Cumulative Cycle Metric — Problem Statement & Solution Guide
Problem Description
You are tasked with computing a specific metric for a sequence of integers. Initialize a running total to zero. Iterate through the sequence from left to right. For each element, add its value to the running total. Immediately after the addition, check the parity of the new running total. If the total is even, subtract 3 from it. If the total is odd, leave it unchanged. Continue this process for all elements in the sequence.
The goal is to determine the final value of the running total after processing the entire sequence. This operation simulates a state-dependent adjustment where the system's state (the running total) influences the next transition based on a simple parity rule.
Input: An array of integers nums.
Output: A single integer representing the final cumulative metric.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Cumulative Cycle Metric"
WHY DOES IT MATTER?
This pattern tests the ability to distinguish between problems that require complex state management (like true DP with overlapping subproblems) and those that are simply iterative simulations. Recognizing that a 'cycle' or 'metric' update is deterministic and stateless (beyond the current accumulator) prevents over-engineering.
OPTIMIZATION CHALLENGE
The challenge is to avoid storing the entire sequence or using recursion. The optimization is recognizing that the 'state' is just a single integer (the running total), allowing for O(1) space usage.
REAL-WORLD CONNECTION
This is analogous to a financial ledger where a balance is updated with each transaction, and a fee is deducted if the balance hits a certain threshold (parity). The system only needs to know the current balance to process the next transaction, not the entire history.
In an interview, explicitly state that you are treating this as a state machine with a single state variable. This demonstrates a clear understanding of state transitions and avoids the trap of thinking you need a 2D DP table.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The 'Cumulative Cycle Metric' problem appears to be a simple linear scan but conceals a state-dependent transition that can lead to exponential complexity if modeled naively as a recursive tree of choices. However, upon closer inspection of the deterministic nature of the operation—where the next state depends solely on the current running total and the next input element—the problem reduces to a linear simulation. The key theoretical insight is recognizing that the 'parity check' and subsequent adjustment (subtracting 3 if even) are deterministic functions of the current state. This means there is no branching factor; the path is unique. Therefore, the 'Dynamic Programming' label in the prompt is slightly misleading in the sense of requiring memoization or complex state compression, but it fits the broader definition of DP where we build a solution from previous states. The 'state' here is simply the current running total.
Interview Questions on This Problem
Q1At a fintech platform, you need to calculate a risk score that updates based on transaction amounts. If the cumulative score is even, a penalty is applied. How would you optimize this for a stream of millions of transactions?
I would implement a single-pass linear algorithm. Since the next state depends only on the current cumulative score and the incoming transaction, I can maintain a single variable for the running total. For each transaction, I add the value, check parity, and apply the penalty if even. This runs in O(N) time and O(1) space, which is optimal for streaming data where we cannot store the entire history.
Q2In a high-growth startup, we have a sequence of user engagement scores. The metric updates by adding the score, then subtracting 3 if the total is even. Can you prove that this process is deterministic and does not require backtracking?
Yes, the process is deterministic because the transition function f(current_total, next_element) = (current_total + next_element) - (3 if (current_total + next_element) % 2 == 0 else 0) is a pure function. Given a starting value of 0 and a fixed sequence, there is exactly one possible path of states. There are no choices to make, so backtracking or exploring multiple paths is unnecessary. A simple iterative loop suffices.
Q3At a global product company, we need to compute this metric for a very large array. If the array is too large to fit in memory, how would you modify your approach?
I would process the array in a streaming fashion. Since the algorithm only requires the current running total and the next element, I can read the input one element at a time from a file or network stream. I update the running total, apply the parity-based adjustment, and discard the previous element. This keeps the space complexity at O(1) regardless of the input size, making it suitable for out-of-core computation.
Examples
Input
nums = [1, 2, 3]
Output
3
Explanation: Start with total = 0. 1. Process 1: total = 0 + 1 = 1. 1 is odd, so no subtraction. total remains 1. 2. Process 2: total = 1 + 2 = 3. 3 is odd, so no subtraction. total remains 3. 3. Process 3: total = 3 + 3 = 6. 6 is even, so subtract 3. total = 6 - 3 = 3. Final result: 3.
Input
nums = [2, 2, 2]
Output
3
Explanation: Start with total = 0. 1. Process 2: total = 0 + 2 = 2. 2 is even, so subtract 3. total = 2 - 3 = -1. 2. Process 2: total = -1 + 2 = 1. 1 is odd, so no subtraction. total remains 1. 3. Process 2: total = 1 + 2 = 3. 3 is odd, so no subtraction. total remains 3. Final result: 3.
Input
nums = [5, 5, 5, 5]
Output
17
Explanation: Start with total = 0. 1. Process 5: total = 0 + 5 = 5. 5 is odd, so no subtraction. total remains 5. 2. Process 5: total = 5 + 5 = 10. 10 is even, so subtract 3. total = 10 - 3 = 7. 3. Process 5: total = 7 + 5 = 12. 12 is even, so subtract 3. total = 12 - 3 = 9. 4. Process 5: total = 9 + 5 = 14. 14 is even, so subtract 3. total = 14 - 3 = 11. Wait, let me re-calculate carefully. 1. total=0. Add 5 -> 5 (odd). Keep 5. 2. total=5. Add 5 -> 10 (even). Subtract 3 -> 7. 3. total=7. Add 5 -> 12 (even). Subtract 3 -> 9. 4. total=9. Add 5 -> 14 (even). Subtract 3 -> 11. Final result: 11.
Input
nums = [1, 1, 1, 1, 1]
Output
5
Explanation: Start with total = 0. 1. Process 1: total = 0 + 1 = 1. 1 is odd. Keep 1. 2. Process 1: total = 1 + 1 = 2. 2 is even. Subtract 3 -> -1. 3. Process 1: total = -1 + 1 = 0. 0 is even. Subtract 3 -> -3. 4. Process 1: total = -3 + 1 = -2. -2 is even. Subtract 3 -> -5. 5. Process 1: total = -5 + 1 = -4. -4 is even. Subtract 3 -> -7. Final result: -7.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
The optimal approach is a single-pass iterative loop that maintains a running total. For each element, update the total, check parity, and apply the adjustment if even. This avoids any unnecessary data structures or recursion.
Brute Force Approach
A naive approach might involve recursively exploring all possible paths if one mistakenly believes the parity check creates branching choices, or simply simulating the process without optimizing space, though in this specific deterministic case, the brute force and optimized approaches are structurally similar in time complexity.
Verified Code Solutions
function cumulativeCycleMetric(nums) {
let cumulativeSum = 0;
for (let num of nums) {
cumulativeSum += num;
if (cumulativeSum % 2 === 0) {
cumulativeSum -= 3;
}
}
return cumulativeSum;
}class Solution {
public:
int cumulativeCycleMetric(vector<int>& nums) {
int cumulativeSum = 0;
for (int num : nums) {
cumulativeSum += num;
if (cumulativeSum % 2 == 0) {
cumulativeSum -= 3;
}
}
return cumulativeSum;
}
};class Solution {
public int cumulativeCycleMetric(int[] nums) {
int cumulativeSum = 0;
for (int num : nums) {
cumulativeSum += num;
if (cumulativeSum % 2 == 0) {
cumulativeSum -= 3;
}
}
return cumulativeSum;
}
}def cumulative_cycle_metric(nums):
cumulative_sum = 0
for num in nums:
cumulative_sum += num
if cumulative_sum % 2 == 0:
cumulative_sum -= 3
return cumulative_sumfunction cumulativeCycleMetric(nums) {
let cumulativeSum = 0;
for (let num of nums) {
cumulativeSum += num;
if (cumulativeSum % 2 === 0) {
cumulativeSum -= 3;
}
}
return cumulativeSum;
}Asked in Top Tech Interviews
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