Can place flowers LeetCode?

Can Place Flowers LeetCode? – A Deep Dive into the Problem

Direct Answer: Yes, the LeetCode problem "Can Place Flowers" can be solved effectively, though the optimal solution requires careful consideration of the input and potential edge cases.

This problem presents a classic application of greedy algorithms and array manipulation, requiring you to assess whether you can plant flowers in a given garden bed under specific constraints.

Understanding the Problem Statement

The "Can Place Flowers" LeetCode problem typically involves a 1D array representing a garden bed where 0 represents an empty plot and 1 represents a plot already occupied. You are given a non-negative integer array flowerbed and an integer n, where:

• flowerbed describes the current state of the garden bed.
• n represents the number of flowers you need to plant.

You need to determine if it’s possible to plant n flowers in an empty plot following a specific rule: no two flowers can be placed in adjacent plots. A plant can only be placed on an empty plot.

Example Scenarios and Constraints

Let’s illustrate with examples:

• Example 1:

  • flowerbed = [1,0,0,0,1]
  • n = 1
  • Output: true (You can plant a flower in the middle)

• Example 2:

  • flowerbed = [1,0,0,0,1]
  • n = 2
  • Output: false (Not enough space between occupied plots)

• Example 3:

  • flowerbed = [0,0,1,0,0]
  • n = 1
  • Output: true (Plant the flower in the beginning)

Important Constraints: You must understand that these constraints will likely be in the problem statement.

• The input array flowerbed will always be of length greater than or equal to 1.
• The input array flowerbed will only contain 0s and 1s.
• n will always be a non-negative integer.

Approaches to Solving the Problem

Several strategies can be employed to solve this problem. While brute-force methods might work for smaller inputs, optimized solutions are required for larger arrays.

1. Iterative Approach using a single loop

This approach is often the most efficient, avoiding unnecessary calculations.

• Initialization: Counter initialized to 0 named placedFlowers. An n value to track the number of flowers to place and an index counter is needed i.

• Iteration: Iterate through the array. If a plot is empty (0), check the adjacent plots:

  • If both adjacent plots are empty (0), plant a flower at the current plot (increment placedFlowers), move the index by 2 (i += 2).
  • Otherwise, move to the next plot (i += 1).

• Comparison: After the loop, compares placedFlowers to n return True if they are equal or larger. False, otherwise.

2. Simplified Iterative Approach

This method avoids the explicit placedFlowers tracker.

• Initialization: Set placedFlowers = 0 and i = 0.
• Iteration: If flowerbed[i] == 0, check the adjacent plots.

  • If both adjacent plots are 0, plant a flower (flowerbed[i] = 1), increment placedFlowers and i by 2.
  • Otherwise, increment i.

• Comparison: Check if placedFlowers >= n and return the result.

Advantages: Efficiency; Easy implementation.

Code Example (using the Simplified Approach)

class Solution {

    public boolean canPlaceFlowers(int[] flowerbed, int n) {

        int count = 0;

        int i = 0;

        while (i < flowerbed.length) {

            if (flowerbed[i] == 0) {

                boolean leftAvailable = (i == 0) || (flowerbed[i - 1] == 0);

                boolean rightAvailable = (i == flowerbed.length - 1) || (flowerbed[i + 1] == 0);



                if (leftAvailable && rightAvailable) {

                    flowerbed[i] = 1;

                    count++;

                    i += 2;

                } else {

                    i++;

                }

            } else {

                i++;

            }

        }

        return count >= n;

    }

}

Time and Space Complexity

• Time Complexity: O(N), where N is the length of the flowerbed. This is because we iterate through the array once.
• Space Complexity: O(1). We are using a constant amount of extra space.

Conclusion

The "Can Place Flowers" problem on LeetCode is a great illustration of how a well-considered iterative approach can be employed to solve array-based problems effectively. Understanding edge cases and using conditional logic efficiently will help you produce solutions that will quickly and correctly deliver the correct response. The iterative method, specifically the simplified approach, prioritizes concise code while maintaining optimal time complexity.