Plant Flowers Without Adjacency

easy array greedy

Problem

You have a row of flower beds where each bed is empty (0) or already planted (1). No two flowers may sit in adjacent beds. Decide whether n additional flowers can be planted. Walk left to right, planting greedily wherever the bed and both neighbours are empty.

Inputbed = [1, 0, 0, 0, 1], n = 1

Outputtrue

Plant in position 2: result becomes [1, 0, 1, 0, 1].

bed (comma-separated 0/1)

def can_place(bed, n):
    i = 0
    while i < len(bed) and n > 0:
        if bed[i] == 0 \
                and (i == 0 or bed[i - 1] == 0) \
                and (i == len(bed) - 1 or bed[i + 1] == 0):
            bed[i] = 1; n -= 1
        i += 1
    return n == 0

function canPlace(bed, n) {
  for (let i = 0; i < bed.length && n > 0; i++) {
    if (bed[i] === 0
        && (i === 0 || bed[i - 1] === 0)
        && (i === bed.length - 1 || bed[i + 1] === 0)) {
      bed[i] = 1; n--;
    }
  }
  return n === 0;
}

class Solution {
    public boolean canPlace(int[] bed, int n) {
        for (int i = 0; i < bed.length && n > 0; i++) {
            if (bed[i] == 0
                    && (i == 0 || bed[i - 1] == 0)
                    && (i == bed.length - 1 || bed[i + 1] == 0)) {
                bed[i] = 1; n--;
            }
        }
        return n == 0;
    }
}

bool canPlace(vector<int>& bed, int n) {
    for (int i = 0; i < (int)bed.size() && n > 0; i++) {
        if (bed[i] == 0
                && (i == 0 || bed[i - 1] == 0)
                && (i == (int)bed.size() - 1 || bed[i + 1] == 0)) {
            bed[i] = 1; n--;
        }
    }
    return n == 0;
}

Time: O(L) Space: O(1)