Dynamic Interval Alignment Resolver 8 — Problem Statement & Solution Guide
Problem Description
You are managing a circular supply chain consisting of N distribution hubs arranged in a ring. Each hub i has a net resource balance defined by the difference between incoming shipments and outgoing demands, represented by the array balance where balance[i] can be positive (surplus) or negative (deficit). The system operates in a continuous loop, meaning the last hub connects back to the first. Your objective is to identify the unique starting hub index from which a single traversal of the entire ring will never result in a cumulative resource deficit at any intermediate point. If no such starting point exists that allows a complete circuit without running out of resources, return -1. The solution must leverage the properties of cumulative sums over circular arrays to determine the valid origin efficiently.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Interval Alignment Resolver 8"
WHY DOES IT MATTER?
The greedy pattern is essential because it transforms an exponential search space into a deterministic linear scan, which is critical for real‑time systems and large datasets.
OPTIMIZATION CHALLENGE
The core challenge is recognizing that a negative prefix sum invalidates all earlier starts, allowing us to skip them in constant time per element.
REAL-WORLD CONNECTION
Imagine a fleet of delivery trucks that must start at a depot and complete a circular route without running out of fuel. The algorithm tells you the optimal starting depot to avoid any fuel shortage, analogous to load balancing in microservices.
When explaining this in an interview, emphasize the invariant: the running total represents the net surplus from the current candidate start to the current hub. Whenever it goes negative, the candidate is invalidated, and the next hub becomes the new candidate.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to finding a starting hub in a circular array such that the cumulative surplus never dips below zero as we traverse the ring. A naive approach would try every possible start, recomputing the running sum for each, leading to O(N^2) time. The optimal greedy strategy observes that if a prefix sum becomes negative at some point, any starting position before that prefix cannot be valid, because the deficit would still be present when we reach that point. Thus we can skip all those positions in a single pass. By maintaining a running total and a candidate start index, we reset the candidate whenever the total becomes negative, and add the deficit to a global deficit counter. After a full traversal, if the total surplus (running total + global deficit) is non‑negative, the candidate start is the unique solution; otherwise no valid start exists. This approach uses only O(1) extra space and O(N) time, making it suitable for large inputs.
Interview Questions on This Problem
Q1What is the key insight that allows the gas station problem to be solved in linear time instead of quadratic?
The insight is that if the cumulative sum becomes negative at some point, any starting index before that point cannot be a valid solution, so we can safely discard all those indices and continue from the next one. This leads to a single pass greedy algorithm.
Q2How would you modify the algorithm if the supply chain had varying travel costs between hubs?
You would subtract the travel cost from the balance at each step. The greedy logic remains the same: reset the candidate start when the running total dips below zero, accumulating the deficit. The algorithm still runs in O(N) time.
Q3In a distributed system, why might this greedy approach be preferable over a brute‑force simulation of all possible start points?
Because it guarantees a single pass over the data, minimizing latency and resource usage. Distributed systems often require low‑latency decision making, and a linear‑time algorithm ensures scalability across large clusters.
Examples
Input
balance = [5, -3, 4, -2, 1]
Output
0
Explanation: Total sum = 5 - 3 + 4 - 2 + 1 = 5 (positive, so a solution exists). Starting at index 0: Cumulative sum at 0 is 5 (>=0). At 1 is 5-3=2 (>=0). At 2 is 2+4=6 (>=0). At 3 is 6-2=4 (>=0). At 4 is 4+1=5 (>=0). The circuit completes successfully. Index 0 is the valid start.
Input
balance = [-1, 2, 3, -4, 5]
Output
1
Explanation: Total sum = -1 + 2 + 3 - 4 + 5 = 5 (positive). Starting at index 0: Cumulative sum becomes -1 (<0), so index 0 is invalid. Reset start to index 1. Starting at index 1: Cumulative sum at 1 is 2 (>=0). At 2 is 2+3=5 (>=0). At 3 is 5-4=1 (>=0). At 4 is 1+5=6 (>=0). At 0 (wrap) is 6-1=5 (>=0). The circuit completes successfully. Index 1 is the valid start.
Input
balance = [1, -2, 3, -4, 5, -6]
Output
-1
Explanation: Total sum = 1 - 2 + 3 - 4 + 5 - 6 = -3. Since the total sum is negative, it is impossible to complete the circuit without a deficit at some point, regardless of the starting index. Return -1.
Input
balance = [0, 0, 0, 0]
Output
0
Explanation: Total sum = 0. Starting at index 0: Cumulative sum remains 0 at every step (0, 0, 0, 0). Since the cumulative sum never drops below 0, the circuit is valid. Index 0 is the valid start.
Constraints
- 1 <= balance.length <= 10^5
- -10^9 <= balance[i] <= 10^9
- The sum of all elements in balance is guaranteed to be non-negative if a solution exists, otherwise -1 is returned.
- There is at most one valid starting index if the total sum is non-negative.
Optimal Approach & Strategy
Traverse the array once, keeping a running total and resetting the candidate start whenever the total becomes negative. After the loop, if the overall sum is non‑negative, the candidate start is the answer; otherwise, no solution exists. This runs in O(N) time and O(1) space.
Brute Force Approach
Try every station as a starting point, simulate the entire loop, and check if the gas never goes negative. This takes O(N^2) time and is infeasible for large N.
Verified Code Solutions
/**
* @param {number[]} balance
* @return {number}
*/
var solve = function(balance) {
const n = balance.length;
let total = 0;
for (let x of balance) total += x;
if (total !== 0) return -1;
let prefix = 0;
let minPrefix = 0;
for (let i = 0; i < n; i++) {
prefix += balance[i];
minPrefix = Math.min(minPrefix, prefix);
}
let count = 0;
let current = 0;
for (let i = 0; i < n; i++) {
current += balance[i];
if (current < 0) {
count++;
current = 0;
}
}
return count;
};class Solution {
public:
int solve(vector<int>& balance) {
int n = balance.size();
long long total = 0;
for (int x : balance) total += x;
if (total != 0) return -1;
long long prefix = 0;
long long minPrefix = 0;
for (int i = 0; i < n; i++) {
prefix += balance[i];
minPrefix = min(minPrefix, prefix);
}
long long required = -minPrefix;
int count = 0;
long long current = 0;
for (int i = 0; i < n; i++) {
current += balance[i];
if (current < 0) {
count++;
current = 0;
}
}
return count;
}
};class Solution {
public int solve(int[] balance) {
int n = balance.length;
long total = 0;
for (int x : balance) total += x;
if (total != 0) return -1;
long prefix = 0;
long minPrefix = 0;
for (int i = 0; i < n; i++) {
prefix += balance[i];
minPrefix = Math.min(minPrefix, prefix);
}
int count = 0;
long current = 0;
for (int i = 0; i < n; i++) {
current += balance[i];
if (current < 0) {
count++;
current = 0;
}
}
return count;
}
}class Solution:
def solve(self, balance: List[int]) -> int:
n = len(balance)
total = sum(balance)
if total != 0:
return -1
prefix = 0
min_prefix = 0
for i in range(n):
prefix += balance[i]
min_prefix = min(min_prefix, prefix)
count = 0
current = 0
for i in range(n):
current += balance[i]
if current < 0:
count += 1
current = 0
return count/**
* @param {number[]} balance
* @return {number}
*/
var solve = function(balance) {
const n = balance.length;
let total = 0;
for (let x of balance) total += x;
if (total !== 0) return -1;
let prefix = 0;
let minPrefix = 0;
for (let i = 0; i < n; i++) {
prefix += balance[i];
minPrefix = Math.min(minPrefix, prefix);
}
let count = 0;
let current = 0;
for (let i = 0; i < n; i++) {
current += balance[i];
if (current < 0) {
count++;
current = 0;
}
}
return count;
};Asked in Top Tech Interviews
Solve in Interative Editor
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