diff --git a/28 October  Distance of nearest cell having 1 b/28 October  Distance of nearest cell having 1
new file mode 100644
index 0000000..5aab5d6
--- /dev/null
+++ b/28 October  Distance of nearest cell having 1	
@@ -0,0 +1,47 @@
+class Solution {
+  public:
+    vector<vector<int>>nearest(vector<vector<int>>grid)
+{
+    int row = grid.size();
+    int col = grid[0].size();
+    
+    queue<pair<int,int>> q;
+    vector<vector<int>> ans(row,vector<int>(col,-1));
+    
+    for(int i=0;i<row;i++)
+    {
+        for(int j=0;j<col;j++)
+        {
+            if(grid[i][j]==1) {
+                q.push({i,j});
+                ans[i][j]=0;
+            }
+        }
+    }
+    int dx[] = {0, 0, 1, -1};
+    int dy[] = {1, -1, 0 , 0};
+    
+    int level=0;
+    
+    while(!q.empty())
+    {
+        int size = q.size();
+        for(int i=0;i<size;i++) {
+            int a = q.front().first;
+            int b = q.front().second;
+            q.pop();
+            
+            for(int k=0;k<4;k++)
+            {
+                int na = a+dx[k];
+                int nb = b+dy[k];
+                
+                if(na<0||nb<0||na>=row||nb>=col||ans[na][nb]!=-1) continue;
+                q.push({na,nb});
+                ans[na][nb]=ans[a][b]+1;
+            }
+        }
+    }
+    return ans;
+}
+};