There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.
Given two integer arrays gas and cost, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique
Example 1:
Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2] Output: 3 Explanation: Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 4. Your tank = 4 - 1 + 5 = 8 Travel to station 0. Your tank = 8 - 2 + 1 = 7 Travel to station 1. Your tank = 7 - 3 + 2 = 6 Travel to station 2. Your tank = 6 - 4 + 3 = 5 Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3. Therefore, return 3 as the starting index.
Example 2:
Input: gas = [2,3,4], cost = [3,4,3] Output: -1 Explanation: You can't start at station 0 or 1, as there is not enough gas to travel to the next station. Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 0. Your tank = 4 - 3 + 2 = 3 Travel to station 1. Your tank = 3 - 3 + 3 = 3 You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3. Therefore, you can't travel around the circuit once no matter where you start.
Constraints:
gas.length == ncost.length == n1 <= n <= 1050 <= gas[i], cost[i] <= 104
Companies:
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Let A[i] = gas[i] - cost[i]. The brute force way is to try each start index and see if the prefix sums starting from this index are always >= 0.
A better way is to loop at most twice over A. Whenever the prefix sum drops under 0, we simply discard this start index and use (i + 1) % N for the next trial.
// OJ: https://leetcode.com/problems/gas-station/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int N = gas.size(), len = 0, start = 0;
for (int i = 0, sum = 0; i < 2 * N && len < N; ++i) {
sum += gas[i % N] - cost[i % N];
if (sum < 0) { // sum drops under 0. Discard this start index and start over.
len = sum = 0;
start = (i + 1) % N;
} else ++len;
}
return len == N ? start : -1; // If we've found a sequence of length `N` whose prefix sums are all `>= 0`, this start index is the answer.
}
};This solution is similar to Solution 1, but it only loops through the arrays once.
The circuit can be completed if the sum of gas[i] - cost[i] is non-negative, which is denoted using total.
Use tank to denote the current available gas in tank, ans to denote the last possible answer.
When tank is negative, this point is a dead point, which means moving from ans to the current point is impossible.
The current point must have gas[i] - cost[i] < 0, so we'll need to try to start from next point -- reset ans to i + 1, and tank to 0.
When we reached the end of the arrays, if total is non-negative, ans is the start point.
Proof:
First of all, assume the start point we found is s. Then this algorithm guarantees we can reach point 0 from s.
How can we prove the last part of the roundtrip -- from 0 to s?
We use proof by contradiction to prove this works.
Assume there is a point k (0 < k <= s) that is unreachable in the roundtrip starting from s.
Let D[i] = gas[i] - cost[i], and SUM(i, j) = D[i] + D[i+1] + ... + D[j].
Since total >= 0, we have SUM(0, N-1) >= 0, i.e.:
SUM(0, k) + SUM(k+1, s-1) + SUM(s, N-1) >= 0
Let them be a, b, c respectively. We have a + b + c >= 0.
Since we can't reach k from s, we have c + a <= 0. And because c >= 0, we have a <= 0. Thus, this algorithm must reset the starting point when we scanned from 0 to k. And since we reset the starting point at s-1 again, we know b <= 0.
Given b <= 0 and a + b + c >= 0, we have a + c >= 0. This contradicts with c + a <= 0. So, there is no such k and we must be able to roundtrip from s.
// OJ: https://leetcode.com/problems/gas-station
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int tank = 0, total = 0, ans = 0, N = gas.size();
for (int i = 0; i < N; ++i) {
total += gas[i] - cost[i];
if ((tank += gas[i] - cost[i]) < 0) {
tank = 0;
ans = i + 1;
}
}
return total < 0 ? -1 : ans;
}
};Let S[i] = sum(gas[0] + ... + gas[i]) - sum(cost[0] + ... + cost[i]).
The circuit can be completed as long as sum(gas) - sum(cost) >= 0, i.e. S[N - 1] >= 0.
The S array forms a polygonal line. We should use x+1 as the start index where x is the index of the lowest point of the polygonal line.
// OJ: https://leetcode.com/problems/gas-station
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int minVal = INT_MAX, tank = 0, ans = 0, N = gas.size();
for (int i = 0; i < N; ++i) {
tank += gas[i] - cost[i];
if (tank < minVal) {
minVal = tank;
ans = i;
}
}
return tank >= 0 ? (ans + 1) % N : -1;
}
};