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    LeetCode link: 605. Can Place Flowers, difficulty: Easy.

    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 * 10^4
    • flowerbed[i] is 0 or 1.
    • There are no two adjacent flowers in flowerbed.
    • 0 <= n <= flowerbed.length

    Intuition
    PreviousNext

    Check each empty plot (0). If both adjacent plots are empty (or boundaries), plant a flower (set to 1) and count. Return true if the final count ≥ n, otherwise false.

    Pattern of "Greedy Algorithm"

    The Greedy Algorithm is a strategy that makes the locally optimal choice at each step with the hope of leading to a "globally optimal" solution. In other words, "local optima" can result in "global optima."

    Step-by-Step Solution

    1. Initialize counter count = 0.
    2. Iterate through each plot:
      • If current is 1, skip.
      • If current is 0 and both adjacent are 0 (or boundaries), plant (1), increment count.
    3. Return count >= n.

    Complexity

    Time complexity

    O(N)

    Space complexity

    O(1)

    Solution in languages: Python Java C++ JavaScript Go C# Ruby

    Python # 

    class Solution:
    def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
        count = 0

        for i in range(len(flowerbed)):
            if flowerbed[i] == 1:
                continue

            if (i == 0 or flowerbed[i - 1] == 0) and \
               (i == len(flowerbed) - 1 or flowerbed[i + 1] == 0):
                flowerbed[i] = 1
                count += 1

        return count >= n

    Java # 

    class Solution {
    public boolean canPlaceFlowers(int[] flowerbed, int n) {
        int count = 0;

        for (int i = 0; i < flowerbed.length; i++) {
            if (flowerbed[i] == 1) {
                continue;
            }

            if ((i == 0 || flowerbed[i - 1] == 0) && 
                (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)) {
                flowerbed[i] = 1;
                count++;
            }
        }

        return count >= n;
    }
}

    C++ # 

    class Solution {
public:
    bool canPlaceFlowers(vector<int>& flowerbed, int n) {
        int count = 0;

        for (int i = 0; i < flowerbed.size(); i++) {
            if (flowerbed[i] == 1) {
                continue;
            }

            if ((i == 0 || flowerbed[i - 1] == 0) && 
                (i == flowerbed.size() - 1 || flowerbed[i + 1] == 0)) {
                flowerbed[i] = 1;
                count++;
            }
        }

        return count >= n;
    }
};

    JavaScript # 

    var canPlaceFlowers = function(flowerbed, n) {
  let count = 0;

  for (let i = 0; i < flowerbed.length; i++) {
    if (flowerbed[i] === 1) {
      continue;
    }

    if ((i === 0 || flowerbed[i - 1] === 0) && 
      (i === flowerbed.length - 1 || flowerbed[i + 1] === 0)) {
      flowerbed[i] = 1;
      count++;
    }
  }

  return count >= n;  
};

    Go # 

    func canPlaceFlowers(flowerbed []int, n int) bool {
    count := 0

    for i := 0; i < len(flowerbed); i++ {
        if flowerbed[i] == 1 {
            continue
        }

        if (i == 0 || flowerbed[i - 1] == 0) && 
           (i == len(flowerbed) - 1 || flowerbed[i + 1] == 0) {
            flowerbed[i] = 1
            count++
        }
    }

    return count >= n
}

    C# # 

    public class Solution
{
    public bool CanPlaceFlowers(int[] flowerbed, int n)
    {
        int count = 0;

        for (int i = 0; i < flowerbed.Length; i++)
        {
            if (flowerbed[i] == 1)
            {
                continue;
            }

            if ((i == 0 || flowerbed[i - 1] == 0) && 
                (i == flowerbed.Length - 1 || flowerbed[i + 1] == 0))
                {
                flowerbed[i] = 1;
                count++;
            }
        }

        return count >= n;
    }
}

    Ruby # 

    def can_place_flowers(flowerbed, n)
  count = 0

  flowerbed.each_with_index do |plot, i|
    next if plot == 1

    if (i == 0 || flowerbed[i - 1] == 0) && 
       (i == flowerbed.length - 1 || flowerbed[i + 1] == 0)
      flowerbed[i] = 1
      count += 1
    end
  end

  count >= n
end
    Solution in languages: Python Java C++ JavaScript Go C# Ruby

    Other languages

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