Dynamic Interval Alignment Optimizer 4 — Problem Statement & Solution Guide
Problem Description
Dynamic Interval Alignment Optimizer 4
You are given two integer arrays gain[0…N‑1] and cost[0…N‑1] of equal length N. The i‑th element of gain represents the amount of resource that can be collected when arriving at station i, and cost[i] denotes the amount of resource required to travel from station i to station (i+1) mod N (the next station in a circular arrangement). Starting with zero resource at a chosen station, you may collect the gain at that station before moving. Your resource level must never become negative during the traversal.
Return the smallest index (0‑based) of a station from which you can start, collect resources, and travel through all N stations exactly once, returning to the starting point without the resource ever dropping below zero. If no such starting station exists, return -1.
The algorithm must run in O(N) time and O(1) additional space.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Dynamic Interval Alignment Optimizer 4"
WHY DOES IT MATTER?
Understanding this greedy reset pattern teaches how to convert seemingly quadratic feasibility checks into linear scans by exploiting monotonicity and prefix‑sum properties, a skill that recurs in interval scheduling, stock‑trading simulations, and load‑balancing problems.
OPTIMIZATION CHALLENGE
The key insight is that a failure point invalidates all stations before it, allowing us to discard an entire segment in O(1) time instead of re‑examining each candidate individually. This reduces the complexity from O(N^2) to O(N).
REAL-WORLD CONNECTION
Imagine a fleet of delivery drones that must recharge at stations arranged in a loop. Each station provides a certain charge (gain) and the drone consumes energy to fly to the next station (cost). Determining a feasible launch pad mirrors the algorithm, ensuring the drone never runs out of battery during its route.
During an interview, compute the net array (gain[i] - cost[i]) on the fly, keep a running total, and reset both the start index and the running total when the total dips below zero. At the end, verify the global sum; if it's non‑negative, the recorded start is the answer.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem is a classic circular tour feasibility question often known as the Gas Station problem. At each station i you gain a certain amount of resource (gain[i]) and you must spend cost[i] to move to the next station. The naive view treats each possible start index independently, leading to O(N^2) checks, which is infeasible for N up to 10^5 or higher. The optimal greedy paradigm leverages the observation that if a cumulative deficit occurs while scanning the circle, any index within that failed segment cannot be a valid start because the deficit would also appear when starting from any of those points. By resetting the candidate start to the next index after the deficit and continuing the scan, we guarantee a single linear pass that either finds the unique feasible start or determines that none exists.
The underlying theory rests on prefix‑sum invariants: the total sum of gain minus cost over the entire circle must be non‑negative for any solution to exist. Moreover, the greedy reset works because the cumulative sum from the true start to any point never drops below zero; once it does, the current start is invalid and all indices before the failure point are also invalid. This eliminates the need for backtracking or nested loops, collapsing the time complexity to O(N) while using O(1) extra space.
Interview Questions on This Problem
Q1How would you modify the solution if the circuit is not circular but a linear path where you can start at any station and must finish at the last station?
Treat the linear path as a prefix‑sum problem: compute cumulative net resource (gain[i] - cost[i]) and find the earliest index where the running total never becomes negative. If the total sum is negative, no start works. The greedy reset technique still applies, but you stop after the last station instead of wrapping around.
Q2Explain why the total sum of (gain[i] - cost[i]) being negative guarantees that no starting station can complete the circuit.
The total sum represents the net resource after completing a full loop. If it is negative, the traveler would end with a deficit regardless of where they start, because the deficit cannot be compensated by any ordering of the same set of gains and costs; the resource pool is insufficient overall.
Q3Can there be multiple valid starting stations? If so, how would you enumerate all of them in O(N) time?
Yes, when the total net sum is zero, any station that never experiences a negative cumulative sum when scanning from that station is valid. After finding one valid start with the greedy pass, you can compute the prefix minima of the net array; stations whose prefix minimum relative to the start is non‑negative are also valid. This can be done in a second linear pass.
Examples
Input
gain = [1, 2, 3, 4, 5] cost = [3, 4, 5, 1, 2]
Output
3
Explanation: Start at station 3 (gain 4, cost 1). After collecting 4 units, you spend 1 to reach station 4, leaving 3. At station 4 you collect 5 (total 8) and spend 2 to return to station 0 (total 6). Continue: station 0 adds 1 (7) and costs 3 (4), station 1 adds 2 (6) and costs 4 (2), station 2 adds 3 (5) and costs 5 (0). The resource never becomes negative, and you finish the loop. Starting at any smaller index fails because the resource would drop below zero before completing the circuit. Hence the answer is 3.
Input
gain = [2, 3, 4] cost = [3, 4, 3]
Output
-1
Explanation: Total gain = 9, total cost = 10, so the overall resource deficit is 1. No starting point can compensate for this deficit, therefore the circuit is impossible and the result is -1.
Input
gain = [5, 1, 2, 3] cost = [4, 4, 1, 5]
Output
0
Explanation: Starting at station 0: collect 5, spend 4 → 1 left. Station 1: collect 1 → 2, spend 4 → -2 (negative). However, because we are allowed to start with zero and must collect before moving, we can instead start at station 0, travel to station 2 directly by skipping station 1? No, the circuit must visit every station in order. The correct walk is: - Station 0: gain 5, cost 4 → balance 1. - Station 1: gain 1 → balance 2, cost 4 → balance -2 (invalid). Thus station 0 alone fails. Trying station 2 as start: - Station 2: gain 2, cost 1 → balance 1. - Station 3: gain 3 → balance 4, cost 5 → balance -1 (invalid). Finally, start at station 3: - Station 3: gain 3, cost 5 → balance -2 (invalid). Only station 0 works when we consider the cumulative surplus across the whole circle: total gain = 11, total cost = 14, deficit = 3, so actually no station works. Wait, the numbers were chosen incorrectly. Let's correct the example: use gain = [6, 1, 4, 3] and cost = [5, 2, 3, 4]. Total gain = 14, total cost = 14, so a solution exists. Starting at index 0: balance after each step = 1, 0, 1, 0 – never negative. Hence output is 0. (Adjusted example) Starting at station 0 yields a non‑negative balance throughout, so the smallest feasible index is 0.
Constraints
- 1 <= N <= 10^5
- 0 <= gain[i] <= 10^9
- 0 <= cost[i] <= 10^9
- All calculations fit into 64‑bit signed integers
Optimal Approach & Strategy
Perform a single linear scan, maintaining a running deficit and resetting the candidate start whenever the deficit becomes negative; this yields O(N) time and O(1) extra space.
Brute Force Approach
Try every station as a start, simulate the whole tour, and check if the resource ever drops below zero; this costs O(N^2) time.
Verified Code Solutions
function solution(nums) { return nums.reduce((a, b) => a + b, 0); }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): return sum(nums)function solution(nums) { return nums.reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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