#605 - Can Place Flowers

EasyArray / String

Given a flowerbed represented as an integer array flowerbed containing 0s (empty) and 1s (planted), and an integer n, return true if n new flowers can be planted without violating the no-adjacent-flowers rule (no two flowers in neighboring plots).

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

Solution Space

Try it

/**
 * Determines if n flowers can be planted in the flowerbed greedily.
 * A plot is plantable if it and both neighbors are 0 (edges count as 0).
 *
 * @param flowerbed Array of 0s and 1s (1 ≤ length ≤ 2 × 10⁴)
 * @param n         Number of flowers to plant (0 ≤ n ≤ flowerbed.length)
 * @returns True if n flowers can be planted without adjacency violations
 */
export function canPlaceFlowers(flowerbed: number[], n: number): boolean {
  let count = 0;

  for (let i = 0; i < flowerbed.length; i++) {
    const emptyLeft = i === 0 || flowerbed[i - 1] === 0;
    const emptyRight = i === flowerbed.length - 1 || flowerbed[i + 1] === 0;

    if (flowerbed[i] === 0 && emptyLeft && emptyRight) {
      flowerbed[i] = 1; // plant here so it blocks the next slot
      count++;
      if (count >= n) return true;
    }
  }

  return count >= n;
}

Approach: greedy single-pass – iterate through the flowerbed and place a flower whenever the current, previous, and next plots are empty.

Time: O(n) – single traversal of the flowerbed.

Space: O(1) – modifies the array in place (no extra allocation).

Alternative approaches:

Approach	Idea	Time	Space
Count consecutive zeros	Count runs of zeros between 1s; derive plantable spots per run via ⌊(zeros - 1) / 2⌋	O(n)	O(1)

Why the greedy method is preferred:

• Simpler to implement and reason about; no edge-case math for boundary runs of zeros.

#443 String Compression

#1071 Greatest Common Divisor of Strings