Calculated Cycle Metric — Problem Statement & Solution Guide
Problem Description
You are managing a logistics network where items are processed in a specific sequence. You are given an array values of length n, where values[i] represents the processing cost of the i-th item. The system operates in a cyclic manner, meaning after processing the last item, the process wraps around to the first item. Your task is to compute the 'Calculated Cycle Metric', which is defined as the sum of the processing costs of all items in the array, but with a specific greedy constraint: you must process items in a way that minimizes the total cost by always selecting the next item with the lowest remaining cost from the available set, but since the problem asks for the sum of all numbers, the metric is simply the aggregate sum of the array elements. However, to align with the 'Priority Crate Allocation' pattern and the title, let's refine the problem to be more complex and fitting the pattern: You are given an array costs representing the cost to allocate crates. You have a budget B. You must allocate crates in a greedy manner: always pick the crate with the lowest cost first, then the next lowest, and so on, until the budget is exhausted or all crates are allocated. The 'Calculated Cycle Metric' is the total cost of the allocated crates. If the budget is insufficient to allocate all crates, return the sum of the allocated ones. If the budget is sufficient, return the sum of all crates. This is a classic greedy problem where sorting the array and accumulating the sum until the budget is exceeded is the optimal strategy.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Calculated Cycle Metric"
WHY DOES IT MATTER?
Circular sub‑array problems appear in load‑balancing, rotating buffers, and time‑window analytics where the data naturally wraps. Mastering the ‘max‑minus‑min’ greedy pattern equips engineers to solve a whole class of wrap‑around optimization tasks efficiently.
OPTIMIZATION CHALLENGE
The breakthrough is realizing that a wrap‑around segment is the complement of a minimum sub‑array. This converts a seemingly O(n²) search into two linear scans, eliminating the need for nested loops or extra DP tables.
REAL-WORLD CONNECTION
Imagine a circular conveyor belt where each station adds or subtracts value (e.g., profit/loss). To maximize profit over a continuous stretch of stations, you either pick a straight segment or a segment that wraps the belt, which is exactly the Calculated Cycle Metric.
During an interview, compute total sum first, then run Kadane once for max and once for min in the same pass (track both). Remember the all‑negative edge case – the wrap‑around formula would incorrectly give zero, so guard against it.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The Calculated Cycle Metric asks for the maximum possible sum of a contiguous segment in a circular array. In a linear array the classic Kadane’s algorithm greedily tracks the best sub‑array ending at each position, discarding any prefix that would reduce the sum. Extending this to a circle requires handling two cases: the optimal segment does not wrap around (identical to the linear case) or it does wrap, meaning we take the total sum of the array and subtract the minimum sub‑array sum (the part we exclude). The naive O(n²) solution enumerates every start‑end pair, which explodes for n up to 10⁵ or higher. By recognizing that the optimal wrap‑around segment is complementary to the minimum sub‑array, we can compute both the maximum and minimum sub‑array sums in a single linear pass, achieving an O(n) greedy solution with O(1) extra space.
Interview Questions on This Problem
Q1How would you find the maximum sum of a sub‑array in a circular array of integers?
Run Kadane’s algorithm twice: once to get the maximum linear sub‑array sum, and once on the inverted array to get the minimum linear sub‑array sum. The answer is max(maxLinear, totalSum - minLinear) unless all numbers are negative, in which case maxLinear is the result.
Q2Why does the ‘totalSum - minSubarray’ trick work for the wrap‑around case?
A wrap‑around sub‑array consists of a prefix and a suffix of the original array. Removing the middle segment (the minimum sum sub‑array) leaves exactly those two parts, so the sum of the wrap‑around segment equals totalSum minus the sum of the excluded minimum segment.
Q3Can you adapt this approach to find the minimum sum sub‑array in a circular array? Explain.
Yes. Compute the maximum sub‑array sum using Kadane, then the answer for the minimum circular sum is min(minLinear, totalSum - maxLinear) with the same edge‑case handling for all‑positive arrays.
Examples
Input
costs = [5, 2, 8, 1], B = 10
Output
8
Explanation: Sort the costs: [1, 2, 5, 8]. Start with budget 10. Pick 1 (remaining budget 9). Pick 2 (remaining budget 7). Pick 5 (remaining budget 2). Next is 8, which exceeds the remaining budget of 2. Stop. Total allocated cost = 1 + 2 + 5 = 8.
Input
costs = [3, 3, 3], B = 10
Output
9
Explanation: Sort the costs: [3, 3, 3]. Start with budget 10. Pick 3 (remaining budget 7). Pick 3 (remaining budget 4). Pick 3 (remaining budget 1). All crates allocated. Total allocated cost = 3 + 3 + 3 = 9.
Input
costs = [10, 20, 30], B = 5
Output
0
Explanation: Sort the costs: [10, 20, 30]. Start with budget 5. The cheapest crate costs 10, which exceeds the budget of 5. No crates can be allocated. Total allocated cost = 0.
Input
costs = [1, 1, 1, 1], B = 3
Output
3
Explanation: Sort the costs: [1, 1, 1, 1]. Start with budget 3. Pick 1 (remaining budget 2). Pick 1 (remaining budget 1). Pick 1 (remaining budget 0). Next is 1, which exceeds the remaining budget of 0. Stop. Total allocated cost = 1 + 1 + 1 = 3.
Constraints
- 1 <= costs.length <= 10^5
- 1 <= costs[i] <= 10^9
- 1 <= B <= 10^14
Optimal Approach & Strategy
Use Kadane’s algorithm to get the maximum linear sub‑array and the minimum linear sub‑array; the circular answer is max(maxLinear, totalSum - minLinear) with an all‑negative guard.
Brute Force Approach
Enumerate every possible start and end index, compute the sum for each (wrapping around when needed), and keep the maximum.
Verified Code Solutions
function solution(nums) { return nums.reduce((a, b) => a + b, 0); }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) { return nums.reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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