【LeetCode OJ】Gas Station

Problem link:

http://oj.leetcode.com/problems/gas-station/

---

We can solve this problem by following algorithm.

 CompleteCircuit(gas, cost):
  let n be the number of stations
  // Check if there exists i such that A[i,n] >= 0
  if (sum of gas) > (sum of cost)
    return -1  // There is no valid i such that A[i,n] is true

  // Find the first valid i
  i = 0  // start station
  tank = 0   // the amount of gas in tank
  x = 0  // initially, it is valid to go 0 steps from any station
  while x < n:
    // Each iteration, we have A[i,x] >= 0, and try to go further
    j = (i+x) mod n
    // refeul at gas[j] and then use cost[j]
    tank += gas[j] - cost[j]
    // If tank < 0, it is impossible for car to travel from j to its next station
　　 if tank < 0:
      // Set j's next station as the new start station
      i = (j+1) mod n
      x = 0
      tank = 0
    else:
      x += 1

   return i

---

Correctness

Now, we would prove the algorithm above can solve the problem correctly.

Define E[i,k] which is the amout of gas earned of the car starting from i and going k steps (k <= n)

E[i,k] = 0, if k=0
E[i,k] = gas_earn[i] + gas_earn[i+1] + ... + gas_earn[i+k-1], if 0 < k <= n
 where gas_earn[i] = gas[i mod n] - cost[i mod n]

So the problem converts to finding some i such that E[i,k]>=0 for k=0,1,...,n.

Firstly we prove that, there exists some i such that E[i,k] >= 0 for k=0,1,...n if and only if sum(gas[0..n-1]) >= sum(cost[0..n-1]).

(=>) If there exists i such that E[i,k] >= 0, for k=0,1,...n, then we have E[i,n] >= 0, that is

gas_earn[i] + gas_earn[i+1] + ... + gas_earn[i+n-1] >= 0
=> gas_earn[0] + gas_earn[1] + ... + gas_earn[n-1] >= 0
=> sum(gas[0..n-1]) - sum(cost[0..n-1]) >= 0

(<=) Assume sum(gas[0..n-1]) >= sum(cost[0..n01]) and we will show that there must exists i such that E[i,k]>=0 for k=0,...,n by induction.

1) For any circuit of n=1 station, we have

E[0,0] = 0
E[0,1] = gas_earn[0] = gas[0]-cost[0] = sum(gas[0..0]) - sum(cost[0..0]) >= 0

2) Now we already know that for any circuit of (n-1) stations, if sum(gas[0..n-2]) >= sum(cost[0..n-2]) then there must exist i, such that E[i,k]>=0 for k=0,...,n-1.

Consider a circuit of n stations with sum(gas[0..n-1]) >= sum(cost[0..n-1]).

Let j be the station with the minimum gas_earn. If gas_earn[j] >= 0, then every gas_earn is non-negative, we have A[i,n]=true for any i=0,1,..,n-1. So we only consider the case of gas_earn[j] < 0. By merging j and j+1 into one station as j' where gas_earn[j'] = gas_earn[j] + gas_earn[j+1], we can obtain a circuit of (n-1) stations with sum(gas) >= sum(cost). Let E'[i,x] denotethe the amout of gas earned of the car starting from i and going k steps (k <= n-1) in this circuit of (n-1) stations, and d is the number of steps to move from i to j (d > 0 since i != j). Then,

E[i,k] = E'[i,x] >= 0, for k = 0, 1, ..., d-1
E[i,k] = E'[i,k-1] >= 0, for k = d+1, ..., n

Now we only need to check E[i,d],

E[i,d] = gas_earn[i] + gas_earn[i+1] + ... + gas_earn[j-1] + gas_earn[j] 
E'[i,d] = gas_earn[i] + gas_earn[i+1] + ... + gas_earn[j-1] + gas_earn[j']
=> E[i,d] = E'[i,d] + gas_earn[j] - gas_earn[j'] = E'[i,d] - gas_earn[j]
=> E[i,d] >= 0 (since E'[i,d]>=0 and gas_earn[j]<0)

By the induction, we can claim that there must exist i such that E[i,k]>=0 for k=0,...,n if sum(gas) >= sum(cost).

Thus, the statement is proved.

Secondly, we will prove that, if E[i,k] >=0 for k=0,1,...,x-1 but E[i,x] < 0, then E[i+j, x-j] < 0 for j = 1, 2, ..., x-1.

According to the definition of E[i,k], we have

E[i,x] = gas_earn[i] + gas_earn[i+1] + ... + gas_earn[i+j] + ... + gas_earn[i+x-1]
E[i,j] = gas_earn[i] + gas_earn[i+1] + ... + gas_earn[i+j-1]
E[i+j,x-j] = gas_earn[i+j] + gas_earn[i+j] + ... + gas_earn[i+j+x-j-1]

So we have:

E[i+j,x-j] = E[i,x] - E[i,j] < 0

This means if we have E[i,k] >=0 for k=0,1,...,x-1 but E[i,x] < 0, then it is impossible for car to go n steps starting from i+1, ..., i+x-1. We only need to check E[i+x, k] for k = 0,...,n.

---

Python Implementation

class Solution:
    # @param gas, a list of integers
    # @param cost, a list of integers
    # @return an integer
    def canCompleteCircuit(self, gas, cost):
        # There exists i such that the car staring from i
        # can travel around the circuit *if and only if*
        # sum(gas) >= sum(cost)
        if sum(gas) < sum(cost):
            return -1

        n = len(gas)
        i = 0  # start station, can be selected randomly
        t = 0  # the amount of gas in the tank, initially 0
        x = 0  # the car go x stations further, initially 0
        # x=n means the car travel around and back to the start station
        while x < n:
            # The car is at station j, going x from i
            j = (i+x) % n
            # refeul gas[j] and spend cost[j] to get next station
            t += gas[j] - cost[j]
            # It is impossible to get next station, reset and start from j's next station
            if t < 0:
                t = 0
                x = 0
                i = (j+1) % n
            else:  # or go further
                x += 1

        return i