Pipeline Grid Validator 39 — Problem Statement & Solution Guide
Problem Description
You are given the head of a singly linked list where each node stores a positive integer value. Also given an integer maxGroups. Determine whether it is possible to split the list into at most maxGroups contiguous groups such that the sum of the values in every group is a prime number. The groups must cover the entire list without overlap and must preserve the original order of nodes. Return true if such a partition exists, otherwise return false. The solution should explore possible group boundaries using a recursive backtracking approach.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Grid Validator 39"
WHY DOES IT MATTER?
Partition‑DP is a core technique for grouping constraints on linear data structures.
OPTIMIZATION CHALLENGE
Reducing exponential cut‑selection to O(n²) by reusing prefix sums and pre‑computed primality.
REAL-WORLD CONNECTION
It mirrors batch processing in pipelines where each batch must satisfy a quality metric (e.g., checksum being prime).
Convert the linked list to an array once, then reuse that immutable structure for fast random access during DP.
COMPLEXITY AT A GLANCE
O(n²)O(n)Core Theory — Why This Approach?
The problem reduces to a partition‑DP on a linear structure: we need to decide cut points so that each segment’s sum is prime and the total number of segments does not exceed maxGroups. By converting the linked list to an array of prefix sums we can test any interval in O(1) and use a DP state dp[i] = minimum groups needed to partition the first i nodes with prime‑sum segments, updating dp[j] = min(dp[j], dp[i] + 1) whenever sum(i+1..j) is prime. Naïve recursion tries every subset of cut positions, leading to O(2^n) blow‑up, which is infeasible for n > 10^4. The optimal paradigm leverages prefix sums, a fast primality sieve up to the total sum, and DP to collapse the exponential search to polynomial time, guaranteeing a solution within the given group limit or proving impossibility.
Interview Questions on This Problem
Q1How can you test whether a segment sum is prime efficiently for many queries?
Pre‑compute a boolean sieve (e.g., Sieve of Eratosthenes) up to the total sum of the list. Then each segment sum lookup becomes O(1).
Q2Why is a greedy “take the longest prime‑sum segment” strategy incorrect?
Greedy may consume nodes that could form smaller prime‑sum groups later, increasing the total group count beyond maxGroups. The optimal solution requires exploring all feasible cut positions via DP.
Q3What DP state formulation solves this problem and how is it updated?
dp[i] stores the minimum groups needed to cover the first i nodes; initialize dp[0]=0. For each i, iterate j>i and if sum(i+1..j) is prime, set dp[j]=min(dp[j], dp[i]+1).
Examples
Input
head = [2, 3, 5, 7], maxGroups = 3
Output
true
Explanation: One feasible partition is: - Group 1: nodes 2 and 3 → sum = 5 (prime) - Group 2: node 5 → sum = 5 (prime) - Group 3: node 7 → sum = 7 (prime) All three groups satisfy the prime‑sum condition and the number of groups (3) does not exceed maxGroups, so the answer is true.
Input
head = [4, 6, 8], maxGroups = 2
Output
false
Explanation: All possible ways to split the list into at most two contiguous groups are: 1) [4] | [6,8] → sums 4 (not prime) and 14 (not prime) 2) [4,6] | [8] → sums 10 (not prime) and 8 (not prime) 3) [4,6,8] → sum 18 (not prime) Every grouping contains at least one non‑prime sum, therefore no valid partition exists and the answer is false.
Input
head = [2, 2, 3, 5], maxGroups = 2
Output
true
Explanation: A valid split is: - Group 1: nodes 2, 2, 3 → sum = 7 (prime) - Group 2: node 5 → sum = 5 (prime) Only two groups are used, which is within the allowed limit, and both sums are prime, so the answer is true.
Constraints
- 1 <= n <= 20 // n is the number of nodes in the linked list
- 1 <= node.val <= 100
- 1 <= maxGroups <= n
Optimal Approach & Strategy
Convert to prefix sums, pre‑sieve primes, and run DP that updates dp[j] from dp[i] whenever the interval sum is prime, achieving polynomial time.
Brute Force Approach
Recursively try every possible cut position, leading to exponential time as each node can be a cut or not.
Verified Code Solutions
function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
let firstLastSum = nums[0] + nums[nums.length - 1];
let secondSecondLastSum = nums[1] + nums[nums.length - 2];
return sum - firstLastSum - secondSecondLastSum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
int firstLastSum = nums[0] + nums[nums.size() - 1];
int secondSecondLastSum = nums[1] + nums[nums.size() - 2];
return sum - firstLastSum - secondSecondLastSum;
}class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
int firstLastSum = nums[0] + nums[nums.length - 1];
int secondSecondLastSum = nums[1] + nums[nums.length - 2];
return sum - firstLastSum - secondSecondLastSum;
}def solution(nums):
sum = sum(nums)
firstLastSum = nums[0] + nums[-1]
secondSecondLastSum = nums[1] + nums[-2]
return sum - firstLastSum - secondSecondLastSumfunction solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
let firstLastSum = nums[0] + nums[nums.length - 1];
let secondSecondLastSum = nums[1] + nums[nums.length - 2];
return sum - firstLastSum - secondSecondLastSum;
}Asked in Top Tech Interviews
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