Monotonic Cycle Metric — Problem Statement & Solution Guide
Problem Description
Given an array of numerical values, compute the monotonic cycle metric as the sum of absolute differences between consecutive elements where the current element is less than the previous one.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Monotonic Cycle Metric"
WHY DOES IT MATTER?
The pattern exemplifies a one‑pass aggregation over adjacent elements, a staple in interview problems that test a candidate’s ability to recognize locality and avoid unnecessary nested loops.
OPTIMIZATION CHALLENGE
The key insight is that each contribution depends only on the immediate predecessor, allowing us to discard all earlier history and achieve O(1) auxiliary space while still capturing the full metric.
REAL-WORLD CONNECTION
Think of a stock price chart: the metric measures total loss during down‑trends, ignoring gains. In distributed systems, it mirrors the calculation of total back‑pressure when a queue size shrinks, which is critical for flow‑control algorithms.
When coding, initialize the accumulator to 0 and store the first element as ‘prev’. Loop from the second element onward, update ‘prev’ at the end of each iteration, and avoid off‑by‑one errors by handling the circular case separately if required.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The monotonic cycle metric captures the total “down‑hill” movement in a numeric sequence. For each adjacent pair (prev, curr) we add |curr‑prev| only when curr < prev, effectively ignoring upward or flat steps. A naive implementation would scan the array and, for every element, recompute the absolute difference with all previous elements, leading to O(n^2) time – infeasible for large n (10^6+). The optimal paradigm leverages a single linear pass: maintain the previous value, compare it to the current, and accumulate the contribution when the monotonic decreasing condition holds. This reduces the problem to a classic “single‑scan aggregation” pattern, which is both time‑optimal and memory‑light because only a constant amount of state is required.
Why the linear scan works stems from the metric’s locality: each term in the sum depends solely on a pair of consecutive elements. There is no need for global information such as prefix minima or segment trees. Consequently, the algorithm fits the “sliding window of size two” model, where the window slides across the array once, performing O(1) work per step. This approach scales gracefully to massive inputs and fits comfortably within typical interview constraints of O(n) time and O(1) auxiliary space.
Interview Questions on This Problem
Q1How would you modify the solution to also return the indices where the monotonic decreases occur?
During the linear scan, whenever curr < prev, push the current index (or the pair (i‑1, i)) into a result vector. The rest of the algorithm stays unchanged, preserving O(n) time and O(k) extra space where k is the number of decreases.
Q2If the input were a circular linked list instead of an array, how would you compute the metric?
Traverse the linked list once, keeping track of the first node’s value as ‘headVal’. For each node, compare its value with the previous node’s value and accumulate when decreasing. After reaching the tail, perform one extra comparison between the tail’s value and headVal to close the cycle. This still runs in O(n) time and O(1) space.
Q3Can you compute the monotonic cycle metric in a streaming fashion where numbers arrive one by one?
Yes. Maintain the last seen value and the running sum. For each incoming number, if it is less than the previous value, add the absolute difference to the sum and update the previous value. This uses O(1) memory and processes each element in O(1) time, suitable for unbounded streams.
Examples
Input
[1, 2, 3, 4, 5]
Output
0
Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5], we observe that it is strictly increasing. Therefore, there are no pairs of consecutive elements where the current element is less than the previous one. Hence, the monotonic cycle metric is 0.
Input
[5, 4, 3, 2, 1]
Output
0
Explanation: Step-by-step: Given the array [5, 4, 3, 2, 1], we observe that it is strictly decreasing. Therefore, there are no pairs of consecutive elements where the current element is less than the previous one. Hence, the monotonic cycle metric is 0.
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
Perform a single linear scan, keeping only the previous value; add |curr‑prev| to the answer when curr < prev. This yields O(n) time and O(1) extra space.
Brute Force Approach
Iterate over every element and, for each, compare it with all previous elements to find decreasing pairs, summing their absolute differences. This results in O(n^2) time.
Verified Code Solutions
function monotonicCycleMetric(nums) {
let metric = 0;
for (let i = 1; i < nums.length; i++) {
if (nums[i] < nums[i - 1]) {
metric += Math.abs(nums[i] - nums[i - 1]);
}
}
return metric;
}class Solution {
public:
int monotonicCycleMetric(vector<int>& nums) {
int metric = 0;
for (int i = 1; i < nums.size(); i++) {
if (nums[i] < nums[i - 1]) {
metric += abs(nums[i] - nums[i - 1]);
}
}
return metric;
}
};class Solution {
public int monotonicCycleMetric(int[] nums) {
int metric = 0;
for (int i = 1; i < nums.length; i++) {
if (nums[i] < nums[i - 1]) {
metric += Math.abs(nums[i] - nums[i - 1]);
}
}
return metric;
}
}def monotonic_cycle_metric(nums):
metric = 0
for i in range(1, len(nums)):
if nums[i] < nums[i - 1]:
metric += abs(nums[i] - nums[i - 1])
return metricfunction monotonicCycleMetric(nums) {
let metric = 0;
for (let i = 1; i < nums.length; i++) {
if (nums[i] < nums[i - 1]) {
metric += Math.abs(nums[i] - nums[i - 1]);
}
}
return metric;
}Asked in Top Tech Interviews
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