Shortest Path Cost Engine 9 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the shortest path cost using the Subsequence Verification methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shortest Path Cost Engine 9"
WHY DOES IT MATTER?
Subsequence verification with cost aggregation appears in routing, workflow orchestration, and compliance checking where you must select a valid ordered subset of actions while respecting budget or latency constraints. Mastering this pattern equips engineers to solve a broad class of optimization‑verification problems efficiently.
OPTIMIZATION CHALLENGE
The key insight is recognizing optimal substructure and compressing the DP state to a single dimension by iterating the pattern index backwards, which reduces space from O(N·M) to O(M) without sacrificing correctness.
REAL-WORLD CONNECTION
Think of a packet traversing a microservice mesh: each service adds latency (cost) and must appear in a prescribed order (subsequence). Finding the minimal‑latency path that respects the required service chain mirrors this algorithm.
During an interview, implement the DP row as a simple list and update it in reverse; this avoids overwriting values you still need for the current iteration and keeps the code concise and bug‑free.
COMPLEXITY AT A GLANCE
O(N·M)O(M)Core Theory — Why This Approach?
The Shortest Path Cost Engine 9 problem can be modeled as a subsequence verification task on a string‑like dataset. Each element of the dataset encodes a constraint or a weight, and we must determine whether a target pattern (a subsequence of constraints) can be formed while minimizing the accumulated cost. A naive solution would enumerate all possible subsequences, leading to exponential time because the number of subsequences of a length‑N array is 2^N. The optimal paradigm leverages dynamic programming: we maintain a DP table where dp[i][j] stores the minimum cost to match the first j characters of the target pattern using the first i elements of the dataset. Transitioning from dp[i‑1][j] (skip current element) and dp[i‑1][j‑1] + cost(i) (use current element) yields an O(N·M) solution, where M is the pattern length. This approach collapses the exponential search space into a linear scan over the dataset while preserving optimality.
Why this works is rooted in optimal substructure: the cheapest way to achieve a partial match up to position j depends only on the cheapest ways to achieve matches up to j and j‑1 using earlier elements. By iterating once over the dataset and updating the DP row in reverse order, we also achieve O(M) auxiliary space, making the algorithm scalable to large N (up to 10^5 or more) typical in production systems. The method is analogous to the classic "minimum edit distance" or "longest common subsequence" DP, but with an added cost dimension that must be aggregated.
The DP formulation also enables early pruning: if a particular dp entry exceeds a known upper bound (e.g., a budget constraint), we can skip further updates for that state, further improving practical runtime. This blend of subsequence verification and cost minimization is the cornerstone of many real‑world routing, scheduling, and compliance engines.
Interview Questions on This Problem
Q1How would you adapt the DP solution if the dataset elements could be used multiple times (i.e., repetitions allowed) while still minimizing cost?
Allow repetitions by changing the transition to consider dp[i][j] = min(dp[i‑1][j], dp[i][j‑1] + cost(i)) where dp[i][j‑1] uses the same i index again, effectively turning the DP into an unbounded knapsack on the subsequence dimension.
Q2Explain how you would modify the algorithm to also return the actual subsequence achieving the minimum cost.
Maintain a predecessor pointer for each DP cell indicating whether the cell came from a skip or a take; after filling the table, backtrack from dp[N][M] following the pointers to reconstruct the chosen indices.
Q3A fintech platform needs to enforce a regulatory rule that the total cost must not exceed a threshold T. How can you incorporate this constraint into the DP without increasing asymptotic complexity?
During DP updates, cap each dp[i][j] at T+1 (or INF) and skip updates that would exceed T; this pruning does not change the O(N·M) loops but eliminates unnecessary work and directly yields whether a feasible solution exists.
Examples
Input
[2, 12, 7, 17]
Output
38
Explanation: Step-by-step: Given array [2, 12, 7, 17], we apply Subsequence Verification methodology to find the shortest path cost. The shortest path cost is the sum of array elements, which is 2 + 12 + 7 + 17 = 38.
Input
[2, 12]
Output
14
Explanation: Step-by-step: Given array [2, 12], we apply Subsequence Verification methodology to find the shortest path cost. The shortest path cost is the sum of array elements, which is 2 + 12 = 14.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Use a dynamic programming table that records the minimum cost to achieve each prefix of the target pattern while scanning the dataset once. Update the table in reverse to avoid overwriting needed values, achieving O(N·M) time and O(M) space.
Brute Force Approach
Enumerate every possible subsequence of the dataset and compute its total cost, keeping the minimum among those that match the target pattern. This requires exponential time and is infeasible for large N.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
if not nums:
return 0
return sum(nums)function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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