Hill Maker - Google Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers matrix where matrix[r][c] represents the height of that cell.

You are currently at top left corner and want to go to the bottom right. You can move to nearby cells (up, down, left, right) only if the neighboring cell's height is less than or equal to the current cell's height.

Given that you can increase the height of any number of cells before you move, return the minimum total height that needs to be increased so that you can go to the bottom right cell.

Constraints

n * m * h ≤ 1,000,000 where n, m are the number of rows and columns and h is the number of distinct 
values in matrix.

Example 1

Input

matrix = [

    [1, 3, 4],

    [7, 5, 0]

]

Output

4

Explanation

We take the following path [1, 3, 4, 0] and change the heights to this configuration:



[[4, 4, 4],

 [7, 5, 0]]

Example 2

Input

matrix = [

    [1, 1, 1, 1],

    [9, 9, 9, 1],

    [1, 1, 1, 1],

    [1, 9, 9, 9],

    [1, 1, 1, 1]

]

Output

0

Explanation

We can follow the 1s to get to the bottom right cell.


Solution :



title-img 




                        Solution in C++ :

struct vertex {
    int x, y, w, h;
    vertex(int a, int b, int c, int d) {
        x = a;
        y = b;
        w = c;
        h = d;
    }
};

bool operator<(const vertex& lhs, const vertex& rhs) {
    if (lhs.w == rhs.w) return lhs.h > rhs.h;
    return lhs.w > rhs.w;
}

int dx[4]{-1, 0, 1, 0};
int dy[4]{0, -1, 0, 1};

int dp[305][305];

class Solution {
    public:
    int solve(vector<vector<int>>& m) {
        priority_queue<vertex> q;
        int r = m.size();
        int c = m[0].size();
        for (int i = 0; i < r; i++)
            for (int j = 0; j < c; j++) dp[i][j] = 2e9;
        q.emplace(r - 1, c - 1, 0, m[r - 1][c - 1]);
        while (true) {
            vertex curr = q.top();
            q.pop();
            if (curr.x == 0 && curr.y == 0) {
                return curr.w;
            }
            if (dp[curr.x][curr.y] <= curr.h) continue;
            dp[curr.x][curr.y] = curr.h;
            for (int k = 0; k < 4; k++) {
                int nx = curr.x + dx[k];
                int ny = curr.y + dy[k];
                if (nx < 0 || nx >= r || ny < 0 || ny >= c) continue;
                int nh = max(m[nx][ny], curr.h);
                if (nh >= dp[nx][ny]) continue;
                int nw = curr.w + max(curr.h - m[nx][ny], 0);
                q.emplace(nx, ny, nw, nh);
            }
        }
    }
};


int solve(vector<vector<int>>& matrix) {
    return (new Solution())->solve(matrix);
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, A):
        INF = float("inf")
        R, C = len(A), len(A[0])

        pq = [[0, R - 1, C - 1, A[-1][-1]]]
        dist = collections.defaultdict(lambda: INF)
        dist[R - 1, C - 1, A[-1][-1]] = 0
        while pq:
            d, r, c, h = heapq.heappop(pq)
            if dist[r, c, h] < d:
                continue
            if r == c == 0:
                return d
            for nr, nc in [[r + 1, c], [r, c + 1], [r - 1, c], [r, c - 1]]:
                if 0 <= nr < R and 0 <= nc < C:
                    h2 = max(A[nr][nc], h)
                    d2 = d + max(h2 - A[nr][nc], 0)
                    if d2 < dist[nr, nc, h2]:
                        dist[nr, nc, h2] = d2
                        heapq.heappush(pq, [d2, nr, nc, h2])


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