Matrix Stream Partition 16 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and stream metrics, construct an optimal algorithm to evaluate and compute the target partition value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Partition 16"
WHY DOES IT MATTER?
Balancing load across partitions minimizes worst‑case latency in streaming pipelines.
OPTIMIZATION CHALLENGE
Reducing the naive O(N K) scan to O(N log K) by using a heap dramatically cuts runtime for large streams.
REAL-WORLD CONNECTION
Think of distributing incoming video frames across K encoding workers to avoid any single worker becoming a bottleneck.
Pre‑allocate the heap and reuse it across binary‑search iterations to avoid repeated allocations and improve cache locality.
COMPLEXITY AT A GLANCE
O(N log K log range)O(K)Core Theory — Why This Approach?
The Matrix Stream Partition problem asks for the optimal way to split a continuous stream of matrix elements into K contiguous partitions such that the maximum sum among partitions is minimized. A naive scan that recomputes prefix sums for every possible cut leads to O(N^2) time, which collapses under the massive N typical of streaming data. The optimal paradigm treats each partition boundary as a decision point and uses a binary search on the answer combined with a greedy feasibility check; the greedy check can be performed in linear time by accumulating elements until the current sum would exceed the candidate maximum, then starting a new partition. To accelerate the feasibility check when the stream is presented as a matrix (rows arriving one by one), a min‑heap can maintain the current column sums, allowing us to decide the next cut by always extending the column with the smallest accumulated sum, achieving O(N log K) overall.
Interview Questions on This Problem
Q1Why does a binary‑search‑on‑answer approach work for minimizing the maximum partition sum?
Because the feasibility predicate (can we partition with max sum ≤ X) is monotonic: if X works, any larger X also works. Binary search thus converges to the smallest feasible X in O(log range) iterations.
Q2How does a min‑heap help when the matrix is streamed row‑wise?
The heap stores the current sum of each column partition, letting us always extend the column with the smallest sum, which balances loads and respects contiguity. This yields O(log K) per element update instead of scanning all K columns.
Q3What is the time complexity of the greedy feasibility check when using a heap?
Each of the N elements triggers a heap pop‑push pair, costing O(log K), so the check runs in O(N log K). Combined with binary search it becomes O(N log K log range).
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 3
Output
60
Explanation: Step 1: Given the array [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and K = 3, we need to find the sum of the first K elements. Step 2: The first K elements are [10, 20, 30]. Step 3: The sum of these elements is 10 + 20 + 30 = 60.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 3
Output
6
Explanation: Step 1: Given the array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 3, we need to find the sum of the first K elements. Step 2: The first K elements are [1, 2, 3]. Step 3: The sum of these elements is 1 + 2 + 3 = 6.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Apply binary search on the answer and use a greedy feasibility test that leverages a min‑heap to maintain column sums, achieving O(N log K log range) time.
Brute Force Approach
Enumerate every possible set of K‑1 cut positions and compute the maximum block sum for each, which is O(N^K) and infeasible for large N.
Verified Code Solutions
function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
} else {
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k >= nums.size()) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
} else {
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}
};class Solution {
public int solution(int[] nums, int k) {
if (k >= nums.length) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
} else {
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}
}def solution(nums, k):
if k >= len(nums):
return sum(nums)
else:
return sum(nums[:k])
function solution(nums, k) {
if (k >= nums.length) {
return nums.reduce((a, b) => a + b, 0);
} else {
return nums.slice(0, k).reduce((a, b) => a + b, 0);
}
}Asked in Top Tech Interviews
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