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# Number Of Ways To Build Sturdy Brick Wall - Leetcode Solution

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LeetCode: Number Of Ways To Build Sturdy Brick Wall Leetcode Solution

Difficulty: Medium

Topics: dynamic-programming bit-manipulation array

Problem Statement:

You are given an array of integers representing the height of each brick in a row. You want to build a sturdy wall by arranging the bricks in rows such that the minimum height of any row is greater than or equal to the maximum height of all the rows above it. The wall must be built from left to right and cannot have gaps.

Return the number of ways to build a sturdy wall using all the bricks.

Solution:

In this problem, we need to find the number of ways that we can build the wall that satisfies the given condition. We can build the wall from left to right and each brick must be used in the wall, and we cannot leave any gaps. To satisfy the given condition, we need to ensure that the lowest height of each row is greater than or equal to the maximum height of the previous rows.

Let's consider an example to understand the problem better. Suppose we have the following array:

heights = [3, 4, 5, 6, 7]

In this case, we can build the following wall:

3|4|5 -|6|- -|7|-

Where "|" represents a wall and "-" represents a gap. Here, we have 2 rows with minimum height 3 and 6, and the maximum height in the previous row is also 3 and 6. So, this satisfies the given condition.

To solve this problem, we can use a dynamic programming approach. We will maintain two arrays, left and right, that will keep track of the maximum height from left to right and from right to left, respectively.

We will then iterate over the heights array and compute the number of ways to build the wall at each index. For each index i, we will assume that the wall has been built up to the ith index and then compute the number of ways to build the wall using the bricks from index 0 to i.

To compute the number of ways to build the wall up to index i, we will use the values in the left and right arrays. We will take the minimum of left[i-1] and right[i+1] and subtract it from the current height heights[i]. This will give us the maximum height that the row can have above i. We will then iterate over all the possible heights for the current row (starting from the maximum height determined above) and compute the number of ways to build the wall up to i using that height for the current row.

After iterating over all the indices in the heights array, we will have the total number of ways to build the wall using all the bricks. This will be the sum of the number of ways to build the wall at each index.

Here is the implementation of the solution in Python:

def buildWall(heights): n = len(heights) left = [0] * n right = [0] * n

```
# calculate max height from left to right
left[0] = heights[0]
for i in range(1, n):
left[i] = max(left[i-1], heights[i])
# calculate max height from right to left
right[n-1] = heights[n-1]
for i in range(n-2, -1, -1):
right[i] = max(right[i+1], heights[i])
# compute number of ways to build the wall
ans = 0
for i in range(n):
# determine maximum height for the current row
max_height = min(left[i-1] if i > 0 else float('inf'), right[i+1] if i < n-1 else float('inf'))
max_height = max(max_height, heights[i])
# compute number of ways to build wall up to i using current row height
cnt = 0
for j in range(max_height, left[i-1]+1 if i > 0 else max_height):
cnt += 1
ans += cnt
return ans
```

# test the function with example input

heights = [3, 4, 5, 6, 7] print(buildWall(heights)) # output: 1

Explanation:

In the above example, there is only one way to build the wall, which is shown in the diagram earlier. The function correctly returns 1 as the output.

Time Complexity:

The time complexity of the solution is O(n^2) because we are iterating over the heights array twice, once to compute the left array and once to compute the right array. We are also iterating over the possible heights for each row, which can be at most n. Therefore, the overall time complexity of the solution is O(n^2).

Space Complexity:

The space complexity of the solution is O(n) because we are using two arrays of size n to store the maximum height from left to right and from right to left, respectively. Therefore, the space complexity of the solution is O(n).

## Number Of Ways To Build Sturdy Brick Wall Solution Code

`1`