1488. Avoid Flood in The City

Problem Description

In this problem, we are given a scenario involving lakes that can collect rainwater and a series of days with varying weather conditions. The task is to devise a strategy to prevent any lake from overflowing, given knowledge of rainfall on specific lakes on each day.

We are provided with an array, rains, where each element corresponds to a day. If rains[i] > 0, it means lake rains[i] receives rain, and if rains[i] == 0, it signifies a sunny day with no rain, allowing us the opportunity to dry one lake of our choosing.

To avoid flooding, if a lake is already full and it begins to rain on it again, we must have dried it on a previous sunny day; otherwise, a flood will occur.

Our objective is to return an array, ans, of the same length as rains that records our actions:

• We set ans[i] to -1 if rains[i] > 0 to denote we can't perform any action other than observing the lake fill.
• On sunny days, i.e., when rains[i] == 0, ans[i] should be the index of the lake we choose to dry, if possible.

If there's no way to avoid a flood with the given sequence of rainy and sunny days, we must return an empty array. Otherwise, we may return any valid sequence of actions that prevents flooding.

Intuition

To solve this puzzle, we need to track when lakes get full and strategically use sunny days to prevent imminent floods. This requires recording the last day any lake received rain and planning for the next rain event on that lake. Since we must dry a lake before it overflows, we must act on a sunny day occurring after the last rainfall but before the next one on that same lake.

Here's how we can approach this problem:

• Keep track of all sunny days in a sorted fashion, to efficiently find future opportunities to dry lakes. We utilize a sorted data structure, like a SortedList, or a priority queue for this purpose.

• Use a hash table to remember the last rainy day for each lake. This will help us link future rainfalls to our records and identify necessary actions.

• Initialize an answer array, ans, filled with -1s, as our default action on rainy days is to do nothing.

As we traverse the rains array:

• For rainy days (where rains[i] > 0), check if the current lake (rains[i]) received rain in the past.
  • If it did, look for the next available sunny day after this past rainy day using binary search in the sorted sunny days list. Then, place the lake's index in the ans array on that sunny day to indicate we dry the lake to prevent a flood. If no suitable sunny day can be found, it means we cannot prevent a flood, and we should return an empty array.
  • Update the hash table with the current day for this lake.
• For sunny days (where rains[i] == 0):
  • Include this day in our sorted sunny days list for future potential use.
  • Set ans[i] to 1, indicating that we could dry any lake if necessary, though the actual choice is arbitrary in the absence of constraints.

After the entire rains array has been processed, return the ans array as the solution. This answer prevents floods through tactical lake drying on sunny days that appropriately follow rainy days. The solution's correctness relies on efficiently pairing sunny days to rainy days just in time to avert floods, which is a characteristic of the Greedy approach combined with Binary Search for optimization.

Learn more about Greedy, Binary Search and Heap (Priority Queue) patterns.

Solution Approach

The implementation is a fine blend of greedy strategy and efficient searching through the use of a binary search algorithm. Here's a step-by-step walkthrough of the approach used in the solution:

• Initialization: We begin by defining our answer array ans filled with -1, a SortedList sunny which will keep track of all sunny days, and a dictionary rainy which maps a lake (v) to the last day it rained on that lake.

• Traversing rains Array: We iterate over each element in the rains array. Each element rains[i] corresponds to a day's forecast; hence we handle two cases:

  • If it's a rainy day (rains[i] > 0):
    • Check if the lake has already been filled before by searching for it in the rainy dictionary.
    • If it's been filled, we need to find a day in sunny to dry it before it rains again. We use the bisect_right method, which effectively performs a binary search to find the index of the first sunny day after the last recorded rain on that lake.
    • If such a day doesn't exist in sunny (i.e., idx == len(sunny)), we have no way to dry the lake before the next rain, and thus, return an empty array to imply a flood is unavoidable.
    • Otherwise, allocate the lake number v to be dried on the found sunny day in ans, and then remove that day from sunny using the discard method since that day is now used up.
    • Update the rainy dictionary to record this latest rain day for the lake.
  • If it's a sunny day (rains[i] == 0):
    • Add the current day i to sunny, which maintains a sorted list of days available for drying lakes.
    • Mark ans[i] as 1, which is a placeholder to show that we can dry any lake of our choice, although the actual choice is made on actual rainy days.

• Return the Result: After processing all days in rains, return the constructed ans array. This array represents a sequence of decisions on each day to either observe rain (marked with -1) or to dry a lake (marked with the lake index) in a way that avoids all potential floods.

The greedy component of our solution assumes that we should always use the closest sunny day after a previous rain to dry a lake. This minimizes the days that lakes remain full and maximizes the flexibility for managing future rains.

The SortedList data structure is crucial here for efficiently managing sunny days and using binary search to quickly find the optimal day to dry a lake. The decision-making process leverages the assumption that it's always best to delay decisions until they are necessary, which is typical in greedy algorithms. In essence, we're deferring the drying of lakes until it's clear that no other option will avoid a flood.

Example Walkthrough

Let's take an example rains array to understand the solution approach practically:

Suppose rains = [1, 2, 0, 0, 2, 1].

Here is how we would process this array:

• Initialization:

  • ans = [-1, -1, -1, -1, -1, -1] because we assume it's raining on every day to begin.
  • sunny = SortedList() to keep track of sunny days.
  • rainy = {} to track the last day it rained on each lake.

• Day 1: rains[0] = 1

  • Lake 1 receives rain. We can't dry any lake today, so ans[0] = -1.
  • Update rainy to show the last rain day for lake 1: rainy[1] = 0.

• Day 2: rains[1] = 2

  • Lake 2 receives rain. We can't do anything, so ans[1] = -1.
  • Update rainy to show the last rain day for lake 2: rainy[2] = 1.

• Day 3: rains[2] = 0

  • No rain today, we can dry a lake if necessary. Add day 3 to sunny: sunny.add(2).
  • Update ans for this first sunny day as a placeholder: ans[2] = 1.

• Day 4: rains[3] = 0

  • Another sunny day, we can dry another lake if necessary. Add day 4 to sunny: sunny.add(3).
  • Update ans for the second sunny day as a placeholder: ans[3] = 1.

• Day 5: rains[4] = 2

  • Rain on lake 2 once again. We need to prevent a flood.
  • Look for the nearest sunny day after lake 2's last rain using bisect_right: bisect_right(sunny, rainy[2]). The nearest day is day 3.
  • Update ans[2] to show lake 2 is dried on day 3: ans[2] = 2.
  • Remove day 3 from sunny: sunny.remove(2).
  • Lake 2 is now filled with rain again, so ans[4] = -1 and update rainy: rainy[2] = 4.

• Day 6: rains[5] = 1

  • It's raining again on lake 1. We then look for the nearest sunny day after lake 1's last rain: bisect_right(sunny, rainy[1]). The nearest day is day 4.
  • Dry lake 1 on day 4: ans[3] = 1, and remove day 4 from sunny: sunny.remove(3).
  • Record the new state of lake 1: ans[5] = -1, and update rainy: rainy[1] = 5.

Our final ans array is [-1, -1, 2, 1, -1, -1], representing that we observed the rainfall on days when rains[i] > 0 and dried up lake 2 on the originally sunny day 3 (third position in ans array) and lake 1 on originally sunny day 4 (fourth position in ans array). This sequence ensures no lake overflow occurs.

The key element of this approach is the timely utilization of sunny days to dry the lakes that will receive more rain soon. We also need to ensure that we keep our records up to date at each step for both rainy and sunny days to maintain a correct and efficient process.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the solution is O(n * log n). This complexity arises due to the following reasons:

1. We iterate over all n elements in the rains list.
2. For each rainy day (non-zero element within rains), we use rainy[v] = i to store the last day a lake was filled which is O(1) for each operation.
3. For each sunny day (zero element within rains), we use sunny.add(i) which is an O(log n) operation since SortedList() maintains a sorted order.
4. When we need to find a sunny day to dry a lake, we perform sunny.bisect_right(rainy[v]) which is an O(log n) operation since it is a binary search within the sorted list of sunny days.
5. We then remove the used sunny day slot with sunny.discard(sunny[idx]). This operation is O(log n) as well, since removal from a SortedList requires a search for the element's index followed by the removal, both contributing to the log n complexity.

Since these operations take place within a loop that runs n times, the overall time complexity combines to O(n * log n).

Space Complexity

The space complexity of the solution is O(n). This is because:

1. We maintain an ans list of size n.
2. We have a SortedList named sunny which could potentially, in the worst case, hold a separate entry for every day which would also be O(n).
3. The rainy dictionary in the worst case will store an entry for every lake that gets filled which is bound by the number of days, so it is also O(n) space complexity.

Thus, the maximum of these space complexities dictates the overall space complexity, which is O(n).

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