# 01 Matrix – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given an `m x n`

binary matrix `mat`

, return *the distance of the nearest *`0`

* for each cell*.

The distance between two adjacent cells is `1`

.

**Example 1:**

Input:mat = [[0,0,0],[0,1,0],[0,0,0]]Output:[[0,0,0],[0,1,0],[0,0,0]]

**Example 2:**

Input:mat = [[0,0,0],[0,1,0],[1,1,1]]Output:[[0,0,0],[0,1,0],[1,2,1]]

**Constraints:**

`m == mat.length`

`n == mat[i].length`

`1 <= m, n <= 10`

^{4}`1 <= m * n <= 10`

^{4}`mat[i][j]`

is either`0`

or`1`

.- There is at least one
`0`

in`mat`

.

# C++ 01 Matrix LeetCode Solution

class Solution { // 48 ms, faster than 99.64%

public:

vector<vector<int>> updateMatrix(vector<vector<int>> &mat) {

int m = mat.size(), n = mat[0].size(), INF = m + n; // The distance of cells is up to (M+N)

for (int r = 0; r < m; r++) {

for (int c = 0; c < n; c++) {

if (mat[r][c] == 0) continue;

int top = INF, left = INF;

if (r – 1 >= 0) top = mat[r – 1][c];

if (c – 1 >= 0) left = mat[r][c – 1];

mat[r][c] = min(top, left) + 1;

}

}

for (int r = m – 1; r >= 0; r–) {

for (int c = n – 1; c >= 0; c–) {

if (mat[r][c] == 0) continue;

int bottom = INF, right = INF;

if (r + 1 < m) bottom = mat[r + 1][c];

if (c + 1 < n) right = mat[r][c + 1];

mat[r][c] = min(mat[r][c], min(bottom, right) + 1);

}

}

return mat;

}

};

# Java 01 Matrix LeetCode Solution

class Solution { // 5 ms, faster than 99.66%

public int[][] updateMatrix(int[][] mat) {

int m = mat.length, n = mat[0].length, INF = m + n; // The distance of cells is up to (M+N)

for (int r = 0; r < m; r++) {

for (int c = 0; c < n; c++) {

if (mat[r][c] == 0) continue;

int top = INF, left = INF;

if (r – 1 >= 0) top = mat[r – 1][c];

if (c – 1 >= 0) left = mat[r][c – 1];

mat[r][c] = Math.min(top, left) + 1;

}

}

for (int r = m – 1; r >= 0; r–) {

for (int c = n – 1; c >= 0; c–) {

if (mat[r][c] == 0) continue;

int bottom = INF, right = INF;

if (r + 1 < m) bottom = mat[r + 1][c];

if (c + 1 < n) right = mat[r][c + 1];

mat[r][c] = Math.min(mat[r][c], Math.min(bottom, right) + 1);

}

}

return mat;

}

}

# Python 3 01 Matrix LeetCode Solution

class Solution: # 520 ms, faster than 96.50%

def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:

m, n = len(mat), len(mat[0])for r in range(m):

for c in range(n):

if mat[r][c] > 0:

top = mat[r – 1][c] if r > 0 else math.inf

left = mat[r][c – 1] if c > 0 else math.inf

mat[r][c] = min(top, left) + 1for r in range(m – 1, -1, -1):

for c in range(n – 1, -1, -1):

if mat[r][c] > 0:

bottom = mat[r + 1][c] if r < m – 1 else math.inf

right = mat[r][c + 1] if c < n – 1 else math.inf

mat[r][c] = min(mat[r][c], bottom + 1, right + 1)return mat

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