# 01 Matrix in C++

Suppose we have a matrix consists of 0 and 1, we have to find the distance of the nearest 0 for each cell. Here the distance between two adjacent cells is 1.

So, if the input is like

 0 0 0 0 1 0 1 1 1

then the output will be

 0 0 0 0 1 0 1 2 1

To solve this, we will follow these steps −

• Define an array dir of size: 4 x 2 := {{1, 0}, { - 1, 0}, {0, - 1}, {0, 1}}

• n := row count, m := column count

• Define one matrix ret of order (n x m) and fill this with inf

• Define one queue q

• for initialize i := 0, when i < n, update (increase i by 1), do −

• for initialize j := 0, when j < m, update (increase j by 1), do −

• if not matrix[i, j] is non-zero, then −

• ret[i, j] := 0

• insert {i, j} into q

• for initialize lvl := 1, when not q is empty, update (increase lvl by 1), do −

• sz := size of q

• while sz is non-zero, decrease sz by 1 in each iteration, do −

• Define one pair curr := front element of q

• delete element from q

• for initialize k := 0, when k < 4, update (increase k by 1), do −

• nx := curr.first + dir[k, 0]

• ny := curr.second + dir[k, 1]

• if nx < 0 or nx >= n or ny < 0 or ny >= m or ret[nx, ny] < lvl, then −

• ret[nx, ny] := lvl

• insert {nx, ny} into q

• return ret

## Example

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<auto> > v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << "[";
for(int j = 0; j <v[i].size(); j++){
cout << v[i][j] << ", ";
}
cout << "],";
}
cout << "]"<<endl;
}
int dir[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
class Solution {
public:
vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
int n = matrix.size();
int m = matrix[0].size();
vector < vector <int> > ret(n, vector <int>(m, INT_MAX));
queue < pair <int, int> > q;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(!matrix[i][j]){
ret[i][j] = 0;
q.push({i, j});
}
}
}
for(int lvl = 1; !q.empty(); lvl++){
int sz = q.size();
while(sz--){
pair <int, int> curr = q.front();
q.pop();
for(int k = 0; k < 4; k++){
int nx = curr.first + dir[k][0];
int ny = curr.second + dir[k][1];
if(nx < 0 || nx >= n || ny < 0 || ny >= m || ret[nx][ny] < lvl) continue;
ret[nx][ny] = lvl;
q.push({nx, ny});
}
}
}
return ret;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{0,0,0},{0,1,0},{1,1,1}};
print_vector(ob.updateMatrix(v));
}

## Input

{{0,0,0},{0,1,0},{1,1,1}}

## Output

[[0, 0, 0, ],[0, 1, 0, ],[1, 2, 1, ],]

Updated on: 17-Nov-2020

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