807. Max Increase to Keep City Skyline

Description

In a 2 dimensional array grid, each value grid[i][j] represents the height of a building located there. We are allowed to increase the height of any number of buildings, by any amount (the amounts can be different for different buildings). Height 0 is considered to be a building as well.

At the end, the “skyline” when viewed from all four directions of the grid, i.e. top, bottom, left, and right, must be the same as the skyline of the original grid. A city’s skyline is the outer contour of the rectangles formed by all the buildings when viewed from a distance. See the following example.

What is the maximum total sum that the height of the buildings can be increased?

Example:
Input: grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
Output: 35
Explanation:
The grid is:
[ [3, 0, 8, 4],
[2, 4, 5, 7],
[9, 2, 6, 3],
[0, 3, 1, 0] ]

The skyline viewed from top or bottom is: [9, 4, 8, 7]
The skyline viewed from left or right is: [8, 7, 9, 3]

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] ]

Notes:

1 < grid.length = grid[0].length <= 50.
All heights grid[i][j] are in the range [0, 100].
All buildings in grid[i][j] occupy the entire grid cell: that is, they are a 1 x 1 x grid[i][j] rectangular prism.

Problem URL

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Solution

给一个二维数组表示每个位置上楼的高度，可以为每一楼增加任意的高度，使得增加后的从上下左右看的天际线和之前一样。

So to maintain the original sky line. We could add it to the min(max of I row, max of j col). We use a for loop to calculate the max value of each row and each col first. Then in another for loop to calculate the difference and add it to result.

Code

class Solution {
    public int maxIncreaseKeepingSkyline(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        int m = grid.length;
        int n = grid[0].length;
        int[] rowMax = new int[m];
        int[] colMax = new int[n];
        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                rowMax[i] = Math.max(rowMax[i], grid[i][j]);
                colMax[j] = Math.max(colMax[j], grid[i][j]);
            }
        }
        int res = 0;
        for (int i = 0; i < m; i++){
            for (int j = 0; j < n; j++){
                res += Math.min(rowMax[i], colMax[j]) - grid[i][j];
            }
        }
        return res;
    }
}

Time Complexity: O(mn)
Space Complexity: O(m+n)

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