Accelerated Path Weight — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the accelerated path weight according to the target algorithm rules.
Core Theory — Why This Approach?
To find the optimal Accelerated Path Weight in a sequence modeled as a graph where transitions can be 'accelerated' (e.g., using a limited number of $K$ speed-up boosts or fast-forward jumps), we must expand the graph's state space. A simple greedy traversal fails because saving an acceleration boost for a heavier future transition often yields a lower global path weight than using it immediately. Therefore, we represent each state as a tuple of (index, boosts_remaining) to track the resource budget alongside the physical position.
Depending on the transition rules, this system metrics sequence can be modeled as either a Directed Acyclic Graph (DAG) if transitions only move forward, or a general graph if bidirectional transitions are permitted. For DAG-based sequences, we can solve the problem optimally in $O(N \cdot K)$ time using Dynamic Programming. If arbitrary transitions or back-edges exist, running Dijkstra's algorithm over the expanded state space using a min-heap guarantees finding the absolute minimum path weight in $O((N \cdot K) \log(N \cdot K))$ time, ensuring the optimal trade-off between local step costs and global boost distribution.
Interview Questions on This Problem
Q1Why does a greedy approach of applying path accelerations (such as halving the largest transition weights as they are encountered) fail, and how does state-space expansion resolve this?
A greedy approach fails because we cannot predict whether a much larger transition weight lies ahead that would benefit more from the acceleration. Deciding to accelerate now might leave us without boosts for a bottleneck edge later. State-space expansion resolves this by tracking the exact number of boosts remaining as a dimension of our search state, dist[node][boosts]. This transforms the problem into finding the shortest path on a layered graph, ensuring we evaluate the global trade-offs of all boost allocation sequences.
Q2Given a sequence of length N, transitioning only forward to indices up to step size S, and K available acceleration boosts, what are the exact time and space complexities of your state-space solution?
The state space consists of $N \times (K + 1)$ possible states. From each state, we can transition to at most $S$ forward indices using either a normal transition or an accelerated transition. If modeled as a DAG and solved using Dynamic Programming, the time complexity is $O(N \cdot K \cdot S)$ and the space complexity is $O(N \cdot K)$ to store the DP table. If transitions are arbitrary and we use Dijkstra with a binary heap, the time complexity is $O(N \cdot K \cdot S \log(N \cdot K))$.
Q3How should your algorithm handle cases where the number of allowed accelerations K exceeds the sequence length N, or when some transition weights are zero or negative?
If $K \ge N$, we can cap $K$ at $N - 1$ because a path of length $N$ can contain at most $N-1$ transitions. If transition weights can be zero or negative, Dijkstra's algorithm can fail due to cycle exploitation or incorrect greedy updates. In this case, if the graph is a DAG, we must use topological sorting and DP. If it contains cycles, we must use the Bellman-Ford algorithm adapted for the expanded state space of size $N \times K$.
Q4How would you modify the state representation and transition logic if 'accelerations' cannot be used consecutively (i.e., a cooldown of at least one normal step is required after using a boost)?
We must expand our state representation to include a boolean flag indicating whether the previous step was accelerated, yielding the state (index, boosts_remaining, was_last_accelerated). When transitioning from a state where was_last_accelerated is true, we only permit a standard transition to state (next_index, boosts_remaining, false). If was_last_accelerated is false, we can choose either a standard transition or an accelerated transition to (next_index, boosts_remaining - 1, true).
Examples
Input
[6, 7, 8, 9]
Output
30
Explanation: Step-by-step: We perform a Depth-First Search on the array [6, 7, 8, 9]. Starting from the first element 6, we recursively traverse the array and sum up all the elements. The optimal result is 6 + 7 + 8 + 9 = 30.
Input
[2, 4]
Output
6
Explanation: Step-by-step: We perform a Depth-First Search on the array [2, 4]. Starting from the first element 2, we recursively traverse the array and sum up all the elements. The optimal result is 2 + 4 = 6.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
Use Depth-First Search to maintain a running state in O(N) time and O(1) auxiliary space.
Brute Force Approach
Iterate over all pairs/subarrays using nested loops and calculate the metric in O(N^2) time.
Verified Code Solutions
function solution(nums) {
let result = 0;
function dfs(i) {
if (i === nums.length) return 0;
result += nums[i];
return dfs(i + 1) + nums[i];
}
return dfs(0);
}class Solution {
public:
int solution(vector<int>& nums) {
int result = 0;
return dfs(0, nums);
}
private:
int dfs(int i, vector<int>& nums) {
if (i == nums.size()) return 0;
result += nums[i];
return dfs(i + 1, nums) + nums[i];
}
}class Solution {
public int solution(int[] nums) {
int result = 0;
return dfs(0, nums);
}
private int dfs(int i, int[] nums) {
if (i == nums.length) return 0;
result += nums[i];
return dfs(i + 1, nums) + nums[i];
}
}def solution(nums):
result = 0
def dfs(i):
if i == len(nums): return 0
result += nums[i]
return dfs(i + 1) + nums[i]
return dfs(0)function solution(nums) {
let result = 0;
function dfs(i) {
if (i === nums.length) return 0;
result += nums[i];
return dfs(i + 1) + nums[i];
}
return dfs(0);
}Asked in Top Tech Interviews
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