Bitmask Subset Energy Evaluator 3 — Problem Statement & Solution Guide
Problem Description
You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the bitmask subset energy using the **Container With Most Water** methodology.
Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Bitmask Subset Energy Evaluator 3"
WHY DOES IT MATTER?
Two‑pointer linear scans are a cornerstone for problems where a global optimum depends on a monotonic relationship between two moving boundaries. They turn quadratic brute‑force into O(N) by exploiting problem‑specific invariants, making them indispensable for high‑scale interview challenges.
OPTIMIZATION CHALLENGE
The key insight is that the container's capacity is limited by the shorter side; therefore discarding the shorter side cannot eliminate a better solution, allowing us to prune half the search space at each step.
REAL-WORLD CONNECTION
Think of a water reservoir formed between two hills; as you slide the hills inward, you continuously evaluate the water volume that can be stored. Engineers designing dams use similar greedy shrink‑age strategies to locate optimal spillway positions.
During an interview, start by visualizing the problem as a physical container, then immediately propose the two‑pointer walk. Keep a running max and remember to update it before moving the pointer; this tiny ordering detail often trips candidates.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to finding two indices i and j (i < j) in an array of heights such that the product of the distance (j‑i) and the minimum of the two heights is maximized. This is exactly the classic "Container With Most Water" problem, which can be solved optimally with a two‑pointer technique. A naive O(N^2) scan enumerates every pair, but for N up to 10^6 this quickly exceeds time limits because the number of pairs grows quadratically. The two‑pointer paradigm works because the area is limited by the shorter side; moving the pointer at the shorter side inward can only potentially increase the height while reducing width, guaranteeing that no better solution is missed. By iteratively discarding the sub‑optimal side, we converge to the global optimum in linear time.
Interview Questions on This Problem
Q1How does the two‑pointer approach guarantee that we never miss the optimal pair in the Container With Most Water problem?
At each step the area is bounded by the shorter height; moving the longer side cannot increase the area because width shrinks while the limiting height stays the same or gets lower. Therefore we only move the pointer at the shorter side, which may find a taller line and possibly a larger area, ensuring all viable candidates are examined.
Q2If the heights array contains duplicate values, does the two‑pointer algorithm need any modification?
No. Duplicate heights are handled naturally; when both pointers point to equal heights we can move either pointer inward because the area contributed by that pair is already considered, and moving either side does not affect correctness.
Q3Explain how you would adapt the algorithm to also return the indices of the optimal container, not just its area.
Maintain two variables bestLeft and bestRight initialized to 0 and N‑1. Whenever a new maximum area is found, update these variables with the current left and right indices. After the loop ends, return both the maximum area and the stored indices.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: To calculate the bitmask subset energy, we simply need to sum up all the elements in the array. For the input [1, 2, 3, 4, 5], we do the following: 1 + 2 + 3 + 4 + 5 = 15.
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Similarly, for the input [10, 20, 30, 40, 50], we calculate the sum: 10 + 20 + 30 + 40 + 50 = 150.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Place two pointers at the array ends, compute area, move the pointer at the smaller height inward, and repeat until pointers meet, updating the maximum each time.
Brute Force Approach
Iterate over all possible pairs (i, j) and compute (j‑i) * min(height[i], height[j]), tracking the maximum.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
return sum(nums)function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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