Dynamic Interval Alignment Resolver — Problem Statement & Solution Guide

GreedyHardGas Station Circuit

TimeO(N)

SpaceO(1)

Problem Description

Given a complex dataset of length N representing system constraints and values, calculate the dynamic interval alignment using the Gas Station Circuit methodology.

DSA Pattern Breakdown

"Dynamic Interval Alignment Resolver"

hard

WHY DOES IT MATTER?

The pattern exemplifies how global feasibility can be decided with a single pass by exploiting cumulative invariants, a technique that recurs in scheduling, circular buffer management, and load‑balancing problems.

OPTIMIZATION CHALLENGE

The key insight is recognizing that a negative running balance invalidates every intermediate start, allowing us to skip large sections of the search space instead of checking each index individually.

REAL-WORLD CONNECTION

Imagine a fleet of delivery trucks arranged in a circular route where each truck can refuel at its depot (gas) and must travel to the next depot (cost). The algorithm determines the depot from which the fleet can start such that no truck runs out of fuel, mirroring real‑world logistics and distributed token‑ring protocols.

During an interview, compute the net array (gas‑cost) first and keep a running total; if it ever dips below zero, immediately move the candidate start pointer forward—this eliminates the need for nested loops and keeps the code clean.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(1)

Core Theory — Why This Approach?

The Dynamic Interval Alignment Resolver is a direct analogue of the classic Gas Station Circuit problem. The core idea is to treat each position in the dataset as a station that provides a certain amount of "resource" (gas) and requires a certain amount of "cost" to move to the next station. A naive solution would simulate every possible starting index, accumulating resources and checking feasibility, which leads to O(N^2) time and quickly becomes infeasible for N up to 10^6 or larger. The optimal paradigm leverages a greedy observation: if a cumulative deficit occurs while scanning from a candidate start, none of the indices between the start and the point of failure can serve as a valid start, because they would inherit the same deficit. By resetting the start to the next index after the failure and resetting the running balance, we guarantee a single linear pass that either finds a feasible start or proves impossibility.

This greedy approach works because the problem exhibits the "prefix‑sum monotonicity" property: the total net resource over the entire circle must be non‑negative for any solution to exist. When the total sum is non‑negative, the algorithm's reset rule ensures that the eventual start index accumulates a non‑negative balance for the entire traversal. The proof hinges on partitioning the circle into a failing prefix and a succeeding suffix; the suffix alone must have a non‑negative cumulative sum, making its first element the only viable start.

Thus, the optimal solution runs in O(N) time with O(1) auxiliary space, dramatically outperforming brute‑force enumeration and scaling to massive datasets typical in modern distributed systems.

Interview Questions on This Problem

Q1How would you modify the Gas Station solution to return all possible starting indices instead of just one?

First compute the total net resource; if it is negative, no start works. Then compute prefix sums of net resource (gas[i] - cost[i]). Any index i where the minimum prefix sum in the circular array (shifted so i becomes start) is non‑negative is a valid start. This can be done in O(N) by scanning for the global minimum prefix sum and noting that all indices after that minimum (mod N) are feasible.

Q2Explain why the greedy reset rule does not miss a valid start when the total net resource is non‑negative.

When a deficit occurs at position j while scanning from start s, the cumulative sum from s to j is negative. Any index k in (s, j] would have an even smaller cumulative sum when starting from k because it excludes the positive contributions before k, so k cannot be a valid start. Resetting to j+1 discards all impossible candidates, preserving correctness.

Q3In a distributed system where each node reports its available capacity and required load to the next node, how would you apply this algorithm to detect a feasible load‑balancing ring?

Treat each node as a station with capacity as "gas" and required load as "cost". Run the linear greedy scan to find a node that can serve as the entry point for a circular load‑balancing pass. If the total capacity >= total load, the algorithm yields a start node guaranteeing that moving clockwise, each node can satisfy its outgoing load using its own capacity plus any surplus from previous nodes.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9]

Output

Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9], we calculate the dynamic interval alignment using the Gas Station Circuit methodology. The tank capacity is initially 0. At index 3, the current sum is 6, which is less than the tank capacity of 0. At index 6, the current sum is 18, which is greater than or equal to the tank capacity of 0. We increment the gas station index and reset the tank capacity to 0 and the current sum to 0. We add the current element 7 to the tank capacity before resetting it. At index 9, the current sum is 28, which is greater than or equal to the tank capacity of 0. We increment the gas station index and reset the tank capacity to 0 and the current sum to 0. We add the current element 9 to the tank capacity before resetting it. The final gas station index is 4.

Example 2

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Output

Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we calculate the dynamic interval alignment using the Gas Station Circuit methodology. The tank capacity is initially 0. At index 2, the current sum is 6, which is less than the tank capacity of 0. At index 5, the current sum is 15, which is greater than or equal to the tank capacity of 0. We increment the gas station index and reset the tank capacity to 0 and the current sum to 0. We add the current element 6 to the tank capacity before resetting it. At index 8, the current sum is 24, which is greater than or equal to the tank capacity of 0. We increment the gas station index and reset the tank capacity to 0 and the current sum to 0. We add the current element 8 to the tank capacity before resetting it. The final gas station index is 3.

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N) or O(N log N)
Space Complexity: O(N) or O(1)

Optimal Approach & Strategy

Perform a single linear scan, resetting the start index whenever the running balance drops below zero, and verify total net sum at the end.

Brute Force Approach

Try every index as a start, simulate the whole circle, and check if the tank ever goes negative.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums) {
   let tankCapacity = 0;
   let currentSum = 0;
   let gasStationIndex = 0;
   for (let i = 0; i < nums.length; i++) {
       currentSum += nums[i];
       tankCapacity += nums[i];
       if (currentSum >= tankCapacity && tankCapacity !== 0) {
           gasStationIndex++;
           tankCapacity = 0;
           currentSum = 0;
       }
   }
   return gasStationIndex;
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       int tankCapacity = 0;
       int currentSum = 0;
       int gasStationIndex = 0;
       for (int i = 0; i < nums.size(); i++) {
           currentSum += nums[i];
           tankCapacity += nums[i];
           if (currentSum >= tankCapacity && tankCapacity != 0) {
               gasStationIndex++;
               tankCapacity = 0;
               currentSum = 0;
           }
       }
       return gasStationIndex;
   }
};

class Solution {
   public int solution(int[] nums) {
       int tankCapacity = 0;
       int currentSum = 0;
       int gasStationIndex = 0;
       for (int i = 0; i < nums.length; i++) {
           currentSum += nums[i];
           tankCapacity += nums[i];
           if (currentSum >= tankCapacity && tankCapacity != 0) {
               gasStationIndex++;
               tankCapacity = 0;
               currentSum = 0;
           }
       }
       return gasStationIndex;
   }
}

def solution(nums):
   tank_capacity = 0
   current_sum = 0
   gas_station_index = 0
   for i in range(len(nums)):
       current_sum += nums[i]
       tank_capacity += nums[i]
       if current_sum >= tank_capacity and tank_capacity != 0:
           gas_station_index += 1
           tank_capacity = 0
           current_sum = 0
   return gas_station_index

function solution(nums) {
   let tankCapacity = 0;
   let currentSum = 0;
   let gasStationIndex = 0;
   for (let i = 0; i < nums.length; i++) {
       currentSum += nums[i];
       tankCapacity += nums[i];
       if (currentSum >= tankCapacity && tankCapacity !== 0) {
           gasStationIndex++;
           tankCapacity = 0;
           currentSum = 0;
       }
   }
   return gasStationIndex;
}

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Dynamic Interval Alignment Resolver Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Greedy

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Dynamic Interval Alignment Resolver — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor