# Peak Heights π¦ - Google Top Interview Questions

### Problem Statement :

```You are given a two-dimensional integer matrix containing 0s and 1s where 0 represents water and 1 represents land.

Return a new matrix of the same dimensions where we increase the height of every land cell as much as possible given that:

The height of any water cell stays 0

Any cell can differ by at most 1 unit of height between neighboring cells (up, down, left, right)

You can assume there is at least one water cell.

Constraints

n, m β€ 250 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [

[0, 1, 0],

[1, 1, 1],

[1, 1, 1]

]
Output

[

[0, 1, 0],

[1, 2, 1],

[2, 3, 2]

]```

### Solution :

```                        ```Solution in C++ :

vector<vector<int>> solve(vector<vector<int>>& g) {
int r = g.size();
int c = g[0].size();
queue<pair<int, int>> q;
for (int i = 0; i < r; i++)
for (int j = 0; j < c; j++) {
if (g[i][j] == 0) {
q.emplace(i, j);
} else {
g[i][j] = 1e9;
}
}
while (q.size()) {
int x, y;
tie(x, y) = q.front();
q.pop();
int dx[4]{-1, 0, 1, 0};
int dy[4]{0, -1, 0, 1};
for (int k = 0; k < 4; k++) {
int nx = x + dx[k];
int ny = y + dy[k];
if (nx >= 0 && nx < r && ny >= 0 && ny < c && g[nx][ny] > 1 + g[x][y]) {
g[nx][ny] = 1 + g[x][y];
q.emplace(nx, ny);
}
}
}
return g;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int[][] solve(int[][] matrix) {
int N = matrix.length;
int M = matrix[0].length;
int[][] ans = new int[N][M];
for (int i = 0; i < N; i++) Arrays.fill(ans[i], Integer.MAX_VALUE);
ArrayDeque<int[]> bfs = new ArrayDeque<int[]>();
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (matrix[i][j] == 0) {
ans[i][j] = 0;
}
}
}

int[][] dirs = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
while (!bfs.isEmpty()) {
int[] cell = bfs.pollFirst();
for (int[] dir : dirs) {
int newR = cell[0] + dir[0];
int newC = cell[1] + dir[1];
if (newR >= 0 && newC >= 0 && newR < N && newC < M
&& ans[newR][newC] == Integer.MAX_VALUE) {
ans[newR][newC] = ans[cell[0]][cell[1]] + 1;
}
}
}

return ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, A):
R, C = len(A), len(A[0])
queue = collections.deque()
seen = set()
for r, row in enumerate(A):
for c, v in enumerate(row):
if v == 0:
queue.append([r, c, v])

while queue:
r, c, v = queue.popleft()
A[r][c] = v
for nr, nc in [[r - 1, c], [r, c - 1], [r + 1, c], [r, c + 1]]:
if 0 <= nr < R and 0 <= nc < C and (nr, nc) not in seen:
queue.append([nr, nc, v + 1])

return A```
```

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