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  1. Prepare
  2. Data Structures
  3. Queues
  4. Truck Tour
  5. Discussions

Truck Tour

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363 Discussions

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  • al_bayona
     + 0 comments

    The problem is misleading, as it does not crearly states that there will always be a solution. You are led to believe to check if you go the full circle. You usually are taught to think of all scenarios, including no solution. So this was a very vague problem description. Finally, I was able to do it by using a prefix list in O(n) time, by using tabulation to optimize checking all the posible circles. 

    def truckTour(petrolpumps):
    _prefix = [ x[0] - x[1] for x in petrolpumps]
    _min = [0] * len(_prefix)
    _min2 = [0] * len(_prefix)

    print(_prefix)

    for i in range(1, len(_prefix)):
        _prefix[i] += _prefix[i-1]

    for i in range(len(_prefix)):
        if i == 0:
            _min[-(i+1)] = _prefix[-(i+1)]
            _min2[i] = _prefix[i]
        else:
            _min[-(i+1)] = min(_min[-i], _prefix[-(i+1)])
            _min2[i] = min(_min2[i-1], _prefix[i])

    # print(_prefix)
    # print(_min)
    # print(_min2)



    if _min[0] > 0:
        return 0

    for i, n in enumerate(_prefix):
        if i > 0:
            if _min[i] - _prefix[i-1] > 0 and _min2[i] + _prefix[-1] - _prefix[i-1] > 0:
                print(i)
                return i

  • manceraaxel
     + 0 comments
    def truckTour(petrol_pumps):
    start_index = 0
    total_liters = 0
    total_distance = 0

    for i in range(len(petrol_pumps)):
        liters, distance = petrol_pumps[i]
        total_liters += liters
        total_distance += distance
        # Arriving to next petrol pump

        if total_liters - total_distance < 0:
            # Initialize start index before restarting in start index
            start_index = i + 1
            total_liters = 0
            total_distance = 0
    
    return start_index

  • zyzz15620
     + 1 comment

    Imo this is very easy

    1. Loop through every pump
    2. Check if a pump has more petrol than the kilometer (1st condition)
    3. Test that pump. If during the tour, the remaining fuel >= 0 (2nd condition)
    4. Return index of the pump that passed 2 conditions note: remaining fuel = fuel - distance (kilometer)

    The challenge is how to test 2nd condition. If you want to test a pump number 2, than it need to go through [2 > 3 > 4 > 5 > 1] if there is 5 pumps. If during testing and that specific pump's remaining fuel get to below 0 than skip that pump

  • sivasubramaniy24
     + 0 comments

    def truckTour(petrolpumps):

    petrolRest = [pump[0]-pump[1] for pump in petrolpumps] * 2

for i in range(len(petrolpumps)):

    tank = 0

    for j in range(i, i + len(petrolpumps)):

        tank += petrolRest[j]

        if tank < 0: break     

    if tank < 0: continue

    else: return i

return -1

  • CSCYS1A_1720014
     + 0 comments

    Python Solution

    • def truckTour(petrolpumps):
    • fuel=0
    • pos=0
    • for i in range(len(petrolpumps)):
    • fuel+=petrolpumps[i][0]-petrolpumps[i][1]
    • if fuel<0:
    • fuel=0
    • pos=i+1
    • return pos