DSA Visualizations | Hello Interview

Gas Station

DESCRIPTION (credit Leetcode.com)

There are n gas stations along a circular route. You are given two integer arrays gas and cost of length n. At each gas station i, gas[i] represents the amount of gas you receive by stopping at this station, and cost[i] represents the amount of gas required to travel from station i to the next station. You begin the journey with an empty tank at one of the gas stations.

Write a function to return the starting gas station's index if you can travel around the circuit once in the clockwise direction; otherwise, return -1. Note that if there exists a solution, it is guaranteed to be unique. Also, you can only travel from station i to station i + 1, and the last station will lead back to the first station.

EXAMPLES

Input:


gas = [5,2,0,3,3]
cost = [1,5,5,1,1]

Output:

Explanation:

Start at station 4 (index 3) and fill up with 3 units of gas. Your tank = 0 + 3 = 3 Travel to station 4 with 1 unit of gas, and fill up with 3 units of gas. Your tank = 3 - 1 + 3 = 5 Travel to station 0 with 1 unit of gas, and fill up with 5 units of gas. Your tank = 5 - 1 + 5 = 9 Travel to station 1 with 5 units of gas, and fill up with 2 units of gas. Your tank = 9 - 5 + 2 = 6 Travel to station 2 with 5 units of gas, and fill up with 0 units of gas. Your tank = 6 - 5 + 0 = 1 Travel back to station 3 with 1 unit of gas to complete the circuit. Therefore, return 3 as the starting index.

Solution


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

gas station

0 / 12

Explanation

If there is more gas along the route than the cost of the route, then there is guaranteed to be a solution to the problem. So the first step is to check if the sum of the gas is greater than or equal to the sum of the cost. If it is not, then we return -1.

Next, we iterate through the gas station to find the starting index of our circuit using a greedy approach: whenever we don't have enough gas to reach the next station, we move our starting gas station to the next station and reset our gas tank.

Walkthrough

We start at the first station, and fill our tank with gas[0] = 5 units of gas. From there, it takes cost[0] = 1 units of gas to travel to the next station, so we arrive at station 2 (index 1) with 4 units of gas.


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

i = 0

0 / 2

Traveling between stations 1 (i = 0) and 2 (i = 2)

At station 2, we fill our tank with gas[1] = 2 units of gas, for a total of 6 units of gas. It takes cost[1] = 5 units of gas to travel to the next station, so we arrive at station 3 with 1 unit of gas.


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

i = 1

0 / 2

Traveling between stations 2 (i = 1) and 3 (i = 2)

Greedy Approach

Now at station 3, we fill our tank with gas[2] = 0 units of gas, for a total of 1 unit of gas. It takes cost[2] = 5 units of gas to travel to the next station, which we don't have.

This is where our greedy approach comes in. We reset our starting station to the next station i + 1 and reset our gas tank to 0. We can do this because all other start indexes between 0 and 2 will also run into the same problem of not having enough gas to reach the next station, so we can rule them out.


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

i = 2

0 / 2

Resetting the start index

If we follow this approach of resetting the start index and gas tank whenever we don't have enough gas to reach the next station, then when we finish iterating, the last start index will be the solution to the problem.


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

update fuel

0 / 3

Complexity Analysis

Time complexity: O(n) because we only iterate through the gas stations once. Space complexity: O(1).

Next: Jump Game

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Gas Station

DESCRIPTION (credit Leetcode.com)

EXAMPLES

Input:


gas = [5,2,0,3,3]
cost = [1,5,5,1,1]

Output:

Explanation:

Solution


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

gas station

0 / 12

Explanation

Walkthrough


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

i = 0

0 / 2

Traveling between stations 1 (i = 0) and 2 (i = 2)


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

i = 1

0 / 2

Traveling between stations 2 (i = 1) and 3 (i = 2)

Greedy Approach

Now at station 3, we fill our tank with gas[2] = 0 units of gas, for a total of 1 unit of gas. It takes cost[2] = 5 units of gas to travel to the next station, which we don't have.


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

i = 2

0 / 2

Resetting the start index


def canCompleteCircuit(gas, cost):
  if sum(gas) < sum(cost):
    return -1

  start, fuel = 0, 0
  for i in range(len(gas)):
    if fuel + gas[i] - cost[i] < 0:
      # can't reach next station:
      # try starting from next station
      start, fuel = i + 1, 0
    else:
      # can reach next station:
      # update remaining fuel
      fuel += gas[i] - cost[i]

  return start

update fuel

0 / 3

Complexity Analysis

Time complexity: O(n) because we only iterate through the gas stations once. Space complexity: O(1).

Next: Jump Game

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