807. Max Increase to Keep City Skyline LeetCode Solution

Last updated: February 20, 2025

In this guide, you will get 807. Max Increase to Keep City Skyline LeetCode Solution with the best time and space complexity. The solution to Max Increase to Keep City Skyline problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Max Increase to Keep City Skyline solution in C++
Max Increase to Keep City Skyline solution in Java
Max Increase to Keep City Skyline solution in Python
Additional Resources

807. Max Increase to Keep City Skyline LeetCode Solution image

Problem Statement of 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

Complexity Analysis

Time Complexity: O(n^2)
Space Complexity: O(n)

807. Max Increase to Keep City Skyline LeetCode Solution in C++

class Solution {
 public:
  int maxIncreaseKeepingSkyline(vector<vector<int>>& grid) {
    const int n = grid.size();
    int ans = 0;
    vector<int> rowMax(n);
    vector<int> colMax(n);

    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) {
        rowMax[i] = max(rowMax[i], grid[i][j]);
        colMax[j] = max(colMax[j], grid[i][j]);
      }

    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        ans += min(rowMax[i], colMax[j]) - grid[i][j];

    return ans;
  }
};
/* code provided by PROGIEZ */

807. Max Increase to Keep City Skyline LeetCode Solution in Java

class Solution {
  public int maxIncreaseKeepingSkyline(int[][] grid) {
    final int n = grid.length;
    int ans = 0;
    int[] rowMax = new int[n];
    int[] colMax = new int[n];

    for (int i = 0; i < n; ++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]);
      }

    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j)
        ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];

    return ans;
  }
}
// code provided by PROGIEZ

807. Max Increase to Keep City Skyline LeetCode Solution in Python

class Solution:
  def maxIncreaseKeepingSkyline(self, grid: list[list[int]]) -> int:
    rowMax = list(map(max, grid))
    colMax = list(map(max, zip(*grid)))
    return sum(min(i, j) for i in rowMax for j in colMax) - sum(map(sum, grid))
 # code by PROGIEZ

Additional Resources

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