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Create DistanceofNearestCellHaving1.cpp
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class Solution | ||
{ | ||
public: | ||
//Function to find distance of nearest 1 in the grid for each cell. | ||
vector<vector<int>>nearest(vector<vector<int>>grid) | ||
{ | ||
// Code here | ||
int n=grid.size(),m=grid[0].size(); | ||
vector<vector<int>>ans(n,vector<int>(m,0)); | ||
vector<vector<int>>visited(n,vector<int>(m,0)); | ||
queue<pair<pair<int,int>,int>>q; | ||
for(int i=0;i<n;i++){ | ||
for(int j=0;j<m;j++){ | ||
if(grid[i][j]==1){ | ||
q.push({{i,j},0}); | ||
visited[i][j]=1; | ||
} | ||
else{ | ||
visited[i][j]=0; | ||
} | ||
} | ||
} | ||
int delrow[]={-1,0,+1,0}; | ||
int delcol[]={0,+1,0,-1}; | ||
while(!q.empty()){ | ||
int row=q.front().first.first; | ||
int col=q.front().first.second; | ||
int steps=q.front().second; | ||
q.pop(); | ||
ans[row][col]=steps; | ||
for(int i=0;i<4;i++){ | ||
int nrow=row+delrow[i]; | ||
int ncol=col+delcol[i]; | ||
if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && visited[nrow][ncol]==0 ){ | ||
visited[nrow][ncol]=1; | ||
q.push({{nrow,ncol},steps+1}); | ||
} | ||
} | ||
} | ||
return ans; | ||
} | ||
}; |