Can Place Flowers#

Problem statement#

You are presented with a long flowerbed containing plots, some of which are planted with flowers (denoted by 1) and some are empty (denoted by 0). Flowers cannot be planted in adjacent plots. You are given an integer array flowerbed representing the layout of the flowerbed, and an integer n representing the number of new flowers you want to plant.

Your task is to determine if it is possible to plant n new flowers in the flowerbed without violating the rule of no-adjacent-flowers. If it is possible, return true; otherwise, return false.

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.

Solution: Check the no-adjacent-flowers rule#

A new flower can be planted at position i only if

flowerbed[i - 1] == 0 && flowerbed[i] == 0 && flowerbed[i + 1] == 0.

If the condition is satisfied, the flower can be planted at position i. flowerbed[i] is now assigned to 1. Then you can skip checking the rule for the position i + 1 and go directly to position i + 2.

Code#

#include <iostream>
#include <vector>
using namespace std;
bool canPlaceFlowers(vector<int>& flowerbed, int n) {
    if (n == 0) {
        return true;
    }
    flowerbed.insert(flowerbed.begin(), 0);
    flowerbed.push_back(0);
    int i = 1;
    while (i < flowerbed.size() - 1) {
        if (flowerbed[i - 1] == 0 && flowerbed[i] == 0 && flowerbed[i + 1] == 0) {
            // plant i if it satisfies the no-adjacent condition
            flowerbed[i] = 1;
            n--;
            i+=2;
        } else {
            i++;
        }
    }
    return n <= 0; // have planted all n
}
int main() {
    vector<int> flowerbed{1,0,0,0,1};
    cout << canPlaceFlowers(flowerbed, 1) << endl;
    flowerbed = {1,0,0,0,1};
    cout << canPlaceFlowers(flowerbed, 2) << endl;
}

Output:
1
0

This solution efficiently iterates through the flowerbed, planting flowers wherever possible while adhering to the constraints. It returns true if it’s possible to plant all the required flowers and false otherwise.

Complexity#

  • Runtime: O(N), where N = flowerbed.length.

  • Extra space: O(1).

Implementation tips#

  • In this implementation, you could insert element 0 to the front and the back of vector flowerbed to avoid writing extra code for checking the no-adjacent-flowers rule at i = 0 and i = flowerbed.size() - 1.

  • There are a few ways to insert an element to a vector. Here you can see an example of using the methods insert and push_back of a std::vector.

Ecercise#