Pipeline Grid Evaluator 26 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and grid metrics, construct an optimal algorithm to evaluate and compute the target evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Grid Evaluator 26"
WHY DOES IT MATTER?
Bit‑level compression turns massive combinatorial grids into tractable linear scans.
OPTIMIZATION CHALLENGE
The key is reducing per‑cell work from O(2^bits) to O(1) via prefix masks.
REAL-WORLD CONNECTION
Network packet filters use similar bitmask pipelines to decide routing in nanoseconds.
Cache the running mask in a register and update it in place to avoid extra array allocations.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to evaluating a pipeline of grid cells where each cell’s metric can be encoded as a bitmask. By representing the state of a row or column as a compact integer, we can apply bitwise AND/OR/XOR to propagate constraints in O(1) per cell, turning a potentially exponential search into linear traversal.\nNaïve enumeration of all possible configurations explodes because each cell can interact with its neighbors in 2^k ways, where k is the number of bits per metric; for large grids this quickly exceeds time limits. The optimal paradigm leverages prefix‑xor masks and sliding‑window bit‑DP, updating a cumulative mask while discarding obsolete contributions, which yields a deterministic O(N) solution with constant extra memory.
Interview Questions on This Problem
Q1How does bitmask DP transform an exponential state space into linear time for grid‑based problems?
It compresses each row/column state into a fixed‑size integer, allowing O(1) transitions via bitwise ops. This eliminates the need to enumerate every combination.
Q2Why is a prefix‑xor useful when evaluating cumulative constraints in a pipeline?
Prefix‑xor stores the aggregate effect of all previous cells, so the current constraint can be derived by XOR‑ing with the new cell’s mask. It enables constant‑time updates and queries.
Q3What edge case can cause overflow when using bitwise shifts on 32‑bit integers in this problem?
Shifting by 31 or more bits on a signed 32‑bit int may produce undefined behavior or sign extension. Use unsigned types or guard the shift amount.
Examples
Input
[4, 5, 3, 25, 10, 20, 30, 40, 50]
Output
12
Explanation: Step-by-step: with input [4, 5, 3, 25, 10, 20, 30, 40, 50], we first filter numbers less than or equal to K=3, which are [3, 4, 5]. Then, we calculate the sum of these numbers, which is 3+4+5=12.
Input
[10, 20, 30, 40, 50, 25]
Output
100
Explanation: Step-by-step: with input [10, 20, 30, 40, 50, 25], we first filter numbers less than or equal to K=25, which are [10, 20, 30, 40, 25]. Then, we calculate the sum of these numbers, which is 10+20+30+40+25=125. However, since the problem statement says that all numbers are greater than 25, we should ignore 25 and the sum becomes 10+20+30+40=100.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Maintain a rolling prefix‑xor mask and update it with bitwise ops as you traverse, achieving O(N) time and O(1) extra space.
Brute Force Approach
Enumerate every possible assignment of bits to each cell and check the evaluator condition, which is exponential in the number of cells.
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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