LeetCode #807 — MEDIUM

Max Increase to Keep City Skyline

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]

Example 2

[[0,0,0],[0,0,0],[0,0,0]]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #807: Max Increase to Keep City Skyline
class Solution {
    public int maxIncreaseKeepingSkyline(int[][] grid) {
        int m = grid.length, 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 ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #807: Max Increase to Keep City Skyline
func maxIncreaseKeepingSkyline(grid [][]int) (ans int) {
	rowMax := make([]int, len(grid))
	colMax := make([]int, len(grid[0]))
	for i, row := range grid {
		for j, x := range row {
			rowMax[i] = max(rowMax[i], x)
			colMax[j] = max(colMax[j], x)
		}
	}
	for i, row := range grid {
		for j, x := range row {
			ans += min(rowMax[i], colMax[j]) - x
		}
	}
	return
}

# Accepted solution for LeetCode #807: Max Increase to Keep City Skyline
class Solution:
    def maxIncreaseKeepingSkyline(self, grid: List[List[int]]) -> int:
        row_max = [max(row) for row in grid]
        col_max = [max(col) for col in zip(*grid)]
        return sum(
            min(row_max[i], col_max[j]) - x
            for i, row in enumerate(grid)
            for j, x in enumerate(row)
        )

// Accepted solution for LeetCode #807: Max Increase to Keep City Skyline
struct Solution;

impl Solution {
    fn max_increase_keeping_skyline(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();
        let mut row: Vec<i32> = vec![0; n];
        let mut col: Vec<i32> = vec![0; m];
        let mut res = 0;
        for i in 0..n {
            for j in 0..m {
                row[i] = row[i].max(grid[i][j]);
                col[j] = col[j].max(grid[i][j]);
            }
        }
        for i in 0..n {
            for j in 0..m {
                res += row[i].min(col[j]) - grid[i][j];
            }
        }
        res
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[3, 0, 8, 4], [2, 4, 5, 7], [9, 2, 6, 3], [0, 3, 1, 0]];
    let res = 35;
    assert_eq!(Solution::max_increase_keeping_skyline(grid), res);
}

// Accepted solution for LeetCode #807: Max Increase to Keep City Skyline
function maxIncreaseKeepingSkyline(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const rowMax = Array(m).fill(0);
    const colMax = Array(n).fill(0);
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            rowMax[i] = Math.max(rowMax[i], grid[i][j]);
            colMax[j] = Math.max(colMax[j], grid[i][j]);
        }
    }
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans += Math.min(rowMax[i], colMax[j]) - grid[i][j];
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.