Shortest Path Cost Protocol 6 — Problem Statement & Solution Guide
Problem Description
You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the shortest path cost using the **Subsequence Verification** methodology.
Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shortest Path Cost Protocol 6"
WHY DOES IT MATTER?
The shortest‑uncommon‑subsequence pattern appears in code‑obfuscation, DNA sequence analysis, and security protocols where you need a minimal distinguishing token. Mastering it demonstrates the ability to convert combinatorial explosion into tractable DP.
OPTIMIZATION CHALLENGE
The key insight is pre‑computing the next occurrence of every character in the target string, turning a linear scan inside the DP into O(1) look‑ups. This reduces the naive O(N·M·Alphabet) to O(N·M).
REAL-WORLD CONNECTION
Think of a network packet header: you want the smallest header fragment that guarantees a router can uniquely identify a flow among all others. The header fragment is a subsequence of the full packet, and the verification step is checking that no other flow shares it.
When coding, build the next‑pos table first, then fill the DP bottom‑up. Use sentinel values (e.g., INF = N+1) to represent impossible states, and break early if you already hit length 1.
COMPLEXITY AT A GLANCE
O(N·M)O(N·M) or O(M) with rolling arrayCore Theory — Why This Approach?
The problem reduces to finding the minimum‑length subsequence of the source string S (length N) that satisfies a verification predicate on a target string T (length M). A subsequence is formed by deleting zero or more characters without changing the order of the remaining characters. The naive way—enumerating all 2^N subsequences and checking each against T—blows up exponentially and cannot handle N up to 10^5. The optimal paradigm uses dynamic programming: dp[i][j] stores the length of the shortest subsequence of S[i…] that is NOT a subsequence of T[j…]. If S[i] does not appear in T[j…], the answer is 1 (just S[i]). Otherwise we either skip S[i] or take it and jump to the next occurrence of S[i] in T. The recurrence dp[i][j] = min(dp[i+1][j], 1 + dp[i+1][nextPos+1]) yields a solution in O(N·M) time. This DP exploits overlapping sub‑problems and optimal substructure, turning an exponential search into a polynomial one.
Interview Questions on This Problem
Q1How would you find the length of the shortest subsequence of string A that is NOT a subsequence of string B?
Use a DP table dp[i][j] where i indexes A and j indexes B. Pre‑compute next occurrence positions of each character in B. The recurrence is dp[i][j] = 1 if A[i] does not appear in B[j…]; otherwise dp[i][j] = min(dp[i+1][j], 1 + dp[i+1][nextPos+1]). The answer is dp[0][0].
Q2Why does a greedy approach (always picking the earliest possible character) fail for this problem?
Greedy picks the first character that appears in B, but the optimal solution may need to skip that character to achieve a shorter overall subsequence. The DP captures the trade‑off between skipping and taking a character, which greedy cannot.
Q3Can the DP be reduced to O(M) space? If so, how?
Yes. Process A from the end to the start and keep only two rows: the current row and the row for i+1. Since dp[i][j] depends only on dp[i+1][*], we can overwrite the previous row, achieving O(M) auxiliary space.
Examples
Input
[2, 12, 7, 17]
Output
38
Explanation: To calculate the sum of the array [2, 12, 7, 17], we iterate over each element and add it to a running total. The correct sum is 2 + 12 + 7 + 17 = 38.
Input
[2, 12]
Output
14
Explanation: To calculate the sum of the array [2, 12], we iterate over each element and add it to a running total. The correct sum 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 DP with a pre‑computed next‑position table to compute the minimal length in O(N·M) time, optionally reducing space to O(M).
Brute Force Approach
Generate every subsequence of the source string (2^N possibilities) and for each check if it appears as a subsequence in the target string.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.