LeetCode #605 — EASY

Can Place Flowers

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You have a long flowerbed in which some of the plots are planted, and some are not. However, flowers cannot be planted in adjacent plots.

Given an integer array flowerbed containing 0's and 1's, where 0 means empty and 1 means not empty, and an integer n, return true if n new flowers can be planted in the flowerbed without violating the no-adjacent-flowers rule and false otherwise.

Example 1:

Input: flowerbed = [1,0,0,0,1], n = 1
Output: true

Example 2:

Input: flowerbed = [1,0,0,0,1], n = 2
Output: false

Constraints:

1 <= flowerbed.length <= 2 * 10⁴
flowerbed[i] is 0 or 1.
There are no two adjacent flowers in flowerbed.
0 <= n <= flowerbed.length

Step 01

Brute Force Baseline

Problem summary: You have a long flowerbed in which some of the plots are planted, and some are not. However, flowers cannot be planted in adjacent plots. Given an integer array flowerbed containing 0's and 1's, where 0 means empty and 1 means not empty, and an integer n, return true if n new flowers can be planted in the flowerbed without violating the no-adjacent-flowers rule and false otherwise.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

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

Example 2

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

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #605: Can Place Flowers
class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        int m = flowerbed.length;
        for (int i = 0; i < m; ++i) {
            int l = i == 0 ? 0 : flowerbed[i - 1];
            int r = i == m - 1 ? 0 : flowerbed[i + 1];
            if (l + flowerbed[i] + r == 0) {
                flowerbed[i] = 1;
                --n;
            }
        }
        return n <= 0;
    }
}

// Accepted solution for LeetCode #605: Can Place Flowers
func canPlaceFlowers(flowerbed []int, n int) bool {
	m := len(flowerbed)
	for i, v := range flowerbed {
		l, r := 0, 0
		if i > 0 {
			l = flowerbed[i-1]
		}
		if i < m-1 {
			r = flowerbed[i+1]
		}
		if l+v+r == 0 {
			flowerbed[i] = 1
			n--
		}
	}
	return n <= 0
}

# Accepted solution for LeetCode #605: Can Place Flowers
class Solution:
    def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
        flowerbed = [0] + flowerbed + [0]
        for i in range(1, len(flowerbed) - 1):
            if sum(flowerbed[i - 1 : i + 2]) == 0:
                flowerbed[i] = 1
                n -= 1
        return n <= 0

// Accepted solution for LeetCode #605: Can Place Flowers
impl Solution {
    pub fn can_place_flowers(flowerbed: Vec<i32>, n: i32) -> bool {
        let (mut flowers, mut cnt) = (vec![0], 0);
        flowers.append(&mut flowerbed.clone());
        flowers.push(0);

        for i in 1..flowers.len() - 1 {
            let (l, r) = (flowers[i - 1], flowers[i + 1]);
            if l + flowers[i] + r == 0 {
                flowers[i] = 1;
                cnt += 1;
            }
        }
        cnt >= n
    }
}

// Accepted solution for LeetCode #605: Can Place Flowers
function canPlaceFlowers(flowerbed: number[], n: number): boolean {
    const m = flowerbed.length;
    for (let i = 0; i < m; ++i) {
        const l = i === 0 ? 0 : flowerbed[i - 1];
        const r = i === m - 1 ? 0 : flowerbed[i + 1];
        if (l + flowerbed[i] + r === 0) {
            flowerbed[i] = 1;
            --n;
        }
    }
    return n <= 0;
}

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(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

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.