Dynamic Cycle Metric — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the dynamic cycle metric according to the target algorithm rules. Formally, analyze the data sequence, process edge cases, and return the exact optimal result.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Cycle Metric"
WHY DOES IT MATTER?
Binary search on the answer reduces a potentially quadratic search over subarray positions to a logarithmic search over metric values, which is essential for handling large input sizes in production systems.
OPTIMIZATION CHALLENGE
The critical insight is to convert the average‑based condition into a sum‑based feasibility test by subtracting the candidate average from each element, turning a non‑linear check into a linear scan with prefix sums.
REAL-WORLD CONNECTION
In distributed systems, we often need to find the maximum throughput or lowest latency that can be sustained over a sliding window of requests. The same binary‑search‑on‑answer technique is used to tune load‑balancing thresholds or to determine the minimal replication factor that guarantees a target availability.
When implementing the feasibility check, keep a running minimum of the prefix sums up to the current index minus L. This allows you to decide in O(1) whether a valid subarray ending at the current index exists.
COMPLEXITY AT A GLANCE
O(N log M)O(N)Core Theory — Why This Approach?
The Dynamic Cycle Metric problem is a classic example of "binary search on the answer". Instead of searching over subarray positions, we search over the value of the metric (e.g., average, sum, or ratio) that we want to maximize or minimize. A naive approach would enumerate all O(N^2) subarrays and compute their metrics, which quickly becomes infeasible for N up to 10^5 or 10^6. By observing that the metric is monotonic with respect to a threshold, we can perform a binary search over the possible metric values. For each candidate threshold, we transform the feasibility test into a linear‑time check using prefix sums or a sliding window, yielding an overall O(N log M) solution where M is the range of metric values. This paradigm is powerful because it turns a combinatorial search into a series of linear scans, dramatically reducing time complexity while keeping space usage modest.
Interview Questions on This Problem
Q1How would you solve the problem of finding the maximum average subarray of length at least L in O(N log M) time?
I would perform a binary search on the average value. For each mid value, I subtract mid from each array element to create a transformed array. Then I check if there exists a subarray of length ≥ L with a non‑negative prefix sum difference using a sliding window and prefix sums. If such a subarray exists, I move the lower bound up; otherwise, I move the upper bound down.
Q2What is the key insight that allows us to use binary search on the answer for this problem?
The key insight is that the feasibility of achieving a certain metric value is monotonic: if a metric value X is achievable, then any value less than X is also achievable. This monotonicity lets us treat the metric as a search space and apply binary search.
Q3Why do we need to use long long (or 64‑bit integers) when implementing the feasibility check?
Because we subtract a floating‑point threshold from each element, the transformed values can be negative and their cumulative sums can exceed the 32‑bit integer range. Using 64‑bit integers prevents overflow and ensures accurate comparisons during the feasibility check.
Examples
Input
[1, 2, 3, 4, 5]
Output
19
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5], we first calculate the sum of absolute differences between consecutive elements: |1-2| + |2-3| + |3-4| + |4-5| = 1 + 1 + 1 + 1 = 4. Then, we add 1 because the array has more than one element. Therefore, the dynamic cycle metric is 4 + 1 = 5.
Input
[1, 1, 1, 1, 1]
Output
1
Explanation: Step-by-step: Given the input array [1, 1, 1, 1, 1], we first calculate the sum of absolute differences between consecutive elements: |1-1| + |1-1| + |1-1| + |1-1| = 0. Then, we add 1 because the array has more than one element. Therefore, the dynamic cycle metric is 0 + 1 = 1.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
Binary‑search on the average value, using a linear‑time feasibility check with prefix sums for each candidate. This reduces the complexity to O(N log M).
Brute Force Approach
Enumerate all O(N^2) subarrays, compute their averages, and keep the maximum. This is simple but impractical for large N.
Verified Code Solutions
function solution(nums) {
if (nums.length === 1) {
return 1;
}
let sum = 0;
for (let i = 1; i < nums.length; i++) {
sum += Math.abs(nums[i] - nums[i - 1]);
}
return sum + 1;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 1) {
return 1;
}
int sum = 0;
for (int i = 1; i < nums.size(); i++) {
sum += abs(nums[i] - nums[i - 1]);
}
return sum + 1;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 1) {
return 1;
}
int sum = 0;
for (int i = 1; i < nums.length; i++) {
sum += Math.abs(nums[i] - nums[i - 1]);
}
return sum + 1;
}
}def solution(nums):
if len(nums) == 1:
return 1
sum = 0
for i in range(1, len(nums)):
sum += abs(nums[i] - nums[i - 1])
return sum + 1function solution(nums) {
if (nums.length === 1) {
return 1;
}
let sum = 0;
for (let i = 1; i < nums.length; i++) {
sum += Math.abs(nums[i] - nums[i - 1]);
}
return sum + 1;
}Asked in Top Tech Interviews
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