LeetCode 1840: Maximum Building Height Solution

Master LeetCode problem 1840 (Maximum Building Height), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1840. Maximum Building Height

Hard

Array Math Sorting

Problem Explanation

Explanation:

To solve this problem, we can use binary search to find the maximum possible height of the tallest building. We need to consider the restrictions and compute the maximum height that can be achieved. The main idea is to iterate through the restrictions and calculate the maximum allowed height at each position based on the adjacent restrictions.

Sort the restrictions by their position.
Initialize two arrays left and right to store the maximum height achievable from the left and right side respectively.
For each restriction, calculate the maximum height that can be achieved at that position based on the adjacent restrictions.
The maximum possible height at a position i is the minimum of the heights calculated from the left and right side.
Use binary search to find the maximum possible height of the tallest building.

Solution Code

class Solution {
    public int maxBuilding(int n, int[][] restrictions) {
        Arrays.sort(restrictions, (a, b) -> a[0] - b[0]);
        
        int m = restrictions.length;
        int[] left = new int[m + 1];
        int[] right = new int[m + 1];
        
        left[0] = 0;
        right[0] = 0;
        
        for (int i = 0; i < m; i++) {
            left[i + 1] = Math.min(restrictions[i][0] - 1, restrictions[i][1]);
        }
        
        for (int i = m - 1; i >= 0; i--) {
            right[i] = Math.min(restrictions[i][0] - 1, restrictions[i][1]);
        }
        
        for (int i = 1; i <= m; i++) {
            left[i] = Math.min(left[i], left[i - 1] + restrictions[i - 1][0] - restrictions[i - 2][0]);
        }
        
        for (int i = 1; i < m; i++) {
            right[i] = Math.min(right[i], right[i + 1] + restrictions[i][0] - restrictions[i - 1][0]);
        }
        
        int maxHeight = 0;
        
        for (int i = 0; i < m; i++) {
            int leftHeight = left[i];
            int rightHeight = right[i];
            int distance = restrictions[i][0] - restrictions[i - 1][0];
            int totalHeight = restrictions[i][1] + Math.max(leftHeight, rightHeight) + (distance - Math.abs(leftHeight - rightHeight)) / 2;
            maxHeight = Math.max(maxHeight, totalHeight);
        }
        
        int lastPosition = restrictions[m - 1][0];
        int lastHeight = restrictions[m - 1][1];
        maxHeight = Math.max(maxHeight, lastHeight + n - lastPosition);
        
        return maxHeight;
    }
}

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