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 noadjacentflowers rule and false
otherwise.
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
Initial Thought Process:
To solve this problem, we need to iterate through the flowerbed and check each position to determine if a flower can be planted. If a position is empty (0) and both adjacent positions are either empty or out of bounds, we can plant a flower there.
Basic Solution:
Code:
function canPlaceFlowersBruteForce(flowerbed: number[], n: number): boolean {
let count = 0;
for (let i = 0; i < flowerbed.length; i++) {
if (flowerbed[i] === 0) {
let prevEmpty = (i === 0)  (flowerbed[i  1] === 0);
let nextEmpty = (i === flowerbed.length  1)  (flowerbed[i + 1] === 0);
if (prevEmpty && nextEmpty) {
flowerbed[i] = 1;
count++;
if (count >= n) {
return true;
}
}
}
}
return count >= n;
}
Time Complexity Analysis:
 Time Complexity: O(n^2), where n is the length of the flowerbed array. The inner loop effectively makes this approach less efficient.
 Space Complexity: O(1), as we are modifying the flowerbed array in place and using only a constant amount of extra space.
Limitations:
The brute force solution is not optimal for larger input sizes due to its higher time complexity.
Optimized Solution:
The optimized solution will still iterate through the flowerbed array but will skip unnecessary checks by moving to the position after the next one once a flower is planted, ensuring we do not plant adjacent flowers.
Code:
function canPlaceFlowersOptimized(flowerbed: number[], n: number): boolean {
let count = 0;
let i = 0;
while (i < flowerbed.length) {
if (flowerbed[i] === 0 &&
(i === 0  flowerbed[i  1] === 0) &&
(i === flowerbed.length  1  flowerbed[i + 1] === 0)) {
flowerbed[i] = 1; // Plant a flower here
count++;
i += 2; // Move to the position after the next one
} else {
i++;
}
if (count >= n) {
return true;
}
}
return count >= n;
}
Time Complexity Analysis:
 Time Complexity: O(n), where n is the length of the flowerbed array. We iterate through the flowerbed array once.
 Space Complexity: O(1), as we are modifying the flowerbed array in place and using only a constant amount of extra space.
Improvements Over Basic Solution:
 The optimized solution skips unnecessary checks by moving to the position after the next one once a flower is planted, ensuring we do not plant adjacent flowers.
Edge Cases and Testing:
Edge Cases:
 The flowerbed is entirely empty.
 The flowerbed has alternating empty and nonempty plots.

n
is 0, meaning no new flowers need to be planted. 
n
is larger than the possible number of plantable positions in the flowerbed.
Test Cases:
console.log(canPlaceFlowersBruteForce([1,0,0,0,1], 1)); // true
console.log(canPlaceFlowersBruteForce([1,0,0,0,1], 2)); // false
console.log(canPlaceFlowersBruteForce([0,0,1,0,0], 1)); // true
console.log(canPlaceFlowersBruteForce([0,0,1,0,0], 2)); // true
console.log(canPlaceFlowersBruteForce([0,0,1,0,1], 1)); // false
console.log(canPlaceFlowersBruteForce([1,0,0,0,0,1], 1)); // true
console.log(canPlaceFlowersBruteForce([1,0,0,0,0,1], 2)); // false
console.log(canPlaceFlowersBruteForce([0,0,0,0,0,0], 3)); // true
console.log(canPlaceFlowersBruteForce([0,0,0,0,0,0], 4)); // false
console.log(canPlaceFlowersOptimized([1,0,0,0,1], 1)); // true
console.log(canPlaceFlowersOptimized([1,0,0,0,1], 2)); // false
console.log(canPlaceFlowersOptimized([0,0,1,0,0], 1)); // true
console.log(canPlaceFlowersOptimized([0,0,1,0,0], 2)); // true
console.log(canPlaceFlowersOptimized([0,0,1,0,1], 1)); // false
console.log(canPlaceFlowersOptimized([1,0,0,0,0,1], 1)); // true
console.log(canPlaceFlowersOptimized([1,0,0,0,0,1], 2)); // false
console.log(canPlaceFlowersOptimized([0,0,0,0,0,0], 3)); // true
console.log(canPlaceFlowersOptimized([0,0,0,0,0,0], 4)); // false
General ProblemSolving Strategies:
 Understand the Problem: Carefully read the problem statement to understand the requirements and constraints.
 Identify Key Operations: Determine the key operations needed, such as checking adjacent plots and planting flowers.
 Optimize for Readability: Use clear and concise logic to ensure the code is easy to follow.
 Test Thoroughly: Test the solution with various cases, including edge cases, to ensure correctness.
Identifying Similar Problems:

Array Manipulation:
 Problems where you need to modify elements of an array based on specific conditions.
 Example: Moving zeros to the end of an array.

Greedy Algorithms:
 Problems where a greedy approach can be used to find an optimal solution by making the best choice at each step.
 Example: Interval scheduling to find the maximum number of nonoverlapping intervals.

Simulation Problems:
 Problems where you need to simulate a process stepbystep based on given rules.
 Example: Simulating the spread of a virus in a population represented by an array.
Conclusion:
 The problem of determining if new flowers can be planted in a flowerbed without violating the noadjacentflowers rule can be efficiently solved using both a brute force approach and an optimized approach.
 Understanding the problem and breaking it down into manageable parts is crucial.
 Using clear logic and optimizing for readability ensures the solution is easy to follow.
 Testing with various edge cases ensures robustness.
 Recognizing patterns in problems can help apply similar solutions to other challenges.
By practicing such problems and strategies, you can improve your problemsolving skills and be better prepared for various coding challenges.
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