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Problem solution · Java

Avoid Flood in The City

Avoid Flood in The City: a Java solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
36 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Hash-based lookup

For Avoid Flood in The City, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 36 lines of Java from the credited upstream file 1488.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 2 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeAvoid Flood in The City · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int[] avoidFlood(int[] rains) {    int[] ans = new int[rains.length];    Arrays.fill(ans, -1);    Map<Integer, Integer> lakeIdToFullDay = new HashMap<>();    TreeSet<Integer> emptyDays = new TreeSet<>(); // indices of rains[i] == 0     for (int i = 0; i < rains.length; ++i) {      final int lakeId = rains[i];      if (lakeId == 0) {        emptyDays.add(i);        continue;      }      if (lakeIdToFullDay.containsKey(lakeId)) {        final int fullDay = lakeIdToFullDay.get(lakeId);        // The lake was full in a previous day. Greedily find the closest day        // to make the lake empty.        Integer emptyDay = emptyDays.higher(fullDay);        if (emptyDay == null) // Not found.          return new int[] {};        // Empty the lake at this day.        ans[emptyDay] = lakeId;        emptyDays.remove(emptyDay);      }      // The lake with `lakeId` becomes full at the day `i`.      lakeIdToFullDay.put(lakeId, i);    }     // Empty an arbitrary lake if there are remaining empty days.    for (final int emptyDay : emptyDays)      ans[emptyDay] = 1;     return ans;  }}

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