Frequency Window Constraint Resolver 5 — Problem Statement & Solution Guide
Problem Description
You are provided with a circular array delta of length N, where delta[i] denotes the net resource fluctuation (inflow minus outflow) at node i. A processing unit begins at a specific node with an initial buffer capacity of zero and must traverse the entire circular topology exactly once, visiting each node in sequential order. The unit can only transition from node i to node (i + 1) % N if the current buffer level remains non-negative after applying the fluctuation at node i. Your objective is to identify the unique starting index from which the unit can successfully complete the full circuit without the buffer dropping below zero at any intermediate step. If the total sum of all fluctuations is negative, no valid starting index exists, and you should return -1. Note that if a valid solution exists, it is guaranteed to be unique.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Frequency Window Constraint Resolver 5"
WHY DOES IT MATTER?
The greedy discard‑and‑reset pattern transforms a seemingly combinatorial search over N possible starts into a deterministic linear scan, which is essential for handling large‑scale circular resource allocation problems under strict time constraints.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that a negative running sum invalidates all indices up to the current position, allowing us to skip them in O(1) time instead of re‑evaluating each individually.
REAL-WORLD CONNECTION
In distributed systems, this mirrors load‑balancing a token ring where each node contributes or consumes capacity; the algorithm quickly identifies a safe entry point for a new job without exhaustive simulation.
During an interview, compute the total sum first; if it’s non‑negative, immediately switch to the greedy scan—this two‑step guard prevents unnecessary work and demonstrates disciplined problem decomposition.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem is a classic circular feasibility check that can be solved with a greedy linear scan. The key insight is that if the total sum of delta over the entire circle is negative, no starting node can satisfy the buffer constraint, because the net resource deficit cannot be compensated regardless of the traversal order. When the total sum is non‑negative, a valid start exists and can be found by iteratively accumulating the buffer; whenever the running sum drops below zero, any index before the current position cannot be a feasible start, so we reset the candidate start to the next node and zero the accumulator. This eliminates the need to test every possible start, collapsing an O(N^2) brute‑force into an O(N) greedy pass.
Naïve solutions attempt to simulate the traversal from each node, recomputing the cumulative buffer each time, which leads to quadratic time and quickly exceeds limits for N up to 10^6 or higher. The optimal paradigm leverages the monotonicity of the cumulative deficit: once a prefix fails, extending it with any earlier start cannot repair the deficit, allowing us to discard whole blocks of candidates in constant time. This greedy discard‑and‑reset technique is the backbone of many circular‑array feasibility problems, including the well‑known Gas Station problem.
Interview Questions on This Problem
Q1How would you determine if there exists a starting node in a circular array of net resource changes such that a processing unit never runs out of buffer?
Compute the total sum; if it is negative, answer is impossible. Otherwise, perform a single pass accumulating the buffer, resetting the candidate start whenever the running sum becomes negative. The final candidate start is the answer.
Q2Why does resetting the start index after a negative cumulative sum guarantee that all earlier indices are invalid?
Because the cumulative sum from the original start to the failure point is negative, any start within that segment would inherit the same deficit before reaching the failure point, making it impossible to stay non‑negative.
Q3Can you adapt the greedy solution to also return the minimum initial buffer needed if the total sum is negative?
Yes. Track the minimum prefix sum during a linear scan; the absolute value of that minimum (plus one if strict positivity is required) gives the smallest extra buffer that would make the traversal feasible.
Examples
Input
delta = [3, -2, -3, 1, 2]
Output
3
Explanation: Total sum = 3 - 2 - 3 + 1 + 2 = 1 (positive, so a solution exists). Start at index 3: Buffer = 0 + 1 = 1. Move to index 4: Buffer = 1 + 2 = 3. Move to index 0: Buffer = 3 + 3 = 6. Move to index 1: Buffer = 6 - 2 = 4. Move to index 2: Buffer = 4 - 3 = 1. Buffer never drops below 0. Return 3.
Input
delta = [-1, -2, -3, -4, -5]
Output
-1
Explanation: Total sum = -1 - 2 - 3 - 4 - 5 = -15. Since the total sum is negative, it is impossible to complete the circuit regardless of the starting position. Return -1.
Input
delta = [1, 1, 1, 1, 1]
Output
0
Explanation: Total sum = 5 (positive). Start at index 0: Buffer = 0 + 1 = 1. Move to index 1: Buffer = 1 + 1 = 2. Move to index 2: Buffer = 2 + 1 = 3. Move to index 3: Buffer = 3 + 1 = 4. Move to index 4: Buffer = 4 + 1 = 5. Buffer remains non-negative throughout. Return 0.
Input
delta = [5, -1, -1, -1, -1, -1]
Output
0
Explanation: Total sum = 5 - 1 - 1 - 1 - 1 - 1 = 0 (non-negative, so a solution exists). Start at index 0: Buffer = 0 + 5 = 5. Move to index 1: Buffer = 5 - 1 = 4. Move to index 2: Buffer = 4 - 1 = 3. Move to index 3: Buffer = 3 - 1 = 2. Move to index 4: Buffer = 2 - 1 = 1. Move to index 5: Buffer = 1 - 1 = 0. Buffer never drops below 0. Return 0.
Constraints
- 1 <= delta.length <= 10^5
- -10^9 <= delta[i] <= 10^9
- The sum of all elements in delta is guaranteed to be non-negative if a valid starting index exists.
Optimal Approach & Strategy
Use a single linear scan with a running sum and a candidate start pointer; reset both when the sum becomes negative, achieving O(N) time and O(1) extra space.
Brute Force Approach
Simulate the traversal from each possible start, resetting the buffer at each node and checking if it ever goes negative; this costs O(N^2) time.
Verified Code Solutions
/**
* @param {number[]} gas
* @return {number}
*/
var canCompleteCircuit = function(gas) {
let n = gas.length;
let totalSum = 0;
let currSum = 0;
let start = 0;
for (let i = 0; i < n; i++) {
totalSum += gas[i];
currSum += gas[i];
if (currSum < 0) {
start = i + 1;
currSum = 0;
}
}
return totalSum >= 0 ? start : -1;
};class Solution {
public:
int canCompleteCircuit(vector<int>& gas) {
int n = gas.size();
int totalSum = 0;
int currSum = 0;
int start = 0;
for (int i = 0; i < n; ++i) {
totalSum += gas[i];
currSum += gas[i];
if (currSum < 0) {
start = i + 1;
currSum = 0;
}
}
return totalSum >= 0 ? start : -1;
}
};class Solution {
public int canCompleteCircuit(int[] gas) {
int n = gas.length;
int totalSum = 0;
int currSum = 0;
int start = 0;
for (int i = 0; i < n; i++) {
totalSum += gas[i];
currSum += gas[i];
if (currSum < 0) {
start = i + 1;
currSum = 0;
}
}
return totalSum >= 0 ? start : -1;
}
}class Solution:
def canCompleteCircuit(self, gas: List[int]) -> int:
n = len(gas)
total_sum = 0
curr_sum = 0
start = 0
for i in range(n):
total_sum += gas[i]
curr_sum += gas[i]
if curr_sum < 0:
start = i + 1
curr_sum = 0
return start if total_sum >= 0 else -1/**
* @param {number[]} gas
* @return {number}
*/
var canCompleteCircuit = function(gas) {
let n = gas.length;
let totalSum = 0;
let currSum = 0;
let start = 0;
for (let i = 0; i < n; i++) {
totalSum += gas[i];
currSum += gas[i];
if (currSum < 0) {
start = i + 1;
currSum = 0;
}
}
return totalSum >= 0 ? start : -1;
};Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.