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Razdalja najbližje celice z 1 v binarni matriki

Preizkusite na GfG Practice ' title=

Glede na dvojiško mreža[][] . Poiščite razdaljo najbližjega 1 v mreži za vsako celico.
Razdalja se izračuna kot  |i 1   - i 2 | + |j 1  - j 2 | kjer sem1j1 sta številka vrstice in številka stolpca trenutne celice in i2j2 sta številka vrstice in številka stolpca najbližje celice z vrednostjo 1. 

Opomba: V mreži mora biti vsaj ena celica z vrednostjo 1.

Primeri:



Vnos: mreža[][] = [[0 1 1 0]
[1 1 0 0]
[0 0 1 1]]
Izhod: [[1 0 0 1]
[0 0 1 1]
[1 1 0 0]]
Pojasnilo:
celica (0 1) ima najbližjo 1 v celici (0 0) - razdalja = |0-0| + |0-1| = 1
celica (0 2) ima najbližjo 1 v celici (0 3) - razdalja = |0-0| + |3-2| = 1
celica (1 0) ima najbližjo 1 v celici (0 0) - razdalja = |1-0| + |0-0| = 1
celica (1 1) ima najbližjo 1 v celici (1 2) - razdalja = |1-1| + |1-2| = 1
celica (2 2) ima najbližjo 1 v celici (2 1) - razdalja = |2-2| + |2-1| = 1
celica (2 3) ima najbližjo 1 v celici (1 3) - razdalja = |2-1| + |3-3| = 1
Vse ostale so celice z 1, zato je njihova razdalja od najbližje celice z 1 enaka 0.

Vnos: mreža[][] = [[1 0 1]
[1 1 0]
[1 0 0]]
Izhod: [[0 1 0]
[0 0 1]
[0 1 2]]
Pojasnilo:
celica (0 0) ima najbližjo 1 v celici (0 1) - razdalja = |0-0| + |0-1| = 1
celica (0 2) ima najbližjo 1 v celici (0 1) - razdalja = |0-0| + |2-1| = 1
celica (1 0) ima najbližjo 1 v celici (0 1) - razdalja = |1-0| + |0-1| = 2
celica (1 1) ima najbližjo 1 v celici (1 2) - razdalja = |1-1| + |1-2| = 1
celica (2 0) ima najbližjo 1 v celici (2 1) - razdalja = |2-2| + |2-1| = 1
celica (2 2) ima najbližjo 1 v celici (2 1) - razdalja = |2-2| + |2-1| = 1
Vse ostale so celice z 1, zato je njihova razdalja od najbližje celice z 1 enaka 0.

Kazalo vsebine

[Naivni pristop] - O((n*m)^2) Čas in O(n * m) Prostor

Ideja je prečkati celotno mrežo in izračunati razdaljo vsake celice na najbližjo 1:

  • Če celica vsebuje 1, je njena razdalja 0.
  • Če celica vsebuje 0, prečkamo celotno mrežo, da poiščemo najbližjo celico, ki vsebuje 1.
  • Za vsako celico 0 izračunajte razdaljo Manhattan do vseh celic z 1 in vzemite najmanjšo razdaljo.

To najmanjšo razdaljo shranite v ustrezno celico matrike rezultatov. Ponovite za vse celice v mreži.

C++
//Driver Code Starts #include  #include    #include  using namespace std; //Driver Code Ends  vector<vector<int>> nearest(vector<vector<int>> &grid) {  int n = grid.size();  int m = grid[0].size();  vector<vector<int>> ans(n vector<int>(m INT_MAX));  // visit each cell of the grid  for (int i = 0; i < n; i++)  {  for (int j = 0; j < m; j++)  {  // if the cell has 1  // then the distance is 0  if (grid[i][j] == 1)  {  ans[i][j] = 0;  continue;  }  // iterate over all the cells  // and find the distance of the nearest 1  for (int k = 0; k < n; k++)  {  for (int l = 0; l < m; l++)  {  if (grid[k][l] == 1)  {  ans[i][j] = min(ans[i][j] abs(i - k) + abs(j - l));  }  }  }  }  }  return ans; }  //Driver Code Starts int main() {  vector<vector<int>> grid = {{0 1 1 0} {1 1 0 0} {0 0 1 1}};  vector<vector<int>> ans = nearest(grid);  for (int i = 0; i < ans.size(); i++)  {  for (int j = 0; j < ans[i].size(); j++)  {  cout << ans[i][j] << ' ';  }  cout << endl;  }  return 0; } //Driver Code Ends 
Java
//Driver Code Starts import java.util.ArrayList; class GFG { //Driver Code Ends   static ArrayList<ArrayList<Integer>>nearest(int[][] grid)  {  int n = grid.length;  int m = grid[0].length;  ArrayList<ArrayList<Integer> > ans  = new ArrayList<>();  // initialize all cells with maximum value  for (int i = 0; i < n; i++) {  ArrayList<Integer> row = new ArrayList<>();  for (int j = 0; j < m; j++) {  row.add(Integer.MAX_VALUE);  }  ans.add(row);  }  // visit each cell of the grid  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // if the cell has 1 distance is 0  if (grid[i][j] == 1) {  ans.get(i).set(j 0);  continue;  }  // iterate over all cells to find nearest 1  for (int k = 0; k < n; k++) {  for (int l = 0; l < m; l++) {  if (grid[k][l] == 1) {  int distance  = Math.abs(i - k)  + Math.abs(j - l);  if (distance  < ans.get(i).get(j)) {  ans.get(i).set(j distance);  }  }  }  }  }  }  return ans;  }  //Driver Code Starts  public static void main(String[] args)  {  int[][] grid = { { 0 1 1 0 }  { 1 1 0 0 }  { 0 0 1 1 } };  ArrayList<ArrayList<Integer> > ans = nearest(grid);  for (ArrayList<Integer> row : ans) {  for (Integer val : row) {  System.out.print(val + ' ');  }  System.out.println();  }  } } //Driver Code Ends 
Python
def nearest(grid): n = len(grid) m = len(grid[0]) ans = [[float('inf')] * m for _ in range(n)] # visit each cell of the grid for i in range(n): for j in range(m): # if the cell has 1 # then the distance is 0 if grid[i][j] == 1: ans[i][j] = 0 continue # iterate over all the cells # and find the distance of the nearest 1 for k in range(n): for l in range(m): if grid[k][l] == 1: ans[i][j] = min(ans[i][j] abs(i - k) + abs(j - l)) return ans   #Driver Code Starts if __name__ == '__main__': grid = [[0 1 1 0] [1 1 0 0] [0 0 1 1]] ans = nearest(grid) for i in range(len(ans)): for j in range(len(ans[i])): print(ans[i][j] end=' ') print() #Driver Code Ends 
C#
//Driver Code Starts using System; using System.Collections.Generic; class GfG { //Driver Code Ends   static List<List<int> > nearest(int[ ] grid)  {  int n = grid.GetLength(0);  int m = grid.GetLength(1);  List<List<int> > ans = new List<List<int> >();  for (int i = 0; i < n; i++) {  List<int> row = new List<int>();  for (int j = 0; j < m; j++) {  row.Add(int.MaxValue);  }  ans.Add(row);  }  // Visit each cell of the grid  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // If the cell has 1 distance is 0  if (grid[i j] == 1) {  ans[i][j] = 0;  continue;  }  // iterate over all the cells  // and find the distance of the nearest 1  for (int k = 0; k < n; k++) {  for (int l = 0; l < m; l++) {  if (grid[k l] == 1) {  int distance  = Math.Abs(i - k)  + Math.Abs(j - l);  if (distance < ans[i][j]) {  ans[i][j] = distance;  }  }  }  }  }  }  return ans;  }  //Driver Code Starts  static void Main()  {  int[ ] grid = { { 0 1 1 0 }  { 1 1 0 0 }  { 0 0 1 1 } };  List<List<int> > ans = nearest(grid);  for (int i = 0; i < ans.Count; i++) {  for (int j = 0; j < ans[i].Count; j++) {  Console.Write(ans[i][j] + ' ');  }  Console.WriteLine();  }  } } //Driver Code Ends 
JavaScript
function nearest(grid) {  let n = grid.length;  let m = grid[0].length;  let ans = new Array(n);  for (let i = 0; i < n; i++) {  ans[i] = new Array(m).fill(Infinity);  }  // visit each cell of the grid  for (let i = 0; i < n; i++) {  for (let j = 0; j < m; j++) {  // if the cell has 1  // then the distance is 0  if (grid[i][j] === 1) {  ans[i][j] = 0;  continue;  }  // iterate over all the cells  // and find the distance of the nearest 1  for (let k = 0; k < n; k++) {  for (let l = 0; l < m; l++) {  if (grid[k][l] === 1) {  ans[i][j] = Math.min(  ans[i][j]  Math.abs(i - k)  + Math.abs(j - l));  }  }  }  }  }  return ans; }  // Driver Code //Driver Code Starts let grid =  [ [ 0 1 1 0 ] [ 1 1 0 0 ] [ 0 0 1 1 ] ]; let ans = nearest(grid); for (let i = 0; i < ans.length; i++) {  console.log(ans[i].join(' ')); } //Driver Code Ends 

Izhod
1 0 0 1 0 0 1 1 1 1 0 0 

[Pričakovan pristop] - Uporaba iskanja najprej po širini - O(n * m) čas in O(n * m) prostor

Težavo je mogoče učinkovito rešiti s pristopom BFS z več viri. Vsaka celica v mreži se obravnava kot vozlišče z robovi, ki povezujejo sosednje celice (gor dol levo desno). Namesto izvajanja ločenega iskanja za vsako celico 0 postavimo v čakalno vrsto vse celice, ki vsebujejo 1 na začetku, in izvedemo en BFS iz teh več virov hkrati. Ko se BFS širi plast za plastjo, posodobimo razdaljo vsake neobiskane celice 0, da je ena večja od razdalje njenega starša. To zagotavlja, da vsaka celica prejme minimalno razdaljo do najbližje 1 na optimalen in učinkovit način.

C++
//Driver Code Starts #include    #include #include #include using namespace std; //Driver Code Ends  vector<vector<int>> nearest(vector<vector<int>> &grid) {  int n = grid.size();  int m = grid[0].size();  vector<vector<int>> ans(n vector<int>(m INT_MAX));  // to store the indices of the cells having 1  queue<pair<int int>> q;  // visit each cell of the grid  for(int i = 0; i<n; i++) {  for(int j = 0; j<m; j++) {  // if the cell has 1   // then the distance is 0  if(grid[i][j] == 1) {  ans[i][j] = 0;  q.push({i j});  }  }  }  // iterate over all the cells  // and find the distance of the nearest 1  while(!q.empty()) {  int len = q.size();    for(int i = 0; i<len; i++) {  int x = q.front().first;  int y = q.front().second;  q.pop();  // check all the four directions  vector<vector<int>> directions =   {{0 1} {0 -1} {1 0} {-1 0}};  for (int j = 0; j < directions.size(); j++) {  int dx = directions[j][0];  int dy = directions[j][1];  // if the cell is within the grid   // and the distance is not calculated yet  if (x+dx >= 0 && x+dx < n && y+dy >= 0 &&   y+dy < m && ans[x+dx][y+dy] == INT_MAX) {  ans[x+dx][y+dy] = ans[x][y] + 1;  q.push({x+dx y+dy});  }  }  }  }  return ans; }  //Driver Code Starts int main() {  vector<vector<int>> grid = {{0110} {1100} {0011}};  vector<vector<int>> ans = nearest(grid);  for (int i = 0; i < ans.size(); i++) {  for (int j = 0; j < ans[i].size(); j++) {  cout << ans[i][j] << ' ';  }  cout << endl;  }  return 0; } //Driver Code Ends 
Java
//Driver Code Starts import java.util.ArrayList; import java.util.Queue; import java.util.LinkedList; import java.util.Arrays; class GfG { //Driver Code Ends   static ArrayList<ArrayList<Integer>> nearest(int[][] grid) {  int n = grid.length;  int m = grid[0].length;  int[][] ans = new int[n][m];  for (int i = 0; i < n; i++) {  Arrays.fill(ans[i] Integer.MAX_VALUE);  }  // to store the indices of the cells having 1  Queue<int[]> q = new LinkedList<>();  // visit each cell of the grid  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // if the cell has 1   // then the distance is 0  if (grid[i][j] == 1) {  ans[i][j] = 0;  q.add(new int[]{i j});  }  }  }  // iterate over all the cells  // and find the distance of the nearest 1  while (!q.isEmpty()) {  int len = q.size();  for (int i = 0; i < len; i++) {  int[] front = q.poll();  int x = front[0];  int y = front[1];  // check all the four directions  int[][] directions = {{0 1} {0 -1} {1 0} {-1 0}};  for (int j = 0; j < directions.length; j++) {  int dx = directions[j][0];  int dy = directions[j][1];  // if the cell is within the grid   // and the distance is not calculated yet  if (x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m  && ans[x + dx][y + dy] == Integer.MAX_VALUE) {  ans[x + dx][y + dy] = ans[x][y] + 1;  q.add(new int[]{x + dx y + dy});  }  }  }  }  ArrayList<ArrayList<Integer>> result = new ArrayList<>();  for (int i = 0; i < n; i++) {  ArrayList<Integer> row = new ArrayList<>();  for (int j = 0; j < m; j++) {  row.add(ans[i][j]);  }  result.add(row);  }  return result;  }  //Driver Code Starts  public static void main(String[] args) {  int[][] grid = {{0110} {1100} {0011}};  ArrayList<ArrayList<Integer>> ans = nearest(grid);  for (ArrayList<Integer> row : ans) {  for (int val : row) {  System.out.print(val + ' ');  }  System.out.println();  }  } } //Driver Code Ends 
Python
#Driver Code Starts from collections import deque import sys #Driver Code Ends  def nearest(grid): n = len(grid) m = len(grid[0]) ans = [[sys.maxsize for _ in range(m)] for _ in range(n)] # to store the indices of the cells having 1 q = deque() # visit each cell of the grid for i in range(n): for j in range(m): # if the cell has 1  # then the distance is 0 if grid[i][j] == 1: ans[i][j] = 0 q.append((i j)) # iterate over all the cells # and find the distance of the nearest 1 while q: len_q = len(q) for _ in range(len_q): x y = q.popleft() # check all the four directions directions = [(0 1) (0 -1) (1 0) (-1 0)] for dx dy in directions: # if the cell is within the grid  # and the distance is not calculated yet if 0 <= x + dx < n and 0 <= y + dy < m and ans[x + dx][y + dy] == sys.maxsize: ans[x + dx][y + dy] = ans[x][y] + 1 q.append((x + dx y + dy)) return ans  #Driver Code Starts if __name__ == '__main__': grid = [[0110] [1100] [0011]] ans = nearest(grid) for row in ans: print(' '.join(map(str row))) #Driver Code Ends 
C#
//Driver Code Starts using System; using System.Collections.Generic; class GFG { //Driver Code Ends   static List<List<int>> nearest(int[] grid)  {  int n = grid.GetLength(0);  int m = grid.GetLength(1);  int[] ans = new int[n m];  for (int i = 0; i < n; i++)  {  for (int j = 0; j < m; j++)  {  ans[i j] = int.MaxValue;  }  }  // to store the indices of the cells having 1  Queue<Tuple<int int>> q = new Queue<Tuple<int int>>();  // visit each cell of the grid  for (int i = 0; i < n; i++)  {  for (int j = 0; j < m; j++)  {  // if the cell has 1   // then the distance is 0  if (grid[i j] == 1)  {  ans[i j] = 0;  q.Enqueue(new Tuple<int int>(i j));  }  }  }  // iterate over all the cells  // and find the distance of the nearest 1  while (q.Count > 0)  {  int len = q.Count;  for (int i = 0; i < len; i++)  {  var node = q.Dequeue();  int x = node.Item1;  int y = node.Item2;  // check all the four directions  int[] directions = new int[]  {  {0 1}  {0 -1}  {1 0}  {-1 0}  };  for (int j = 0; j < 4; j++)  {  int dx = directions[j 0];  int dy = directions[j 1];  // if the cell is within the grid   // and the distance is not calculated yet  if (x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m && ans[x + dx y + dy] == int.MaxValue)  {  ans[x + dx y + dy] = ans[x y] + 1;  q.Enqueue(new Tuple<int int>(x + dx y + dy));  }  }  }  }  // Convert 2D array to List> before returning  List<List<int>> result = new List<List<int>>();  for (int i = 0; i < n; i++)  {  List<int> row = new List<int>();  for (int j = 0; j < m; j++)  {  row.Add(ans[i j]);  }  result.Add(row);  }  return result;  }  //Driver Code Starts  static void Main()  {  int[] grid = new int[]  {  {0 1 1 0}  {1 1 0 0}  {0 0 1 1}  };  List<List<int>> ans = nearest(grid);  for (int i = 0; i < ans.Count; i++)  {  for (int j = 0; j < ans[i].Count; j++)  {  Console.Write(ans[i][j] + ' ');  }  Console.WriteLine();  }  } } //Driver Code Ends 
JavaScript
//Driver Code Starts const Denque = require('denque'); //Driver Code Ends  function nearest(grid) {  let n = grid.length;  let m = grid[0].length;  // Initialize answer matrix with Infinity  let ans = [];  for (let i = 0; i < n; i++) {  ans.push(new Array(m).fill(Infinity));  }  // to store the indices of the cells having 1  let q = new Denque();  // visit each cell of the grid  for (let i = 0; i < n; i++) {  for (let j = 0; j < m; j++) {  // if the cell has 1   // then the distance is 0  if (grid[i][j] === 1) {  ans[i][j] = 0;  q.push([i j]);  }  }  }  // iterate over all the cells  // and find the distance of the nearest 1  while (!q.isEmpty()) {  let [x y] = q.shift();  // check all the four directions  let directions = [  [0 1]  [0 -1]  [1 0]  [-1 0]  ];  for (let dir of directions) {  let dx = dir[0];  let dy = dir[1];  // if the cell is within the grid   // and the distance is not calculated yet  if (x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m && ans[x + dx][y + dy] === Infinity) {  ans[x + dx][y + dy] = ans[x][y] + 1;  q.push([x + dx y + dy]);  }  }  }  return ans; }  //Driver Code Starts // Driver Code let grid = [  [0 1 1 0]  [1 1 0 0]  [0 0 1 1] ]; let ans = nearest(grid); for (let i = 0; i < ans.length; i++) {  console.log(ans[i].join(' ')); } //Driver Code Ends 

Izhod
1 0 0 1 0 0 1 1 1 1 0 0 
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