Can Place Flowers

• If n is 0, return true immediately.
• Create a copy of the flowerbed array to avoid modifying the original array while iterating.
• Initialize a counter count to 0, which will track the number of flowers planted.
• Iterate through the copied flowerbed from the first plot to the last.
• For each plot i, if it's empty (flowerbed[i] == 0):
  • a. Check if the previous plot is empty. This condition is met if i is the first plot (i == 0) or if flowerbed[i - 1] == 0.
  • b. Check if the next plot is empty. This condition is met if i is the last plot (i == flowerbed.length - 1) or if flowerbed[i + 1] == 0.
  • c. If both neighboring plots are empty, plant a flower by setting flowerbed[i] = 1 in the copy and increment count.
• After planting a flower, check if count has reached n. If it has, we can stop early and return true.
• If the loop completes, return true if count >= n and false otherwise.

The core idea is to simulate the process of planting flowers without altering the original array during the decision-making process for each plot. We create a temporary copy of the flowerbed array. Then, we iterate through the flowerbed from left to right. For each empty plot flowerbed[i] == 0, we check its neighbors. The neighbors are at i-1 and i+1. We must handle the edge cases where i is 0 or the last index. If a plot at index i is empty and both its neighbors are also empty (or it's an edge), we can plant a flower there. We increment a counter for planted flowers and update the copy of the array at index i to 1. This ensures that when we check the next plot, our decision is based on the updated state of the flowerbed. Finally, we check if the total count of flowers we could plant is greater than or equal to n.

1class Solution {2    public boolean canPlaceFlowers(int[] flowerbed, int n) {3        if (n == 0) {4            return true;5        }6        int[] temp = new int[flowerbed.length];7        System.arraycopy(flowerbed, 0, temp, 0, flowerbed.length);8        9        int count = 0;10        for (int i = 0; i < temp.length; i++) {11            if (temp[i] == 0) {12                // Check previous plot13                boolean prevEmpty = (i == 0) || (temp[i - 1] == 0);14                // Check next plot15                boolean nextEmpty = (i == temp.length - 1) || (temp[i + 1] == 0);16                17                if (prevEmpty && nextEmpty) {18                    temp[i] = 1; // Plant the flower in the copy19                    count++;20                    if (count >= n) {21                        return true;22                    }23                }24            }25        }26        return count >= n;27    }28}

• Initialize a counter count to 0.
• Iterate through the flowerbed array from i = 0 to length - 1.
• At each position i, check if a flower can be planted:
  • a. The current plot flowerbed[i] must be empty (0).
  • b. The previous plot must be empty. This is true if i is the first plot (i == 0) or flowerbed[i - 1] == 0.
  • c. The next plot must be empty. This is true if i is the last plot (i == flowerbed.length - 1) or flowerbed[i + 1] == 0.
• If all three conditions are met, it's a valid spot. Plant a flower by modifying the array (flowerbed[i] = 1) and increment count.
• After incrementing count, check if count is already greater than or equal to n. If so, we have found enough spots, and we can return true immediately.
• If the loop finishes, it means we have checked all plots. We return true if we were able to plant at least n flowers (count >= n), and false otherwise.

The greedy strategy is to iterate through the flowerbed and plant a flower at the first available spot. An available spot at index i is one where flowerbed[i] is 0, and its adjacent plots (i-1 and i+1) are also 0. We must handle the boundaries carefully: for the first plot (i=0), we only need to check the next plot (i+1), and for the last plot, we only need to check the previous one (i-1).

When we find such a spot, we plant a flower by setting flowerbed[i] = 1 and increment our count of planted flowers. This modification correctly affects the check for the next plot i+1, as its preceding plot is now occupied. We continue this until we have planted n flowers (at which point we can return true early) or have scanned the entire flowerbed.

1class Solution {2    public boolean canPlaceFlowers(int[] flowerbed, int n) {3        int count = 0;4        for (int i = 0; i < flowerbed.length; i++) {5            // Check if current plot is available6            if (flowerbed[i] == 0) {7                // Check left and right plots8                boolean emptyLeftPlot = (i == 0) || (flowerbed[i - 1] == 0);9                boolean emptyRightPlot = (i == flowerbed.length - 1) || (flowerbed[i + 1] == 0);10                11                if (emptyLeftPlot && emptyRightPlot) {12                    flowerbed[i] = 1; // Plant flower13                    count++;14                    if (count >= n) {15                        return true;16                    }17                }18            }19        }20        return count >= n;21    }22}