Leetcode problem 88 - Max Increase to Keep City Skyline

There is a city composed of n x n blocks, where each block contains a single building shaped like a vertical square prism. You are given a 0-indexed n x n integer matrix grid where grid[r][c] represents the height of the building located in the block at row r and column c.

A city’s skyline is the outer contour formed by all the building when viewing the side of the city from a distance. The skyline from each cardinal direction north, east, south, and west may be different.

We are allowed to increase the height of any number of buildings by any amount (the amount can be different per building). The height of a 0-height building can also be increased. However, increasing the height of a building should not affect the city’s skyline from any cardinal direction.

Return the maximum total sum that the height of the buildings can be increased by without changing the city’s skyline from any cardinal direction.

Example 1:

Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]] Output: 35 Explanation: The building heights are shown in the center of the above image. The skylines when viewed from each cardinal direction are drawn in red. The grid after increasing the height of buildings without affecting skylines is: gridNew = [ [8, 4, 8, 7], [7, 4, 7, 7], [9, 4, 8, 7], [3, 3, 3, 3] ] Example 2:

Input: grid = [[0,0,0],[0,0,0],[0,0,0]] Output: 0 Explanation: Increasing the height of any building will result in the skyline changing.

Constraints:

n == grid.length n == grid[r].length 2 <= n <= 50 0 <= grid[r][c] <= 100

/**
 * @param {number[][]} grid
 * @return {number}
 */
var maxIncreaseKeepingSkyline = function (grid) {
  // Initialize arrays to store maximum heights for each row and column
  let rows = [];
  let cols = [];

  // Iterate through each row of the grid
  for (let i = 0; i < grid.length; i++) {
    // Calculate and store the maximum height for the current row
    rows.push(Math.max(...grid[i]));

    // Nested loop to update the maximum heights in the cols array
    for (let j = 0; j < grid[i].length; j++) {
      // If cols[j] is already defined, update it to the maximum of the current value and the grid cell value
      if (cols[j]) {
        cols[j] = Math.max(cols[j], grid[i][j]);
      } else {
        // If cols[j] is not defined, set it to the grid cell value
        cols[j] = grid[i][j];
      }
    }
  }

  // Initialize a variable to store the total increased heights
  let result = 0;

  // Iterate through each cell of the grid
  for (let i = 0; i < grid.length; i++) {
    for (let j = 0; j < grid[i].length; j++) {
      // Calculate the minimum height between the maximum row height and the maximum column height
      // Subtract the original height of the cell to get the increased height
      result += Math.min(rows[i], cols[j]) - grid[i][j];
    }
  }

  // Return the total sum of increased heights
  return result;
};

The time complexity is O(n^2) because the function iterates through the entire grid twice: once to find the maximum heights for each row and column and again to calculate the total increased heights. The space complexity is O(n) because the function stores the maximum heights for each row and column in arrays of length n.