How to solve Leetcode 605. Can Place Flowers

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.

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.

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] + flowerbed[i] + flowerbed[i + 1] == 0) {
            flowerbed[i] = 1;
            n--;
            if (n == 0) {
                return true;
            }
        }
        i++;
    }
    return false;
}
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

Complexity

Runtime: O(N), where N = flowerbed.length.
Extra space: O(1).

Implementation note

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.

References

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