Frequency Window Constraint Optimizer 8 — Problem Statement & Solution Guide
Problem Description
You are tasked with optimizing the throughput of a circular data pipeline consisting of N processing nodes. Each node i is characterized by two integer values: gain[i], representing the net resource accumulation (positive) or depletion (negative) at that node, and cost[i], representing the fixed overhead required to transition from node i to the next node in the sequence. The pipeline operates in a continuous loop, meaning the node following the last one is the first one.
A valid execution path must start at a specific node and traverse the entire circle exactly once without the cumulative resource balance dropping below zero at any point during the traversal. The resource balance is updated by adding the gain of the current node and then subtracting the cost to move to the next node. Your objective is to identify the unique starting index that allows for a complete, uninterrupted circuit. If no such starting position exists, return -1.
Input: Two integer arrays, gain and cost, both of length N.
Output: An integer representing the starting index of the valid circuit, or -1 if no valid circuit exists.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Frequency Window Constraint Optimizer 8"
WHY DOES IT MATTER?
The greedy elimination pattern transforms a seemingly quadratic search into a linear scan by discarding whole blocks of impossible starts in O(1) time, a technique that recurs in many circular or prefix‑sum problems.
OPTIMIZATION CHALLENGE
The key insight is that a negative cumulative balance at position j proves that every index between the current start and j cannot be a solution, allowing the start pointer to jump directly to j+1 and reset the balance.
REAL-WORLD CONNECTION
Think of a circular assembly line where each station adds or consumes inventory; the algorithm finds the rotation where the line never runs out of parts, analogous to load‑balancing in ring‑topology networks.
During an interview, compute the total net gain first; if it's negative, return -1 immediately. Otherwise, run the single‑pass greedy loop—this early exit saves time and demonstrates you understand feasibility conditions.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem is a variant of the classic circular tour (gas‑station) problem, which can be solved in linear time using a greedy sweep. The naive view treats each starting node independently, summing gain‑cost along the circle and resetting when the cumulative balance becomes negative; this leads to O(N^2) time because each failure discards only the current start. The optimal paradigm observes that if a prefix of the tour fails at position j, none of the nodes between the original start and j can serve as a valid start, because they would inherit the same negative deficit before reaching j. By advancing the candidate start directly to j+1 and resetting the running balance, we guarantee a single pass over the array, achieving O(N) time while preserving correctness. This greedy elimination works only because the total net gain (sum(gain) – sum(cost)) determines feasibility: if the global sum is negative, no start works; otherwise, the algorithm will locate the unique feasible start.
Interview Questions on This Problem
Q1How would you modify the greedy solution if the pipeline were allowed to skip at most K nodes during traversal?
Introduce a sliding window that tracks the cumulative deficit and allow a reset only after K consecutive negative steps; use a deque to maintain the best candidate start within the allowed skips, resulting in O(N) time with O(K) extra space.
Q2Explain why the total sum of gain[i]‑cost[i] being non‑negative is a necessary but not sufficient condition for every node to be a valid start.
A non‑negative total ensures at least one feasible start, but individual prefixes can still dip below zero, causing early failure for many nodes; only the node immediately after the deepest deficit can serve as a valid start.
Q3In a distributed system where each node represents a microservice with latency (cost) and throughput gain, how would you use this algorithm to rebalance request routing?
Model each service's net capacity as gain‑cost, run the linear greedy pass to find a rotation where cumulative capacity never goes negative, then reorder the request pipeline accordingly so that load never exceeds service capacity, guaranteeing stable throughput.
Examples
Input
gain = [5, -3, 4, -2], cost = [2, 1, 3, 1]
Output
0
Explanation: Total net change = (5-2) + (-3-1) + (4-3) + (-2-1) = 3 + (-4) + 1 + (-3) = -3. Wait, let's re-calculate. Net per step: 3, -4, 1, -3. Sum = -3. Since total sum is negative, no valid start exists? Let's check the logic. If total sum < 0, return -1. Let's adjust the example to be valid. Revised Input: gain = [5, -3, 4, 2], cost = [2, 1, 3, 1]. Net: 3, -4, 1, 1. Sum = 1. Start 0: Bal=3, Next=3-4=-1 (Fail). Start 1: Bal=-4 (Fail). Start 2: Bal=1, Next=1+1=2, Next=2+3=5, Next=5-4=1 (Success). Output: 2.
Input
gain = [1, 2, 3, 4], cost = [1, 1, 1, 1]
Output
0
Explanation: Net changes: 0, 1, 2, 3. Total sum = 6 > 0. Start 0: Bal=0, Next=0+1=1, Next=1+2=3, Next=3+3=6. Min balance 0. Valid. Output: 0.
Input
gain = [-1, -2, -3, -4], cost = [1, 1, 1, 1]
Output
-1
Explanation: Net changes: -2, -3, -4, -5. Total sum = -14 < 0. No valid start exists. Output: -1.
Constraints
- 1 <= gain.length == cost.length <= 10^5
- -10^9 <= gain[i] <= 10^9
- 0 <= cost[i] <= 10^9
Optimal Approach & Strategy
Perform a single linear scan, resetting the candidate start whenever the cumulative balance becomes negative, which yields O(N) time and O(1) extra space.
Brute Force Approach
Try each node as a start, simulate the whole circle, and stop when you find a feasible one; this costs O(N^2) time.
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):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
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return sum;
}Asked in Top Tech Interviews
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