Matrix Stream Analyzer 21 — 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 analyzer value under given operational constraints. The optimal algorithm should return the maximum element if K is greater than or equal to the maximum element in the array, otherwise it should return the sum of all elements greater than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Stream Analyzer 21"
WHY DOES IT MATTER?
Single‑pass aggregation is essential for high‑throughput streaming workloads.
OPTIMIZATION CHALLENGE
Eliminate the extra O(n) pass or O(n log n) sort by merging both calculations into one loop.
REAL-WORLD CONNECTION
Think of monitoring sensor data where you need the peak reading and total consumption without buffering the entire history.
Initialize max to -∞ and sum to 0, then update both in the same iteration to keep the code tight and cache‑friendly.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to two fundamental operations on a linear collection: finding the global maximum and computing the total sum. Both can be achieved in a single pass using the classic scan technique, which maintains running aggregates while iterating, guaranteeing linear time regardless of input size.
Naïve solutions might first sort the array (O(n log n)) or compute the sum and maximum in separate traversals, doubling the constant factor. In large streams, these extra passes or sorting become prohibitive, especially when memory is constrained. The optimal paradigm leverages a single traversal with constant auxiliary space, aligning with the streaming model where each element is processed exactly once.
Interview Questions on This Problem
Q1How would you handle the case when the input array is empty?
Return a sentinel value such as 0 or raise an exception based on the specification. An empty stream has no maximum or sum, so the contract must define the expected behavior.
Q2Can you compute the result without storing the entire array in memory?
Yes, by maintaining only two variables: current max and running sum, you can process elements on the fly. This satisfies O(1) auxiliary space.
Q3What is the time complexity if you first sort the array before applying the rule?
Sorting costs O(n log n), which dominates the subsequent O(n) scan. Hence the overall complexity becomes O(n log n), which is suboptimal.
Examples
Input
[3, 4, 5, 2, 1], 3
Output
12
Explanation: Step-by-step: Given the input array [3, 4, 5, 2, 1] and K = 3, we first find the maximum element in the array, which is 5. Since 3 is less than 5, we then find all elements greater than or equal to 3, which are 3, 4, and 5. The sum of these elements is 3 + 4 + 5 = 12.
Input
[1, 2, 3, 4, 5], 5
Output
5
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and K = 5, we find that 5 is the maximum element in the array. Therefore, we return the maximum element, which is 5.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain max and sum variables during one linear traversal, achieving O(n) time and O(1) space.
Brute Force Approach
Sort the array to get the maximum, then compute the sum in a second pass, leading to O(n log n) time.
Verified Code Solutions
function solution(nums, k) {
let maxElement = Math.max(...nums);
if (k >= maxElement) {
return maxElement;
} else {
return nums.filter(x => x >= k).reduce((a, b) => a + b, 0);
}
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int maxElement = *max_element(nums.begin(), nums.end());
if (k >= maxElement) {
return maxElement;
} else {
int sum = 0;
for (int x : nums) {
if (x >= k) {
sum += x;
}
}
return sum;
}
}
};class Solution {
public int solution(int[] nums, int k) {
int maxElement = Arrays.stream(nums).max().getAsInt();
if (k >= maxElement) {
return maxElement;
} else {
return Arrays.stream(nums).filter(x -> x >= k).sum();
}
}
}def solution(nums, k):
max_element = max(nums)
if k >= max_element:
return max_element
else:
return sum(x for x in nums if x >= k)function solution(nums, k) {
let maxElement = Math.max(...nums);
if (k >= maxElement) {
return maxElement;
} else {
return nums.filter(x => x >= k).reduce((a, b) => a + b, 0);
}
}Asked in Top Tech Interviews
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