Tank Water Allocation — Problem Statement & Solution Guide
Problem Description
Given two arrays, capacities representing the capacities of water tanks and requirements representing the water requirements, calculate a boolean array where each element at index i is true if the capacity of the tank at index i can fulfill the requirement at index i, and false otherwise.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tank Water Allocation"
WHY DOES IT MATTER?
Identifying that each tank’s capacity can be evaluated independently allows us to avoid unnecessary computations, leading to linear time complexity. This pattern is essential for scaling solutions to large datasets and for meeting strict performance constraints in production systems.
OPTIMIZATION CHALLENGE
The critical insight is that no cross‑index dependency exists; thus, a single pass suffices. Eliminating nested loops or auxiliary data structures reduces both time and space overhead.
REAL-WORLD CONNECTION
In distributed systems, a similar pattern appears when checking node health: each node’s status is evaluated against a threshold independently, and the system aggregates the results to decide routing or failover actions. Just as we compare capacities to requirements, a load balancer compares node metrics to thresholds to make routing decisions.
When explaining this in an interview, emphasize the O(n) time and O(n) space trade‑off, and mention that the result array can be built in place if the caller allows mutation, further reducing space to O(1).
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The Tank Water Allocation problem reduces to a simple element‑wise comparison between two equally sized arrays: for each index i, determine whether capacities[i] is greater than or equal to requirements[i]. While a naive solution might involve nested loops or repeated scans of the arrays, such approaches would unnecessarily inflate the time complexity to O(n^2) or worse, making them impractical for large inputs where n can reach millions. The optimal paradigm leverages the fact that each comparison is independent; a single linear pass suffices, yielding an O(n) time complexity and O(n) auxiliary space for the result array.
In algorithmic terms, this is a classic example of the "element‑wise comparison" pattern, which is a subset of the broader "array traversal" family. By recognizing that no cross‑dependency exists between indices, we avoid any need for sorting, hashing, or other heavy data structures. The key insight is that the problem can be solved with a single for‑loop, performing a constant‑time comparison at each step.
This pattern is often overlooked in interviews because it appears trivial, yet it tests a candidate’s ability to identify the minimal necessary work. It also serves as a building block for more complex problems where the comparison condition might involve additional constraints (e.g., thresholds, cumulative sums, or dynamic updates). Mastery of this pattern ensures efficient code and demonstrates a solid grasp of algorithmic optimization.
Interview Questions on This Problem
Q1During a recent interview at a fintech platform, I was asked to explain how you would handle a scenario where the capacities array contains negative values. What would be your answer?
I would clarify that negative capacities are physically impossible in this context, so the function should treat them as zero or raise an exception. If the requirement is to still perform the comparison, I would compare the negative capacity against the requirement, resulting in false unless the requirement is also negative, which would be a logical error. The key point is to validate inputs before processing.
Q2A high‑growth startup asked: "Can you modify the algorithm to return the indices of tanks that can meet their requirements instead of a boolean array?" How would you respond?
Yes, I would iterate once over both arrays, and whenever capacities[i] >= requirements[i], I would append i to a result list. This maintains O(n) time and O(k) space, where k is the number of satisfying tanks. The logic is identical; only the output format changes.
Q3A global product company inquired: "What if the arrays are extremely large and cannot fit into memory? How would you adapt the solution?"
I would process the arrays in a streaming fashion, reading chunks of data from disk or a network source. For each chunk, I would perform the comparison and write the boolean results to an output stream or file. This approach keeps memory usage bounded to the chunk size, achieving O(1) auxiliary space beyond the output.
Examples
Input
[1, 2, 3], [1, 2, 4]
Output
[true, true, false]
Explanation: Step-by-step: with input capacities = [1, 2, 3] and requirements = [1, 2, 4], we compare each capacity with its corresponding requirement. For the first tank, capacity 1 >= requirement 1, so it's true. For the second tank, capacity 2 >= requirement 2, so it's true. For the third tank, capacity 3 < requirement 4, so it's false. Thus, the output is [true, true, false].
Input
[5, 5, 5], [1, 2, 3]
Output
[true, true, true]
Explanation: Step-by-step: with input capacities = [5, 5, 5] and requirements = [1, 2, 3], we compare each capacity with its corresponding requirement. Since all capacities (5) are greater than or equal to their respective requirements (1, 2, 3), all tanks can fulfill their requirements, resulting in the output [true, true, true].
Constraints
- 1 <= capacities.length <= 100
- 1 <= requirements.length <= 100
- 1 <= capacities[i] <= 1000
- 1 <= requirements[i] <= 1000
- The sum of all capacities does not exceed 10000
Optimal Approach & Strategy
The optimal solution performs a single linear pass over both arrays, comparing capacities[i] to requirements[i] and storing the boolean result. This achieves O(n) time and O(n) space.
Brute Force Approach
A naive approach might involve nested loops or repeatedly scanning the arrays for each index, leading to O(n^2) time. It would also unnecessarily allocate extra memory for intermediate results.
Verified Code Solutions
function solution(capacities, requirements) { return capacities.map((capacity, index) => capacity >= requirements[index]); }class Solution { public: vector<bool> solution(vector<int>& capacities, vector<int>& requirements) { vector<bool> result; for (int i = 0; i < capacities.size(); i++) { result.push_back(capacities[i] >= requirements[i]); } return result; } }class Solution { public boolean[] solution(int[] capacities, int[] requirements) { boolean[] result = new boolean[capacities.length]; for (int i = 0; i < capacities.length; i++) { result[i] = capacities[i] >= requirements[i]; } return result; } }def solution(capacities, requirements): return [cap >= req for cap, req in zip(capacities, requirements)]function solution(capacities, requirements) { return capacities.map((capacity, index) => capacity >= requirements[index]); }Asked in Top Tech Interviews
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