# 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 :

```                        ```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])```
```

## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =