Calculated Capacity Window — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the throughput of a logistics network by calculating the total payload capacity across a sequence of transport nodes. The system provides an array capacity of length N, where each element represents the maximum weight (in kilograms) that a specific node can handle during a single operational cycle. Your objective is to compute the 'Calculated Capacity Window,' defined strictly as the aggregate sum of all individual node capacities within the provided sequence.
This metric serves as the baseline for determining the total system load before applying dynamic allocation strategies. The calculation must be performed in a single pass to ensure minimal latency in real-time monitoring systems. You must handle both positive and negative values, where negative values may represent maintenance deductions or calibration offsets in the raw sensor data.
Given the array capacity, return the integer sum of all elements. The result represents the net effective capacity available for the Priority Crate Allocation algorithm to distribute incoming shipments.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Calculated Capacity Window"
WHY DOES IT MATTER?
Maximum sub‑array (Kadane) is a cornerstone greedy pattern that appears in profit‑maximization, signal processing, and resource allocation problems. Mastering it equips engineers to recognize when a global optimum can be built from optimal local decisions, a skill that scales to many real‑world optimization tasks.
OPTIMIZATION CHALLENGE
The key insight is that the optimal window either extends the previous window or starts fresh at the current node. By tracking only the best ending‑here sum, we avoid recomputing every possible window, collapsing O(N²) work into O(N).
REAL-WORLD CONNECTION
Imagine a logistics pipeline where each node’s capacity fluctuates due to maintenance or weather. The ‘calculated capacity window’ is the longest stretch of nodes that can collectively move the most cargo in a single cycle – akin to finding the most profitable contiguous segment of a stock price chart.
During an interview, write Kadane in a single pass, keep variables named clearly (curr, best), and immediately discuss edge cases (all negatives, empty array) before coding. This shows you think about correctness before implementation.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Calculated Capacity Window problem asks for the maximum total payload that can be handled by any contiguous sequence of transport nodes. Formally, given an integer array capacity[0…N‑1], we must find the maximum possible sum of a sub‑array. A naïve solution enumerates every possible start and end index, computes each sub‑array sum, and keeps the best – this costs O(N²) time and quickly becomes infeasible for N up to 10⁵ or 10⁶, which is typical in interview constraints. The optimal paradigm is Kadane’s algorithm, a greedy dynamic‑programming technique that scans the array once while maintaining two variables: the best sum ending at the current position (currentMax) and the overall best seen so far (globalMax). At each step we decide whether to extend the previous window (currentMax + capacity[i]) or start a new window at i (capacity[i]). This local‑optimal choice guarantees a global optimum because any sub‑array that yields the maximum sum must end at some index, and the algorithm records the best possible ending sum for every index. The result is achieved in linear time with constant extra space, satisfying the constraints of large inputs.
Interview Questions on This Problem
Q1How would you modify Kadane’s algorithm to also return the start and end indices of the maximum‑capacity window?
Maintain two additional pointers: a temporary start index that updates whenever you start a new window (i.e., when capacity[i] > currentMax + capacity[i]), and bestStart/bestEnd that update whenever globalMax is improved. At the end, return capacity[bestStart…bestEnd] as the window.
Q2If the capacity array can contain negative numbers, how does that affect the solution and what edge case must you handle?
Kadane already handles negatives; however, if all numbers are negative, the algorithm should return the largest (least negative) single element. This requires initializing globalMax to -∞ (or the first element) and ensuring currentMax is reset to capacity[i] when it becomes smaller than capacity[i].
Q3Explain how you would compute the maximum‑capacity window when you are allowed to delete at most one element from the sub‑array.
Perform two passes: forward Kadane to compute max sub‑array ending at each index, and backward Kadane for max sub‑array starting at each index. For each position i, consider deleting capacity[i] and combine forward[i‑1] + backward[i+1]; the answer is the maximum of these combined values and the standard Kadane result.
Examples
Input
capacity = [12, 45, -3, 8, 20]
Output
82
Explanation: Step 1: Initialize sum to 0. Step 2: Add 12 -> sum = 12. Step 3: Add 45 -> sum = 57. Step 4: Add -3 -> sum = 54. Step 5: Add 8 -> sum = 62. Step 6: Add 20 -> sum = 82. Final Output: 82.
Input
capacity = [100, 100, 100]
Output
300
Explanation: Step 1: Initialize sum to 0. Step 2: Add 100 -> sum = 100. Step 3: Add 100 -> sum = 200. Step 4: Add 100 -> sum = 300. Final Output: 300.
Input
capacity = [-5, -10, -15]
Output
-30
Explanation: Step 1: Initialize sum to 0. Step 2: Add -5 -> sum = -5. Step 3: Add -10 -> sum = -15. Step 4: Add -15 -> sum = -30. Final Output: -30.
Input
capacity = [0, 0, 0]
Output
0
Explanation: Step 1: Initialize sum to 0. Step 2: Add 0 -> sum = 0. Step 3: Add 0 -> sum = 0. Step 4: Add 0 -> sum = 0. Final Output: 0.
Constraints
- 1 <= capacity.length <= 10^5
- -10^9 <= capacity[i] <= 10^9
- The sum of all elements will fit within a 64-bit signed integer.
Optimal Approach & Strategy
Use Kadane’s algorithm: iterate once, maintaining the best sum ending at the current index and the overall best sum.
Brute Force Approach
Enumerate every possible start and end index, compute the sum for each sub‑array, and keep 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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