2021. Brightest Position on Street

Problem Description

In this problem, we are provided with a representation of a street and its street lamps. The street is visualized as a number line, and street lamps are given as a 2D integer array lights. Each entry in this array, lights[i], contains two integers: position_i and range_i. The first integer position_i indicates the location of the street lamp on the number line, while the second integer range_i tells us how far to the left and right the lamp illuminates from its position. Consequently, the area that each street lamp lights up is from position_i - range_i to position_i + range_i inclusive.

The problem defines the brightness of a position on the street as the number of street lamps that illuminate that particular position. Our task is to determine the brightest position on the street, which is the position that is lit by the highest number of street lamps. If there happens to be more than one position with the same maximum brightness, the problem asks us to return the smallest of these brightest positions.

Intuition

The intuition behind solving this problem involves treating the street as a range of positions and each lamp as contributing a unit of brightness to a specific range. We can use sweep line algorithm concepts to efficiently determine the brightness of each position. Here's how we can approach the problem:

1. Consider each lamp as triggering an 'event' at the start and end of its range.
2. When a lamp starts at position_i - range_i, it adds to the brightness (+1), and when its influence ends at position_i + range_i + 1, it subtracts from the brightness (-1). This is because the range is inclusive, so the end of the influence is technically one position beyond the range.
3. We can sweep over all the positions influenced by any lamp, and at each point, we can add or subtract the brightness accordingly.
4. We use a dictionary to keep track of all these 'events' of starting and ending of a lamp's influence, then sort this dictionary by the key (which represents positions on the street).
5. We sweep the sorted positions and maintain a running sum (s) of the brightness, updating the maximum brightness (mx) seen so far and the position (ans) where this maximum brightness occurs.
6. At the end of the sweep, ans will hold the smallest position with the maximum brightness because we sorted the positions and the calculation respects the order.

The implementation translates this intuition into a working solution, utilizing Python's defaultdict to facilitate the management of the brightness changes and keeping track of the running sum and maximum brightness dynamically as we sweep through the positions.

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Solution Approach

The implementation of the solution uses a variety of data structures and an intelligent algorithm to achieve an efficient approach.

• Data Structures Used:

  • defaultdict(int): A Python defaultdict with int type is used to track the starting and ending influence points of each lamp's light on the street. If an index is not already in the dictionary, it defaults to 0 which is convenient because it represents the initial state of brightness for any position on the street.

• Algorithms and Patterns:

  • Sweep Line Algorithm: This algorithm is used to simulate the process of 'sweeping' through the street from left to right. As we 'sweep' through the street, we encounter events that start or end a lamp's range.
  • Event Processing: Each lamp generates two events: one where its light begins (position_i - range_i) and one where it ends (position_i + range_i + 1). In the defaultdict, we increase the brightness by 1 at the start and decrease it by 1 at the end of the range.
  • Sorting the Events: Sorting the keys of the defaultdict ensures that we process these events in the order of their occurrence along the street.
  • Running Sum and Maximization: In the sweep process, a running sum s is maintained which updates at each event, increasing or decreasing according to the lamp's influence. We then compare this running sum to the maximum brightness mx encountered so far. If the current sum is greater, we update the mx and record the position ans at which this new maximum occurs.

The code reflects these concepts as follows:

• We iterate through each lamp in lights and calculate the starting (l) and ending (r) points of its range, updating the defaultdict accordingly.
• We then sort the keys of the defaultdict (the positions where brightness changes occur) and perform the 'sweep' by iteratively adding or subtracting from the running sum s.
• Whenever the running sum s exceeds the previously tracked maximum brightness mx, we update mx with s and set ans to the current position k.

As a result, the code does not require us to calculate the brightness for every single position on the street. Instead, it efficiently tracks changes in brightness at specific points, which allows us to find the brightest position without redundant calculations.

Example Walkthrough

Let's go through the solution approach with a small example. Suppose we are given the following lights array representing the street lamps:

1lights = [[1, 1], [4, 1]]

Each of the subarrays represents a street lamp with [position_i, range_i]. Now, we will walk through the algorithm step by step.

1. Initialization of Events:

   • For the first lamp [1, 1], it influences the street in the range [0, 2] (from position_i - range_i to position_i + range_i).
   • For the second lamp [4, 1], it influences the street in the range [3, 5].

2. Creating and Populating Event Points:

   • We treat the start and end of each range as events. We use a defaultdict to store the brightness increase and decrease at specific points:

   1events = defaultdict(int)
   2events[0] += 1
   3events[3] += 1
   4events[2+1] -= 1  # We add 1 because the range end is inclusive
   5events[5+1] -= 1  # Same as above

3. Sorting Events:

   • We now have the events {0: 1, 3: 1, 3: -1, 6: -1}.
   • Sorting them by key gives us the order [0, 3, 3, 6].

4. Sweeping Through Sorted Event Points:

   • We begin with a running sum s = 0, maximum brightness mx = 0, and the answer ans = 0.
   • At position 0, we encounter the start of the first lamp’s influence, so s += events[0] (now s = 1).
   • At position 3, we have two events. The start of the second lamp’s influence (s += events[3], now s = 2) and the end of the first lamp’s influence (s += events[3], now s = 1).
   • At position 6, we encounter the end of the second lamp’s influence, so s -= events[6] (now s = 0).
   • After processing each event, we update mx and ans whenever we find a new maximum brightness.

5. Finding the Brightest Position:

   • During the process, we found the maximum brightness to be mx = 2 at positions 3, but since there are two events at 3, this number decreases back to 1 by the end of processing all events at 3.
   • ans during the occurrence of mx is set to the position at which this happened, which in this case, initially, is 3.

Hence, by the end of the sweep, we determined that the brightest position on the street is 3 with a brightness of 2, and that is our final answer.

By following this approach, we only calculate the brightness at specific event points instead of for every position on the street, which makes the algorithm efficient and effective for solving this kind of problem.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(N log N) where N is the number of light ranges in the lights list. This complexity arises because we sort the keys of our dictionary d, which contains at most 2N keys (each light contributes two keys: the start and end of its illumination range). Sorting these keys dominates the runtime complexity.

The space complexity of the code is O(N) since we use a dictionary to store the changes to brightness at each key point. In the worst case, if every light has a unique range, the dictionary could have as many as 2N keys, where N is the number of light ranges in the lights list.

Learn more about how to find time and space complexity quickly using problem constraints.