LeetCode #1488 — MEDIUM

Avoid Flood in The City

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Your country has 10⁹ lakes. Initially, all the lakes are empty, but when it rains over the n^th lake, the n^th 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 must 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

Step 01

Brute Force Baseline

Problem summary: Your country has 109 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 must 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search · Greedy

Example 1

[1,2,3,4]

Example 2

[1,2,0,0,2,1]

Example 3

[1,2,0,1,2]

Step 02

Core Insight

What unlocks the optimal approach

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

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1488: Avoid Flood in The City
class Solution {
    public int[] avoidFlood(int[] rains) {
        int n = rains.length;
        int[] ans = new int[n];
        Arrays.fill(ans, -1);
        TreeSet<Integer> sunny = new TreeSet<>();
        Map<Integer, Integer> rainy = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            int v = rains[i];
            if (v > 0) {
                if (rainy.containsKey(v)) {
                    Integer t = sunny.higher(rainy.get(v));
                    if (t == null) {
                        return new int[0];
                    }
                    ans[t] = v;
                    sunny.remove(t);
                }
                rainy.put(v, i);
            } else {
                sunny.add(i);
                ans[i] = 1;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1488: Avoid Flood in The City
func avoidFlood(rains []int) []int {
	n := len(rains)
	ans := make([]int, n)
	for i := range ans {
		ans[i] = -1
	}

	sunny := redblacktree.New[int, struct{}]()
	rainy := map[int]int{}

	for i, v := range rains {
		if v > 0 {
			if last, ok := rainy[v]; ok {
				node, found := sunny.Ceiling(last + 1)
				if !found {
					return []int{}
				}
				t := node.Key
				ans[t] = v
				sunny.Remove(t)
			}
			rainy[v] = i
		} else {
			sunny.Put(i, struct{}{})
			ans[i] = 1
		}
	}
	return ans
}

# Accepted solution for LeetCode #1488: Avoid Flood in The City
class Solution:
    def avoidFlood(self, rains: List[int]) -> List[int]:
        n = len(rains)
        ans = [-1] * n
        sunny = SortedList()
        rainy = {}
        for i, v in enumerate(rains):
            if v:
                if v in rainy:
                    idx = sunny.bisect_right(rainy[v])
                    if idx == len(sunny):
                        return []
                    ans[sunny[idx]] = v
                    sunny.discard(sunny[idx])
                rainy[v] = i
            else:
                sunny.add(i)
                ans[i] = 1
        return ans

// Accepted solution for LeetCode #1488: Avoid Flood in The City
use std::collections::{BTreeSet, HashMap};

impl Solution {
    pub fn avoid_flood(rains: Vec<i32>) -> Vec<i32> {
        let n = rains.len();
        let mut ans = vec![-1; n];
        let mut sunny = BTreeSet::new();
        let mut rainy = HashMap::new();

        for (i, &v) in rains.iter().enumerate() {
            if v > 0 {
                if let Some(&last) = rainy.get(&v) {
                    if let Some(&t) = sunny.range((last + 1) as usize..).next() {
                        ans[t] = v;
                        sunny.remove(&t);
                    } else {
                        return vec![];
                    }
                }
                rainy.insert(v, i as i32);
            } else {
                sunny.insert(i);
                ans[i] = 1;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #1488: Avoid Flood in The City
import { AvlTree } from 'datastructures-js';

function avoidFlood(rains: number[]): number[] {
    const n = rains.length;
    const ans = Array(n).fill(-1);
    const sunny = new AvlTree<number>((a, b) => a - b);
    const rainy = new Map<number, number>();

    for (let i = 0; i < n; ++i) {
        const v = rains[i];
        if (v > 0) {
            if (rainy.has(v)) {
                const last = rainy.get(v)!;
                const node = sunny.ceil(last + 1);
                if (!node) return [];
                const t = node.getValue();
                ans[t] = v;
                sunny.remove(t);
            }
            rainy.set(v, i);
        } else {
            sunny.insert(i);
            ans[i] = 1;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.