Accelerated Stream Minimum — Problem Statement & Solution Guide
Problem Description
Given an array of integers, find the accelerated stream minimum by calculating the minimum of the sum of the sequence and the minimum value in the sequence plus the sum of the rest of the sequence.
Core Theory — Why This Approach?
The 'Accelerated Stream Minimum' problem can be elegantly solved by modeling the decision-making process over the stream as a shortest-path problem on a state-space Directed Acyclic Graph (DAG). At each element in the sequence, we transition through states: Accumulating Sum (State 0), Identifying the Minimum (State 1), and Accumulating the Rest (State 2). By constructing a DAG where vertices represent these states at each index and weighted edges represent the elements' values, the problem reduces to finding the shortest path from the start state to the end state.
This graph-theoretic formulation is highly optimal because a DAG has a natural topological ordering (left-to-right processing of the stream). This allows us to find the shortest path in $O(V + E)$ time, which translates to $O(N)$ time and $O(1)$ auxiliary space if we only track the active states of the previous index. This bypasses the need for nested iterations or complex heap tracking, matching the performance of a greedy scan while providing a robust mathematical model that easily scales to more complex stream routing rules.
Interview Questions on This Problem
Q1How does modeling the stream as a 3-state Directed Acyclic Graph (DAG) simplify finding the accelerated minimum compared to a brute-force approach?
Brute-forcing requires selecting each element as the potential minimum and summing the rest, leading to an $O(N^2)$ time complexity. By modeling this as a DAG with three states per step—State 0 (summing before the minimum), State 1 (identifying the minimum), and State 2 (summing after the minimum)—we transform the problem into finding the shortest path in a DAG. Since the stream flows in one direction, the graph is acyclic, allowing us to find the optimal path in a single $O(N)$ pass using dynamic programming over the states.
Q2What are the exact time and space complexities of the DAG transition approach, and how can we optimize space to $O(1)$?
The state graph contains $3N$ vertices and approximately $4N$ edges. Thus, finding the shortest path takes $O(V + E) = O(N)$ time. While a naive implementation of the DP table or adjacency list takes $O(N)$ space, we can optimize space to $O(1)$ because transitions at step $i$ only depend on the state values at step $i-1$. We only need to maintain three variables representing the minimum path costs to reach State 0, State 1, and State 2 at the current step.
Q3How does your graph-based solution handle negative integers in the stream, and why does the standard DAG shortest path algorithm remain valid?
Negative integers are naturally handled because the state transitions are acyclic. While Dijkstra's algorithm fails on general graphs with negative edge weights, a DAG shortest path algorithm relies on topological sorting. Since our transitions only move forward in the stream (from index $i$ to $i+1$), no cycles can exist. Thus, negative weights do not create infinite negative cycles, and the topological order DP guarantees correctness.
Q4If the problem is modified such that we can select up to $k$ elements to be excluded (accelerated) from the standard sum, how does the state graph scale?
We would scale the state-space of our DAG by increasing the number of tracking layers. Instead of 3 states, we would define $2k + 1$ states, where each layer represents the number of elements excluded so far (from $0$ to $k$). Transitions either keep us in the current layer (adding the element to the sum) or move us to the next layer (excluding the element/transitioning with cost 0). The time complexity would scale to $O(N \cdot k)$ and space complexity to $O(k)$ to keep track of the previous step's state layer values.
Examples
Input
[5, 8, 7, 6]
Output
26
Explanation: Step-by-step: with input [5, 8, 7, 6], we calculate the sum of the sequence (5 + 8 + 7 + 6 = 26) and the minimum value in the sequence (5) plus the sum of the rest of the sequence (8 + 7 + 6 = 21). Then, we find the minimum of these two values: min(26, 5 + 21) = min(26, 26) = 26.
Input
[10, 10]
Output
10
Explanation: Step-by-step: with input [10, 10], we calculate the sum of the sequence (10 + 10 = 20) and the minimum value in the sequence (10) plus the sum of the rest of the sequence (10). Then, we find the minimum of these two values: min(20, 10 + 10) = min(20, 20) = 10.
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 sum = nums.reduce((a, b) => a + b, 0);
let min = Math.min(...nums);
let restSum = sum - min;
return Math.min(sum, min + restSum);
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
int min = INT_MAX;
for (int num : nums) {
sum += num;
min = std::min(min, num);
}
int restSum = sum - min;
return std::min(sum, min + restSum);
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
int min = Integer.MAX_VALUE;
for (int num : nums) {
sum += num;
min = Math.min(min, num);
}
int restSum = sum - min;
return Math.min(sum, min + restSum);
}
}def solution(nums):
sum_seq = sum(nums)
min_val = min(nums)
rest_sum = sum_seq - min_val
return min(sum_seq, min_val + rest_sum)function solution(nums) {
let sum = nums.reduce((a, b) => a + b, 0);
let min = Math.min(...nums);
let restSum = sum - min;
return Math.min(sum, min + restSum);
}Asked in Top Tech Interviews
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