1 = 5. The maximum sum is 5.
Constraints
- 1 <= weights.length <= 10^5
- 1 <= weights[i] <= 10^4
- 1 <= K <= weights.length
- The sum of weights[i] will not exceed 10^9
Optimal Approach & Strategy
Binary search the answer between max element and total sum, using a greedy linear scan to check if a candidate maximum allows at most K segments. This yields O(N log S) time and O(1) extra space.
Brute Force Approach
Enumerate all ways to place K-1 cuts in the array and compute the maximum segment sum for each partition, then take the minimum. This is exponential in N and impractical for large arrays.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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