Accelerated Subsequence Sum — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the maximum sum of a subsequence according to the target algorithm rules.
Core Theory — Why This Approach?
To solve the Accelerated Subsequence Sum problem within strict $O(N)$ time and $O(1)$ auxiliary space, we model the sequence as an implicit Directed Acyclic Graph (DAG) of state transitions. Each element in the sequence represents a state node, and the allowable decisions (e.g., including or excluding a metric) represent directed edges. Finding the maximum subsequence sum under specific rules is mathematically equivalent to finding the longest path in this state-transition DAG.
While a traditional Depth-First Search (DFS) on a graph yields $O(N)$ space due to the call stack, we can exploit the linear, forward-only dependency of our DAG. By utilizing tail-recursion or an iterative simulation of the DFS path, we propagate the running maximum sum as state parameters. This allows the compiler to optimize the stack frames (Tail Call Optimization) or allows us to maintain the active path using simple scalar variables. Consequently, we traverse the state graph in a single linear pass, achieving the optimal $O(1)$ auxiliary space complexity.
Interview Questions on This Problem
Q1How do we model this linear subsequence sum problem as a Graph DFS problem, and what is the core state-transition trick?
We represent the decision space as an implicit Directed Acyclic Graph (DAG) where each index $i$ corresponds to a node. The edges from node $i$ represent the valid transitions to subsequent indices based on our selection rules. The core trick is that because the graph transitions only move forward (from $i$ to $i+1$), the DFS has a single-branch active path. We can propagate the accumulated subsequence sum as a state variable along this path, transforming the longest-path DAG search into a continuous, linear state update.
Q2A standard DFS traversal on a graph of size $N$ requires $O(N)$ auxiliary space. How does your DFS implementation achieve $O(1)$ auxiliary space?
Standard DFS uses $O(N)$ space due to the recursion stack. We achieve $O(1)$ auxiliary space by implementing the DFS using tail recursion, where the recursive call is the absolute final operation and the accumulated state is passed forward. This allows compilers to perform Tail Call Optimization (TCO), reusing the same stack frame. Alternatively, we can iteratively simulate the DFS traversal by traversing the implicit graph using a loop and a fixed number of state variables to track the maximum path sum, eliminating stack overhead entirely.
Q3How should the DFS state transition behave if the sequence consists entirely of negative system metrics, assuming we must select a non-empty subsequence?
If all metrics are negative and an empty subsequence is disallowed, resetting our running state to 0 on a negative transition is incorrect. The DFS state transition must track the maximum single element encountered during the traversal. If the running sum tracker drops below the value of the current state-node, the DFS transition must prioritize the single maximum metric node, preventing the accumulator from masking the optimal 'least-negative' single-element path.
Q4How would you modify the DFS state graph if we add a constraint that no two adjacent metrics can be included in the subsequence sum?
We expand the state space of our implicit graph. At each step $i$, instead of a single node, we have two states: State A (subsequence ends at $i$ and includes metric $i$) and State B (subsequence ends at $i$ but excludes metric $i$). The DFS transitions from step $i-1$ to $i$ would update State A using State B's previous value plus the current metric, and State B using the maximum of State A and State B's previous values. By keeping only these two state variables active during the linear DFS traversal, we maintain $O(N)$ time and $O(1)$ auxiliary space.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we find the maximum sum of a subsequence by considering all possible subsequences: [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]. The maximum sum is 15, which is the sum of the subsequence [1, 2, 3, 4, 5].
Input
[5, 4, 3, 2, 1]
Output
15
Explanation: Step-by-step: with input [5, 4, 3, 2, 1], we find the maximum sum of a subsequence by considering all possible subsequences: [5], [5, 4], [5, 4, 3], [5, 4, 3, 2], [5, 4, 3, 2, 1]. The maximum sum is 15, which is the sum of the subsequence [5, 4, 3, 2, 1].
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 maxSum = -Infinity;
let currentSum = 0;
for (let num of nums) {
currentSum = Math.max(num, currentSum + num);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public:
int solution(vector<int>& nums) {
int maxSum = INT_MIN;
int currentSum = 0;
for (int num : nums) {
currentSum = max(num, currentSum + num);
maxSum = max(maxSum, currentSum);
}
return maxSum;
}
};class Solution {
public int solution(int[] nums) {
int maxSum = Integer.MIN_VALUE;
int currentSum = 0;
for (int num : nums) {
currentSum = Math.max(num, currentSum + num);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}def solution(nums):
max_sum = float('-inf')
current_sum = 0
for num in nums:
current_sum = max(num, current_sum + num)
max_sum = max(max_sum, current_sum)
return max_sumfunction solution(nums) {
let maxSum = -Infinity;
let currentSum = 0;
for (let num of nums) {
currentSum = Math.max(num, currentSum + num);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}Asked in Top Tech Interviews
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