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Optimizing Space Station Oxygen Levels Solution

Problem Statement

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.

Example 1
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.

Example 2
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
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Optimizing Space Station Oxygen Levels — Problem Statement & Solution Guide

ArraysHardDynamic
TimeO(n*m)
|
SpaceO(n)

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.

Examples

Example 1

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.

Example 2

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

The optimized approach uses dynamic programming to store the maximum oxygen levels achievable at each step, reducing the time complexity to O(n * 2^n).

Brute Force Approach

The brute-force approach involves trying all possible sequences of adjustments and calculating the total oxygen level for each sequence, resulting in a time complexity of O(n^2 * 2^n).

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