Given two graphs **G1 **and **G2**, the task is to find the **union** and **intersection** of the two given graphs, i.e. **(G1 ∪ G2)** and **(G1 ∩ G2)**.

**Examples:**

Input:G1 = { (“e1”, 1, 2), (“e2”, 1, 3), (“e3”, 3, 4), (“e4”, 2, 4) }, G2 = = { (“e4”, 2, 4), (“e5”, 2, 5), (“e6”, 4, 5) }Output:

G1 union G2 is

e1 1 2

e2 1 3

e3 3 4

e4 2 4

e5 2 5

e6 4 5G1 intersection G2 is

e4 2 4Explanation:

Union of the graphs G1 and G2:Intersection of the graphs G1 and G2:

**Approach:** Follow the steps below to solve the problem:

- Define a function, say
**Union(G1, G2)**, to find the union of the**G1**and**G2**:- Initialize a map, say
**added**, that stores if an edge is already been added or not. - Iterate over the edges of the graph
**G1**and push all the edges in a graph, say**G**, and mark all the edges visited in**added**. - Now, again traverse over the edges of the graph
**G2**and push the edge in the**G**if the edge is not already been added, and then mark the edge added in the map**added.**

- Initialize a map, say
- Define a function say
**Intersection(G1, G2)**to find the Intersection of the**G1**and**G2**:- Initialize a map, say
**added**, that stores if an edge is already been added or not. - Traverse over the edges of the graph
**G1**and marked all the edges visited in the map**added.** - Now, again traverse over the edges of the graph
**G2**and push the edge in the graph**G**, if the edge is already been added. Then, mark the edge added in the map.

- Initialize a map, say
- Now, print the graphs obtained after the function call of
**Union(G1, G2)**and**Intersection(G1, G2)**.

Below is the implementation of the above approach:

## C++14

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to find union of two graphs` `void` `find_union(` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G1,` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G2)` `{` ` ` `// Stores an edge of the graph G1` ` ` `map<string, pair<` `int` `, ` `int` `> > added;` ` ` ` ` `// Stores the unioun graph G1` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G;` ` ` ` ` `// Iterate over the edges` ` ` `// of the graph G1` ` ` `for` `(` `auto` `p : G1) {` ` ` ` ` `string a = get<0>(p);` ` ` ` ` `// Get the edges` ` ` `int` `b = get<1>(p);` ` ` `int` `c = get<2>(p);` ` ` ` ` `// Insert the current` ` ` `// edges into graph G` ` ` `G.push_back(` ` ` `make_tuple(a, b, c));` ` ` `added[a] = { b, c };` ` ` `}` ` ` ` ` `// Iterate over the edges` ` ` `// of the graph G1` ` ` `for` `(` `auto` `p : G2) {` ` ` ` ` `string a = get<0>(p);` ` ` `int` `b = get<1>(p);` ` ` `int` `c = get<2>(p);` ` ` ` ` `pair<` `int` `, ` `int` `> x = { b, c };` ` ` `pair<` `int` `, ` `int` `> y = { c, b };` ` ` ` ` `// If either edge x or` ` ` `// y is already added` ` ` `if` `(added[a] == x || added[a] == y)` ` ` `continue` `;` ` ` ` ` `// Otherwise` ` ` `G.push_back(make_tuple(a, b, c));` ` ` `}` ` ` ` ` `// Print the unioun` ` ` `cout << ` `"G1 union G2 is\n"` `;` ` ` ` ` `for` `(` `auto` `p : G) {` ` ` ` ` `string a = get<0>(p);` ` ` `int` `b = get<1>(p);` ` ` `int` `c = get<2>(p);` ` ` `cout << a << ` `" "` `<< b << ` `" "` ` ` `<< c << endl;` ` ` `}` `}` ` ` `// Function to find intersection of two graphs` `void` `find_intersection(` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G1,` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G2)` `{` ` ` `// Stores an edge` ` ` `map<string, pair<` `int` `, ` `int` `> > added;` ` ` ` ` `// Stores the graph of intersection` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G;` ` ` ` ` `// Iterate over edges of graph G1` ` ` `for` `(` `auto` `p : G1) {` ` ` `string a = get<0>(p);` ` ` `int` `b = get<1>(p);` ` ` `int` `c = get<2>(p);` ` ` ` ` `added[a] = { b, c };` ` ` `}` ` ` ` ` `// Iterate over edges of graph G2` ` ` `for` `(` `auto` `p : G2) {` ` ` ` ` `string a = get<0>(p);` ` ` `int` `b = get<1>(p);` ` ` `int` `c = get<2>(p);` ` ` ` ` `pair<` `int` `, ` `int` `> x = { b, c };` ` ` `pair<` `int` `, ` `int` `> y = { c, b };` ` ` ` ` `// If either edge x or` ` ` `// y is already added` ` ` `if` `(added[a] == x || added[a] == y)` ` ` `G.push_back(make_tuple(a, b, c));` ` ` `}` ` ` ` ` `// Print the graph G` ` ` `cout << ` `"G1 intersection G2 is\n"` `;` ` ` ` ` `for` `(` `auto` `p : G) {` ` ` ` ` `string a = get<0>(p);` ` ` `int` `b = get<1>(p);` ` ` `int` `c = get<2>(p);` ` ` ` ` `cout << a << ` `" "` `<< b` ` ` `<< ` `" "` `<< c << endl;` ` ` `}` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G1` ` ` `= { make_tuple(` `"e1"` `, 1, 2),` ` ` `make_tuple(` `"e2"` `, 1, 3),` ` ` `make_tuple(` `"e3"` `, 3, 4),` ` ` `make_tuple(` `"e4"` `, 2, 4) };` ` ` ` ` `vector<tuple<string, ` `int` `, ` `int` `> > G2` ` ` `= { make_tuple(` `"e4"` `, 2, 4),` ` ` `make_tuple(` `"e5"` `, 2, 5),` ` ` `make_tuple(` `"e6"` `, 4, 5) };` ` ` ` ` `// Function call for finding the` ` ` `// Union of the given graph` ` ` `find_union(G1, G2);` ` ` ` ` `// Function call for finding the` ` ` `// Intersection of the given graph` ` ` `find_intersection(G1, G2);` ` ` ` ` `return` `0;` `}` |

**Output:**

G1 union G2 is e1 1 2 e2 1 3 e3 3 4 e4 2 4 e5 2 5 e6 4 5 G1 intersection G2 is e4 2 4

**Time Complexity:** O(N * log(N))**Auxiliary Space:** O(N)

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