Peak Heights πŸ¦– - 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 ≀ 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 :



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                        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;
                    bfs.add(new int[] {i, j});
                }
            }
        }

        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;
                    bfs.add(new int[] {newR, newC});
                }
            }
        }

        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])
                    seen.add((r, c))

        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:
                    seen.add((nr, nc))
                    queue.append([nr, nc, v + 1])

        return A
                    


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