Optimizing Space Station Oxygen Levels — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Optimizing Space Station Oxygen Levels problem optimally.
O(m log m)O(m)Problem Description
Given an array of oxygen levels in different modules of a space station and a list of possible adjustments, where each adjustment is represented as [duration, module_start_index, adjustment_amount], find the maximum total oxygen level achievable by applying a sequence of adjustments. Each adjustment modifies the oxygen level of a contiguous subset of modules starting from module_start_index.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimizing Space Station Oxygen Levels"
WHY DOES IT MATTER?
Weighted interval scheduling captures a wide range of resource‑allocation scenarios where tasks compete for exclusive use of a limited span—think CPU time slots, ad placements, or, in this case, oxygen adjustments. Mastering this pattern equips engineers to turn combinatorial explosion into tractable DP.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the optimal substructure depends only on the last non‑conflicting interval. By pre‑computing this predecessor with binary search (or a segment tree), we avoid recomputing overlapping checks, collapsing an O(m^2) DP into O(m log m).
REAL-WORLD CONNECTION
In distributed systems, allocating bandwidth to non‑overlapping time windows for data replication mirrors the same constraints: each window (adjustment) yields a benefit (throughput) but cannot clash with another. Optimizing the schedule maximizes overall system performance.
When coding, first convert each adjustment to (start, end, weight) and sort by end. Write a helper that returns the index of the last interval ending before a given start using bisect_right; then fill a 1‑D DP array. Keep the DP array as long as needed—no need for a full 2‑D table.
COMPLEXITY AT A GLANCE
O(m log m)O(m)Core Theory — Why This Approach?
The problem can be modeled as a weighted interval scheduling task. Each adjustment defines an interval [l, r] where l = module_start_index and r = l + duration - 1, and contributes a weight equal to adjustment_amount * duration to the total oxygen level if it is selected. The goal is to pick a subset of non‑overlapping intervals that yields the maximum sum of weights, which is a classic dynamic‑programming paradigm. A naïve solution would enumerate every subset of adjustments (2^m) or try all possible orderings, leading to exponential time and quickly blowing up for large inputs (n, m up to 10^5). The optimal approach leverages the fact that intervals can be ordered by their right endpoint; for each interval we either take it (adding its weight to the best solution that ends before its start) or skip it. By pre‑computing the “previous compatible interval” with binary search, we achieve O(m log m) time, which is optimal for this class of problems. The DP can also be implemented with a segment tree or Fenwick tree to achieve the same complexity while handling dynamic queries, reinforcing the importance of prefix‑based data structures in interval‑selection problems.
Interview Questions on This Problem
Q1How would you modify the solution if adjustments could overlap but the total oxygen in any module cannot exceed a given capacity?
Introduce a segment tree with lazy propagation that tracks the current oxygen level per module. Process adjustments sorted by descending benefit, and before applying an adjustment, query the minimum remaining capacity over its range; apply only the feasible portion. This turns the problem into a greedy‑plus‑range‑query task, still O(m log n).
Q2Explain how the weighted interval scheduling DP can be transformed into a bottom‑up iterative solution using a Fenwick tree.
Sort intervals by end index. For each interval i, compute dp[i] = max(dp[i‑1], weight_i + query(maxIndex < start_i) ) where query retrieves the maximum dp value for intervals ending before start_i. The Fenwick tree stores dp values keyed by end positions, supporting O(log n) updates and prefix‑max queries, yielding O(m log n) overall.
Q3Why is binary search sufficient to find the previous compatible interval after sorting by end, and what is its time complexity?
Because the intervals are sorted by their right endpoint, the set of intervals that end before a given start forms a contiguous prefix. Binary searching this prefix for the largest end ≤ start‑1 returns the index of the previous compatible interval in O(log m) time.
Examples
Input
[10, 20, 30, 40, 50], [[3, 1, 8], [1, 2, 5]]
Output
188
Explanation: Step-by-step: Given oxygen levels [10, 20, 30, 40, 50] and adjustments [[3, 1, 8], [1, 2, 5]], we first apply the first adjustment [3, 1, 8] to modules 1-3 (10, 20, 30), resulting in oxygen levels [18, 28, 38, 40, 50]. Then, we apply the second adjustment [1, 2, 5] to modules 2-3 (28, 38), resulting in oxygen levels [18, 33, 38, 40, 50]. Finally, we increase the oxygen level of module 4 by 5 because the adjustment is applied to a contiguous subset of modules starting from module_start_index 2, so the correct oxygen levels are [18, 33, 38, 45, 50]. The maximum total oxygen level achievable is 188.
Input
[10, 20, 30, 40, 50], [[2, 2, 5], [1, 3, 10]]
Output
70
Explanation: Step-by-step: Given oxygen levels [10, 20, 30, 40, 50] and adjustments [[2, 2, 5], [1, 3, 10]], we first apply the first adjustment [2, 2, 5] to modules 2-3 (20, 30), resulting in oxygen levels [10, 25, 35, 40, 50]. Then, we apply the second adjustment [1, 3, 10] to modules 3 (35), resulting in oxygen levels [10, 25, 45, 40, 50]. The maximum total oxygen level achievable is 70.
Constraints
- 1 <= length of oxygenLevels <= 20
- 1 <= number of adjustments <= 10
- 1 <= duration <= 5
- 1 <= module_start_index <= length of oxygenLevels
- -10 <= adjustment_amount <= 10
Optimal Approach & Strategy
Sort adjustments by end index, compute for each the last non‑overlapping adjustment via binary search, and apply DP: dp[i] = max(dp[i‑1], weight_i + dp[prev_i]).
Brute Force Approach
Enumerate every subset of adjustments, check if the chosen intervals overlap, and compute the total oxygen gain; keep the maximum.
Code Solutions
/**
* @param {number[]} levels
* @param {number[][]} adjustments
* @return {number}
*/
var maxOxygenLevel = function(levels, adjustments) {
const n = levels.length;
const m = adjustments.length;
if (m === 0) return levels.reduce((a, b) => a + b, 0);
let maxSum = 0;
for (let mask = 0; mask < (1 << m); ++mask) {
const temp = [...levels];
for (let i = 0; i < m; ++i) {
if (mask & (1 << i)) {
const duration = adjustments[i][0];
const start = adjustments[i][1];
const amount = adjustments[i][2];
for (let j = start; j < n; ++j) {
temp[j] += amount;
}
}
}
const sum = temp.reduce((a, b) => a + b, 0);
maxSum = Math.max(maxSum, sum);
}
return maxSum;
};class Solution {
public:
int maxOxygenLevel(vector<int>& levels, vector<vector<int>>& adjustments) {
int n = levels.size();
int m = adjustments.size();
if (m == 0) return accumulate(levels.begin(), levels.end(), 0);
vector<int> dp(1 << m, 0);
int maxSum = 0;
for (int mask = 0; mask < (1 << m); ++mask) {
vector<int> temp = levels;
for (int i = 0; i < m; ++i) {
if (mask & (1 << i)) {
int duration = adjustments[i][0];
int start = adjustments[i][1];
int amount = adjustments[i][2];
for (int j = start; j < n; ++j) {
temp[j] += amount;
}
}
}
int sum = accumulate(temp.begin(), temp.end(), 0);
dp[mask] = sum;
maxSum = max(maxSum, sum);
}
return maxSum;
}
};class Solution {
public int maxOxygenLevel(int[] levels, int[][] adjustments) {
int n = levels.length;
int m = adjustments.length;
if (m == 0) {
int sum = 0;
for (int val : levels) sum += val;
return sum;
}
int maxSum = 0;
for (int mask = 0; mask < (1 << m); ++mask) {
int[] temp = levels.clone();
for (int i = 0; i < m; ++i) {
if ((mask & (1 << i)) != 0) {
int duration = adjustments[i][0];
int start = adjustments[i][1];
int amount = adjustments[i][2];
for (int j = start; j < n; ++j) {
temp[j] += amount;
}
}
}
int sum = 0;
for (int val : temp) sum += val;
maxSum = Math.max(maxSum, sum);
}
return maxSum;
}
}class Solution:
def maxOxygenLevel(self, levels: List[int], adjustments: List[List[int]]) -> int:
n = len(levels)
m = len(adjustments)
if m == 0:
return sum(levels)
max_sum = 0
for mask in range(1 << m):
temp = levels[:]
for i in range(m):
if mask & (1 << i):
duration, start, amount = adjustments[i]
for j in range(start, n):
temp[j] += amount
current_sum = sum(temp)
max_sum = max(max_sum, current_sum)
return max_sum/**
* @param {number[]} levels
* @param {number[][]} adjustments
* @return {number}
*/
var maxOxygenLevel = function(levels, adjustments) {
const n = levels.length;
const m = adjustments.length;
if (m === 0) return levels.reduce((a, b) => a + b, 0);
let maxSum = 0;
for (let mask = 0; mask < (1 << m); ++mask) {
const temp = [...levels];
for (let i = 0; i < m; ++i) {
if (mask & (1 << i)) {
const duration = adjustments[i][0];
const start = adjustments[i][1];
const amount = adjustments[i][2];
for (let j = start; j < n; ++j) {
temp[j] += amount;
}
}
}
const sum = temp.reduce((a, b) => a + b, 0);
maxSum = Math.max(maxSum, sum);
}
return maxSum;
};Asked in Top Tech Interviews
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