Question

1765. Map of Highest Peak

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Description

You are given an integer matrix isWater of size m x n that represents a map of land and water cells.

• If isWater[i][j] == 0, cell (i, j) is a land cell.
• If isWater[i][j] == 1, cell (i, j) is a water cell.

You must assign each cell a height in a way that follows these rules:

• The height of each cell must be non-negative.
• If the cell is a water cell, its height must be 0.
• Any two adjacent cells must have an absolute height difference of at most 1. A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).

Find an assignment of heights such that the maximum height in the matrix is maximized.

Return _an integer matrix _height_ of size _m x n_ where _height[i][j]_ is cell _(i, j)_'s height. If there are multiple solutions, return any of them_.

 

Example 1:

Input: isWater = [[0,1],[0,0]]
Output: [[1,0],[2,1]]
Explanation: The image shows the assigned heights of each cell.
The blue cell is the water cell, and the green cells are the land cells.

Example 2:

Input: isWater = [[0,0,1],[1,0,0],[0,0,0]]
Output: [[1,1,0],[0,1,1],[1,2,2]]
Explanation: A height of 2 is the maximum possible height of any assignment.
Any height assignment that has a maximum height of 2 while still meeting the rules will also be accepted.

 

Constraints:

• m == isWater.length
• n == isWater[i].length
• 1 <= m, n <= 1000
• isWater[i][j] is 0 or 1.
• There is at least one water cell.

Solutions

Solution 1

class Solution:
  def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:
    m, n = len(isWater), len(isWater[0])
    ans = [[-1] * n for _ in range(m)]
    q = deque()
    for i, row in enumerate(isWater):
      for j, v in enumerate(row):
        if v:
          q.append((i, j))
          ans[i][j] = 0
    while q:
      i, j = q.popleft()
      for a, b in pairwise((-1, 0, 1, 0, -1)):
        x, y = i + a, j + b
        if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:
          ans[x][y] = ans[i][j] + 1
          q.append((x, y))
    return ans

class Solution {
  public int[][] highestPeak(int[][] isWater) {
    int m = isWater.length, n = isWater[0].length;
    int[][] ans = new int[m][n];
    Deque<int[]> q = new ArrayDeque<>();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans[i][j] = isWater[i][j] - 1;
        if (ans[i][j] == 0) {
          q.offer(new int[] {i, j});
        }
      }
    }
    int[] dirs = {-1, 0, 1, 0, -1};
    while (!q.isEmpty()) {
      var p = q.poll();
      int i = p[0], j = p[1];
      for (int k = 0; k < 4; ++k) {
        int x = i + dirs[k], y = j + dirs[k + 1];
        if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
          ans[x][y] = ans[i][j] + 1;
          q.offer(new int[] {x, y});
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  const int dirs[5] = {-1, 0, 1, 0, -1};

  vector<vector<int>> highestPeak(vector<vector<int>>& isWater) {
    int m = isWater.size(), n = isWater[0].size();
    vector<vector<int>> ans(m, vector<int>(n));
    queue<pair<int, int>> q;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans[i][j] = isWater[i][j] - 1;
        if (ans[i][j] == 0) {
          q.emplace(i, j);
        }
      }
    }
    while (!q.empty()) {
      auto [i, j] = q.front();
      q.pop();
      for (int k = 0; k < 4; ++k) {
        int x = i + dirs[k], y = j + dirs[k + 1];
        if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
          ans[x][y] = ans[i][j] + 1;
          q.emplace(x, y);
        }
      }
    }
    return ans;
  }
};

func highestPeak(isWater [][]int) [][]int {
  m, n := len(isWater), len(isWater[0])
  ans := make([][]int, m)
  type pair struct{ i, j int }
  q := []pair{}
  for i, row := range isWater {
    ans[i] = make([]int, n)
    for j, v := range row {
      ans[i][j] = v - 1
      if v == 1 {
        q = append(q, pair{i, j})
      }
    }
  }
  dirs := []int{-1, 0, 1, 0, -1}
  for len(q) > 0 {
    p := q[0]
    q = q[1:]
    i, j := p.i, p.j
    for k := 0; k < 4; k++ {
      x, y := i+dirs[k], j+dirs[k+1]
      if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 {
        ans[x][y] = ans[i][j] + 1
        q = append(q, pair{x, y})
      }
    }
  }
  return ans
}

function highestPeak(isWater: number[][]): number[][] {
  const m = isWater.length;
  const n = isWater[0].length;
  let ans: number[][] = [];
  let q: number[][] = [];
  for (let i = 0; i < m; ++i) {
    ans.push(new Array(n).fill(-1));
    for (let j = 0; j < n; ++j) {
      if (isWater[i][j]) {
        q.push([i, j]);
        ans[i][j] = 0;
      }
    }
  }
  const dirs = [-1, 0, 1, 0, -1];
  while (q.length) {
    let tq: number[][] = [];
    for (const [i, j] of q) {
      for (let k = 0; k < 4; k++) {
        const [x, y] = [i + dirs[k], j + dirs[k + 1]];
        if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
          tq.push([x, y]);
          ans[x][y] = ans[i][j] + 1;
        }
      }
    }
    q = tq;
  }
  return ans;
}

use std::collections::VecDeque;

impl Solution {
  #[allow(dead_code)]
  pub fn highest_peak(is_water: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
    let n = is_water.len();
    let m = is_water[0].len();
    let mut ret_vec = vec![vec![-1; m]; n];
    let mut q: VecDeque<(usize, usize)> = VecDeque::new();
    let vis_pair: Vec<(i32, i32)> = vec![(-1, 0), (1, 0), (0, -1), (0, 1)];

    // Initialize the return vector
    for i in 0..n {
      for j in 0..m {
        if is_water[i][j] == 1 {
          // This cell is water, the height of which must be 0
          ret_vec[i][j] = 0;
          q.push_back((i, j));
        }
      }
    }

    while !q.is_empty() {
      // Get the front X-Y Coordinates
      let (x, y) = q.front().unwrap().clone();
      q.pop_front();
      // Traverse through the vis pair
      for d in &vis_pair {
        let (dx, dy) = *d;
        if Self::check_bounds((x as i32) + dx, (y as i32) + dy, n as i32, m as i32) {
          if ret_vec[((x as i32) + dx) as usize][((y as i32) + dy) as usize] == -1 {
            // This cell hasn't been visited, update its height
            ret_vec[((x as i32) + dx) as usize][((y as i32) + dy) as usize] =
              ret_vec[x][y] + 1;
            // Enqueue the current cell
            q.push_back((((x as i32) + dx) as usize, ((y as i32) + dy) as usize));
          }
        }
      }
    }

    ret_vec
  }

  #[allow(dead_code)]
  fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
    i >= 0 && i < n && j >= 0 && j < m
  }
}

Solution 2

class Solution:
  def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:
    m, n = len(isWater), len(isWater[0])
    ans = [[-1] * n for _ in range(m)]
    q = deque()
    for i, row in enumerate(isWater):
      for j, v in enumerate(row):
        if v:
          q.append((i, j))
          ans[i][j] = 0
    while q:
      for _ in range(len(q)):
        i, j = q.popleft()
        for a, b in pairwise((-1, 0, 1, 0, -1)):
          x, y = i + a, j + b
          if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:
            ans[x][y] = ans[i][j] + 1
            q.append((x, y))
    return ans

class Solution {
  public int[][] highestPeak(int[][] isWater) {
    int m = isWater.length, n = isWater[0].length;
    int[][] ans = new int[m][n];
    Deque<int[]> q = new ArrayDeque<>();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans[i][j] = isWater[i][j] - 1;
        if (ans[i][j] == 0) {
          q.offer(new int[] {i, j});
        }
      }
    }
    int[] dirs = {-1, 0, 1, 0, -1};
    while (!q.isEmpty()) {
      for (int t = q.size(); t > 0; --t) {
        var p = q.poll();
        int i = p[0], j = p[1];
        for (int k = 0; k < 4; ++k) {
          int x = i + dirs[k], y = j + dirs[k + 1];
          if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
            ans[x][y] = ans[i][j] + 1;
            q.offer(new int[] {x, y});
          }
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  const int dirs[5] = {-1, 0, 1, 0, -1};

  vector<vector<int>> highestPeak(vector<vector<int>>& isWater) {
    int m = isWater.size(), n = isWater[0].size();
    vector<vector<int>> ans(m, vector<int>(n));
    queue<pair<int, int>> q;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans[i][j] = isWater[i][j] - 1;
        if (ans[i][j] == 0) {
          q.emplace(i, j);
        }
      }
    }
    while (!q.empty()) {
      for (int t = q.size(); t; --t) {
        auto [i, j] = q.front();
        q.pop();
        for (int k = 0; k < 4; ++k) {
          int x = i + dirs[k], y = j + dirs[k + 1];
          if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
            ans[x][y] = ans[i][j] + 1;
            q.emplace(x, y);
          }
        }
      }
    }
    return ans;
  }
};

func highestPeak(isWater [][]int) [][]int {
  m, n := len(isWater), len(isWater[0])
  ans := make([][]int, m)
  type pair struct{ i, j int }
  q := []pair{}
  for i, row := range isWater {
    ans[i] = make([]int, n)
    for j, v := range row {
      ans[i][j] = v - 1
      if v == 1 {
        q = append(q, pair{i, j})
      }
    }
  }
  dirs := []int{-1, 0, 1, 0, -1}
  for len(q) > 0 {
    for t := len(q); t > 0; t-- {
      p := q[0]
      q = q[1:]
      i, j := p.i, p.j
      for k := 0; k < 4; k++ {
        x, y := i+dirs[k], j+dirs[k+1]
        if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 {
          ans[x][y] = ans[i][j] + 1
          q = append(q, pair{x, y})
        }
      }
    }
  }
  return ans
}