Merge branch '7oSkaaa:main' into main

04- April/28- Similar String Groups/28- Similar String Groups (Mahmoud Aboelsoud).cpp

@@ -0,0 +1,79 @@
+// Author: Mahmoud Aboelsoud
+
+class Solution {
+public:
+    // Disjoint Set Union (DSU) data structure
+
+    vector<int> parent, set_size;
+
+    int find_set(int v){
+        if(v == parent[v]) return v;
+
+        return parent[v] = find_set(parent[v]);
+    }
+
+    void union_sets(int a, int b){
+        a = find_set(a);
+        b = find_set(b);
+        if(a != b){
+            if(set_size[a] < set_size[b])
+                swap(a, b);
+            parent[b] = a;
+            set_size[a] += set_size[b];
+        }
+    }
+
+    bool same_set(int a, int b){
+        a = find_set(a);
+        b = find_set(b);
+
+        return a == b;
+    }
+
+    // in this problem we need to know how many similar groups are there
+    // we can do that by brute force and check each pair of strings if they are similar or not
+    // and if they are similar we can merge them in the same group which will help us to know the number of groups at the end
+
+
+    int numSimilarGroups(vector<string>& strs) {
+        // n: number of strings
+        int n = strs.size();
+        // parent: parent of each string
+        parent.assign(n, 0);
+        // set_size: size of each set
+        set_size.assign(n, 1);
+
+        // initially each string is in its own set (group) and its parent is itself
+        for(int i = 0; i < n; i++)
+            parent[i] = i;
+
+
+        // we can check each pair of strings if they are similar or not and merge them in the same group if they are similar
+        for(int i = 0; i < n; i++){
+            for(int j = i + 1; j < n; j++){
+                // if they are not in the same group then we check if they are similar or not
+                if(!same_set(i, j)){
+                    // cnt: number of different characters between the two strings
+                    int cnt = 0;
+                    // loop over the two strings and count the number of different characters and add a condition to break the loop if the number of different characters is more than 2
+                    // because if the number of different characters is more than 2 then the two strings are not similar and we don't need to check the rest of the characters
+                    for(int k = 0; k < strs[i].size() && cnt <= 2; k++)
+                        cnt += (strs[i][k] != strs[j][k]);
+
+                    // if the number of different characters is 0 or 2 then the two strings are similar and we merge them in the same group
+                    if(cnt == 0 || cnt == 2) union_sets(i, j);
+                }
+            }
+        }
+
+        // st: set of parents of all strings
+        set<int> st;
+
+        // loop over all strings and insert their parents in the set
+        for(int i = 0; i < n ; i++)
+            st.insert(find_set(i));
+
+        // return the size of the set which is the number of groups
+        return st.size();
+    }
+};

04- April/28- Similar String Groups/28- Similar String Groups (Osama Ayman).java

@@ -0,0 +1,57 @@
+// Author: Osama Ayman
+// Time: O(n*n*m)
+// Space: O(n*n)
+class Solution {
+    public int numSimilarGroups(String[] strs) {
+        // adjacency map to represent the graph
+        Map<Integer, List<Integer>> adj = new HashMap<>();
+        int n = strs.length;
+        // adding edges if they are similar
+        for(int i=0; i<n; i++){
+            for(int j=i+1; j<n; j++){
+                if(isSimilar(strs[i], strs[j])){
+                    // adding undirected edge (bi-directional)
+                    adj.computeIfAbsent(i, k -> new ArrayList<>()).add(j);
+                    adj.computeIfAbsent(j, k -> new ArrayList<>()).add(i);
+                }
+            }
+        }
+        int connectedComponents = 0;
+        Set<Integer> vis = new HashSet<>();
+        for(int i=0; i<n; i++){
+            if(!vis.contains(i)){
+                connectedComponents++;
+                dfs(i, adj, vis);
+            }
+        }
+        return connectedComponents;
+    }
+    private void dfs(int node, Map<Integer, List<Integer>> adj, Set<Integer> vis){
+        if(vis.contains(node) || !adj.containsKey(node)) return;
+        vis.add(node);
+        for(int neigh: adj.get(node)){
+            dfs(neigh, adj, vis);
+        }
+    }
+    private boolean isSimilar(String s1, String s2){
+        char c1='.',c2='.';
+
+        for(int i=0; i<s1.length(); i++){
+            // if they are equal, continue
+            if(s1.charAt(i) == s2.charAt(i)) continue;
+            // if they are not equal and c1 is not set, set c1 and c2
+            if(c1=='.'){
+                c1=s1.charAt(i);
+                c2=s2.charAt(i);
+            }
+           // if they are equal and c1 and c2 are set, but they are not equal to the
+           //char of the other string, return false
+            else if(!(s1.charAt(i) == c2 && s2.charAt(i) == c1) ){
+                return false;
+            }
+
+
+        }
+        return true;
+    }
+}

04- April/30- Remove Max Number of Edges to Keep Graph Fully Traversable/30- Remove Max Number of Edges to Keep Graph Fully Traversable (Ahmed Hossam).cpp

@@ -0,0 +1,76 @@
+// Author: Ahmed Hossam
+
+// Define a generic disjoint-set union data structure with an optional base node
+// parameter for 1-based indexing, and a template parameter for data type T.
+template <typename T = int, int Base = 1>
+struct DSU {
+    // The parent vector stores the parent of each node, and the Gsize vector
+    // stores the size of each disjoint set.
+    vector<T> parent, Gsize;
+
+    // Constructor initializes the parent and Gsize vectors for all nodes.
+    DSU(int MaxNodes) {
+        parent = Gsize = vector<T>(MaxNodes + 5);
+        for (int i = Base; i <= MaxNodes; i++)
+            parent[i] = i, Gsize[i] = 1;
+    }
+
+    // Recursive function to find the leader of the disjoint set that the node
+    // belongs to, and updates the parent of all visited nodes to the leader.
+    T find_leader(int node) {
+        return parent[node] = (parent[node] == node ? node : find_leader(parent[node]));
+    }
+
+    // Returns true if the two nodes belong to the same disjoint set.
+    bool is_same_sets(int u, int v) {
+        return find_leader(u) == find_leader(v);
+    }
+
+    // Merges the disjoint sets containing the two nodes u and v.
+    void union_sets(int u, int v) {
+        int leader_u = find_leader(u), leader_v = find_leader(v);
+        if (leader_u == leader_v) return;
+        if (Gsize[leader_u] < Gsize[leader_v]) swap(leader_u, leader_v);
+        Gsize[leader_u] += Gsize[leader_v], parent[leader_v] = leader_u;
+    }
+
+    // Returns the size of the disjoint set that the node belongs to.
+    int get_size(int u) {
+        return Gsize[find_leader(u)];
+    }
+};
+
+class Solution {
+public:
+    int maxNumEdgesToRemove(int n, vector<vector<int>>& edges) {
+        // Variable to store the number of edges removed.
+        int removed_edges = 0;
+        // Create two disjoint-set union objects, one for Alice and one for Bob.
+        DSU<int> alice(n), bob(n);
+        // Sort edges in decreasing order of type so that type 3 edges are
+        // processed last.
+        sort(edges.rbegin(), edges.rend());
+        // Loop through each edge in the sorted list.
+        for (auto& edge : edges) {
+            int t = edge[0], u = edge[1], v = edge[2];
+            // Process the edge based on its type.
+            if (t == 1) {
+                if (alice.is_same_sets(u, v)) removed_edges++;
+                else alice.union_sets(u, v);
+            } else if (t == 2) {
+                if (bob.is_same_sets(u, v)) removed_edges++;
+                else bob.union_sets(u, v);
+            } else {
+                if (alice.is_same_sets(u, v) && bob.is_same_sets(u, v)) removed_edges++;
+                else alice.union_sets(u, v), bob.union_sets(u, v);
+            }
+        }
+        // Check if all nodes belong to the same disjoint set for both Alice and
+        // Bob. If not, return -1 as it is impossible to remove enough edges to
+        // create a spanning tree for both players.
+        if (alice.get_size(1) != n || bob.get_size(1) != n)
+            return -1;
+        // Return the number of edges removed.
+        return removed_edges;
+    }
+};

04- April/README.md

@@ -51,6 +51,7 @@
 1. **[Add Digits](#26--add-digits)**
 1. **[Bulb Switcher](#27--bulb-switcher)**
 1. **[Similar String Groups](#28--similar-string-groups)**
+1. **[Remove Max Number of Edges to Keep Graph Fully Traversable](#30--remove-max-number-of-edges-to-keep-graph-fully-traversable)**
 
 <hr>
 <br><br>
@@ -1631,4 +1632,97 @@ public:
 
 };
 ```
+    
+<hr>
+<br><br>
+
+## 30)  [Remove Max Number of Edges to Keep Graph Fully Traversable](https://leetcode.com/problems/remove-max-number-of-edges-to-keep-graph-fully-traversable/)
+
+### Difficulty
+
+![](https://img.shields.io/badge/Hard-red?style=for-the-badge)
+
+### Related Topic
+
+`Union Find` `Graph`
+
+### Code
+
+
+```cpp
+// Define a generic disjoint-set union data structure with an optional base node
+// parameter for 1-based indexing, and a template parameter for data type T.
+template <typename T = int, int Base = 1>
+struct DSU {
+    // The parent vector stores the parent of each node, and the Gsize vector
+    // stores the size of each disjoint set.
+    vector<T> parent, Gsize;
+
+    // Constructor initializes the parent and Gsize vectors for all nodes.
+    DSU(int MaxNodes) {
+        parent = Gsize = vector<T>(MaxNodes + 5);
+        for (int i = Base; i <= MaxNodes; i++)
+            parent[i] = i, Gsize[i] = 1;
+    }
+
+    // Recursive function to find the leader of the disjoint set that the node
+    // belongs to, and updates the parent of all visited nodes to the leader.
+    T find_leader(int node) {
+        return parent[node] = (parent[node] == node ? node : find_leader(parent[node]));
+    }
+
+    // Returns true if the two nodes belong to the same disjoint set.
+    bool is_same_sets(int u, int v) {
+        return find_leader(u) == find_leader(v);
+    }
+
+    // Merges the disjoint sets containing the two nodes u and v.
+    void union_sets(int u, int v) {
+        int leader_u = find_leader(u), leader_v = find_leader(v);
+        if (leader_u == leader_v) return;
+        if (Gsize[leader_u] < Gsize[leader_v]) swap(leader_u, leader_v);
+        Gsize[leader_u] += Gsize[leader_v], parent[leader_v] = leader_u;
+    }
+
+    // Returns the size of the disjoint set that the node belongs to.
+    int get_size(int u) {
+        return Gsize[find_leader(u)];
+    }
+};
+
+class Solution {
+public:
+    int maxNumEdgesToRemove(int n, vector<vector<int>>& edges) {
+        // Variable to store the number of edges removed.
+        int removed_edges = 0;
+        // Create two disjoint-set union objects, one for Alice and one for Bob.
+        DSU<int> alice(n), bob(n);
+        // Sort edges in decreasing order of type so that type 3 edges are
+        // processed last.
+        sort(edges.rbegin(), edges.rend());
+        // Loop through each edge in the sorted list.
+        for (auto& edge : edges) {
+            int t = edge[0], u = edge[1], v = edge[2];
+            // Process the edge based on its type.
+            if (t == 1) {
+                if (alice.is_same_sets(u, v)) removed_edges++;
+                else alice.union_sets(u, v);
+            } else if (t == 2) {
+                if (bob.is_same_sets(u, v)) removed_edges++;
+                else bob.union_sets(u, v);
+            } else {
+                if (alice.is_same_sets(u, v) && bob.is_same_sets(u, v)) removed_edges++;
+                else alice.union_sets(u, v), bob.union_sets(u, v);
+            }
+        }
+        // Check if all nodes belong to the same disjoint set for both Alice and
+        // Bob. If not, return -1 as it is impossible to remove enough edges to
+        // create a spanning tree for both players.
+        if (alice.get_size(1) != n || bob.get_size(1) != n)
+            return -1;
+        // Return the number of edges removed.
+        return removed_edges;
+    }
+};
+```