Day 59 by admin

Day 59/Geek's Village and Wells_GFG/Geek's_Village_and_Wells.cpp

@@ -0,0 +1,46 @@
+class Solution{
+public:
+    vector<vector<int>> chefAndWells(int n,int m,vector<vector<char>> &c){
+        // Code here
+        vector<vector<int>> vis(n,vector<int>(m,0));
+        vector<vector<int>> ans(n,vector<int>(m,-1));
+        queue<pair<int,pair<int,int>>> q;
+        for(int i=0;i<n;i++)
+        {
+            for(int j=0;j<m;j++)
+            {
+                if(c[i][j]=='N') {
+                    ans[i][j]=0;
+                    vis[i][j]=1;
+                }
+                if(c[i][j]=='W'){
+                    vis[i][j]=1;
+                    q.push({0,{i,j}});
+                    ans[i][j]=0;
+                }
+                if(c[i][j]=='.') ans[i][j]=0;
+            }
+        }
+        int dx[4] = {0,0,1,-1};
+        int dy[4] = {1,-1,0,0};
+        while(!q.empty())
+        {
+            int row = q.front().second.first;
+            int col = q.front().second.second;
+            int dis = q.front().first;
+            q.pop();
+            for(int i=0;i<4;i++)
+            {
+                int newi = row + dx[i];
+                int newj = col + dy[i];
+                if(newi>=0&&newi<n&&newj>=0&&newj<m&&vis[newi][newj]==0&&(c[newi][newj]=='H'||c[newi][newj]=='.'))
+                {
+                    if(c[newi][newj]=='H') ans[newi][newj] = 2*(dis+1);
+                    q.push({dis+1,{newi,newj}});
+                    vis[newi][newj]=1;
+                }
+            }
+        }
+        return ans;
+    }
+};

Day 59/Similar String Groups_Leetcode/Similar_String_Groups.cpp

@@ -0,0 +1,32 @@
+class Solution {
+ public:
+  int numSimilarGroups(vector<string>& A) {
+    int ans = 0;
+    vector<bool> seen(A.size());
+
+    for (int i = 0; i < A.size(); ++i)
+      if (!seen[i]) {
+        dfs(A, i, seen);
+        ++ans;
+      }
+
+    return ans;
+  }
+
+ private:
+  // Dfs on string A[i]
+  void dfs(const vector<string>& A, int i, vector<bool>& seen) {
+    seen[i] = true;
+    for (int j = 0; j < A.size(); ++j)
+      if (!seen[j] && isSimilar(A[i], A[j]))
+        dfs(A, j, seen);
+  }
+
+  bool isSimilar(const string& X, const string& Y) {
+    int diff = 0;
+    for (int i = 0; i < X.length(); ++i)
+      if (X[i] != Y[i] && ++diff > 2)
+        return false;
+    return true;
+  }
+};