1488. Avoid Flood in The City

1488. Avoid Flood in The City

Difficulty: Medium

Topics: Array, Hash Table, Binary Search, Greedy, Heap (Priority Queue), Weekly Contest 194

Your country has an infinite number of lakes. Initially, all the lakes are empty, but when it rains over the nᵗʰ lake, the nᵗʰ lake becomes full of water. If it rains over a lake that is full of water, there will be a flood. Your goal is to avoid floods in any lake.

Given an integer array rains where:

• rains[i] > 0 means there will be rains over the rains[i] lake.
• rains[i] == 0 means there are no rains this day, and you can choose one lake this day and dry it.

Return an array ans where:

• ans.length == rains.length
• ans[i] == -1 if rains[i] > 0.
• ans[i] is the lake you choose to dry in the iᵗʰ day if rains[i] == 0.

If there are multiple valid answers return any of them. If it is impossible to avoid flood return an empty array.

Notice that if you chose to dry a full lake, it becomes empty, but if you chose to dry an empty lake, nothing changes.

Example 1:

• Input: rains = [1,2,3,4]
• Output: [-1,-1,-1,-1]
• Explanation: After the first day full lakes are [1]
  • After the second day full lakes are [1,2]
  • After the third day full lakes are [1,2,3]
  • After the fourth day full lakes are [1,2,3,4]
  • There's no day to dry any lake and there is no flood in any lake.

Example 2:

• Input: rains = [1,2,0,0,2,1]
• Output: [-1,-1,2,1,-1,-1]
• Explanation: After the first day full lakes are [1]
  • After the second day full lakes are [1,2]
  • After the third day, we dry lake 2. Full lakes are [1]
  • After the fourth day, we dry lake 1. There is no full lakes.
  • After the fifth day, full lakes are [2].
  • After the sixth day, full lakes are [1,2].
  • It is easy that this scenario is flood-free. [-1,-1,1,2,-1,-1] is another acceptable scenario.

Example 3:

• Input: rains = [1,2,0,1,2]
• Output: []
• Explanation: After the second day, full lakes are [1,2]. We have to dry one lake in the third day.
  • After that, it will rain over lakes [1,2]. It's easy to prove that no matter which lake you choose to dry in the 3rd day, the other one will flood.

Constraints:

• 1 <= rains.length <= 10⁵
• 0 <= rains[i] <= 10⁹

Hint:

1. Keep An array of the last day there was rains over each city.
2. Keep an array of the days you can dry a lake when you face one.
3. When it rains over a lake, check the first possible day you can dry this lake and assign this day to this lake.

Solution:

We are given an array rains where:

• rains[i] > 0 means it rains on lake rains[i]
• rains[i] == 0 means we can dry one lake on that day

We need to return an array where:

• For rain days, we output -1
• For dry days, we output the lake we choose to dry

The goal is to avoid flooding any lake. If a lake is already full and it rains again, a flood occurs. We can use dry days to empty lakes before they receive rain again.

Approach

We can solve this problem using a Greedy Strategy with a Min-Heap:

1. Track Last Rain Day: Use a hash map to record the last day each lake received rain.
2. Use Min-Heap for Dry Days: When we encounter a dry day, we store its index in a min-heap so we can quickly access the earliest available dry day.
3. Handle Rain Days: When it rains on a lake:
   • If the lake was previously rained on (exists in our hash map), we need to find a dry day that occurred after the last rain on this lake to dry it before the current rain.
   • We use the min-heap to find the earliest dry day after the last rain day for this lake.
   • If no such dry day exists, return an empty array (flood is inevitable).
4. Update Last Rain Day: After processing rain on a lake, update the last rain day for that lake.

Let's implement this solution in PHP: 1488. Avoid Flood in The City

<?php
/**
 * @param Integer[] $rains
 * @return Integer[]
 */
function avoidFlood($rains) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
print_r(avoidFlood([1,2,3,4]));        // [-1,-1,-1,-1]
print_r(avoidFlood([1,2,0,0,2,1]));    // [-1,-1,2,1,-1,-1]
print_r(avoidFlood([1,2,0,1,2]));      // []
?>

Explanation:

1. Initialization:

   • $ans array initialized with -1 for all positions
   • $lastRain hash map to track last rain day for each lake
   • $dryDays min-heap to store indices of dry days

2. Processing Each Day:

   • Dry Day (rains[i] == 0): Store the day index in the min-heap and temporarily set ans[i] = 1
   • Rain Day (rains[i] > 0):
     • If the lake has rained before, find the earliest dry day after the last rain using the min-heap
     • If no suitable dry day is found, return empty array (flood inevitable)
     • Update the last rain day for the lake

3. Heap Management:

   • We extract dry days from the heap until we find one that occurs after the last rain
   • Dry days that are too early are stored back in the heap for future use

Complexity Analysis

• Time Complexity: O(n log n) - Each dry day is pushed and popped from the heap at most once
• Space Complexity: O(n) - For the heap and hash map storage

This approach efficiently ensures we use dry days optimally to prevent floods while handling the constraints effectively.

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