grid47 Exploring patterns and algorithms
Aug 18, 2024
6 min read
Solution to LeetCode 807: Max Increase to Keep City Skyline Problem
In a city grid, buildings are represented by an n x n matrix where each element corresponds to the height of a building at a specific location. You can increase the height of any building in the city grid, but the increased height should not alter the skyline of the city when viewed from any of the four cardinal directions (north, south, east, and west). Your task is to calculate the maximum sum of the heights that can be added to the buildings without changing the skyline.
Problem
Approach
Steps
Complexity
Input: The input consists of an n x n grid where each grid[i][j] represents the height of a building at row i and column j.
Output: Return the maximum sum that the heights of the buildings can be increased by without affecting the skyline.
Example: Output: 10
Constraints:
• The sum must be calculated based on the heights of the buildings without affecting the skyline from any direction.
Goal: To maximize the height of buildings without altering the skyline, we need to calculate the maximum possible height increase for each building by comparing it to the skyline views from each direction.
Steps:
• Find the maximum building heights for each row and column (north-south and west-east).
• For each building, calculate the maximum height it can reach based on the minimum of the corresponding row and column maximums.
• Sum up the differences between the calculated maximum heights and the original building heights to get the total increase in height.
Goal: Ensure that the grid meets the problem constraints for the grid size and values.
Steps:
• n is between 2 and 50.
• Each building height is between 0 and 100.
Assumptions:
• The height of any building can be increased, but the changes must respect the skyline constraints.
• Explanation: In this grid, the maximum sum that the heights of buildings can be increased by is 10. The calculated skyline views from all four directions (north, south, east, and west) are taken into account to ensure that increasing the height of any building does not alter the skyline.
Approach: To solve this problem, we will compute the maximum possible height for each building that maintains the city's skyline from all four cardinal directions. This involves finding the maximum building heights for each row and column, then using these to determine the permissible increase for each building.
Observations:
• We can maintain the skyline by respecting the maximum height from each direction.
• The problem requires iterating over the rows and columns to calculate the maximum heights from each direction and then determining the feasible height increase for each building.
Steps:
• Compute the maximum height in each row and each column.
• For each building, compute the maximum height it can reach by taking the minimum of the maximum row height and column height for that building.
• Sum the difference between the new maximum height and the current height of each building to get the total possible height increase.
Empty Inputs:
• If the grid has no buildings (n=0), the result will be 0.
Large Inputs:
• For large grids (n=50), ensure that the solution efficiently handles the size and performs within time limits.
Special Values:
• If all buildings are at the maximum possible height, no increase can be made.
Constraints:
• The solution must account for the constraints of grid size and building heights.
intmaxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
int n = grid.size();
vector<int> row(n, 0), col(n, 0);
for(int i =0; i < n; i++) {
for(int j =0; j < n; j++) {
row[i] = max(row[i], grid[i][j]);
col[j] = max(col[j], grid[i][j]);
}
}
int res =0;
for(int i =0; i < n; i++)
for(int j =0; j < n; j++) {
res += min(row[i], col[j]) - grid[i][j];
}
return res;
}
This defines the function `maxIncreaseKeepingSkyline` which takes a 2D vector `grid` as input and returns an integer. The function aims to calculate the maximum increase in grid height while keeping the skyline unchanged.
2 : Variable Initialization
int n = grid.size();
Here, we initialize the variable `n` which stores the size of the grid (i.e., the number of rows or columns, assuming the grid is square).
3 : Array Initialization
vector<int> row(n, 0), col(n, 0);
We create two vectors `row` and `col`, both of size `n`, initialized to 0. These will store the maximum values for each row and column in the grid, respectively.
4 : Outer Loop - Rows
for(int i =0; i < n; i++) {
We start an outer loop that iterates over each row of the grid.
5 : Inner Loop - Columns
for(int j =0; j < n; j++) {
This inner loop iterates over each column in the current row.
6 : Row Maximum Calculation
row[i] = max(row[i], grid[i][j]);
This line updates the maximum value for the current row `i` by comparing the current element `grid[i][j]` with the existing value stored in `row[i]`.
7 : Column Maximum Calculation
col[j] = max(col[j], grid[i][j]);
Similarly, this line updates the maximum value for the current column `j` by comparing the current element `grid[i][j]` with the existing value stored in `col[j]`.
8 : Result Initialization
int res =0;
We initialize a variable `res` to store the result, which will accumulate the total increase in height for the grid.
9 : Nested Loop for Calculation
for(int i =0; i < n; i++)
We start a loop to iterate over each row again to calculate the increase for each element.
10 : Inner Loop for Calculation
for(int j =0; j < n; j++) {
An inner loop is used to process each element in the grid for a given row.
11 : Grid Height Increase Calculation
res += min(row[i], col[j]) - grid[i][j];
For each element in the grid, we compute the maximum possible increase in height by subtracting the current value from the minimum of the maximum row and column heights.
12 : Return Result
return res;
Return the total calculated increase in grid heights.
Best Case: O(n^2), where n is the size of the grid.
Average Case: O(n^2), as each building must be processed individually.
Worst Case: O(n^2), since the grid size is bounded by n = 50, leading to a manageable number of operations.
Description: The time complexity is quadratic in the size of the grid due to the need to process both rows and columns.
Best Case: O(n), since we only need additional space for row and column maximums.
Worst Case: O(n), as space is needed to store the row and column maximums.
Description: The space complexity is linear with respect to the grid size, as we only need to store the row and column maximum heights.
