# Find if an array of strings can be chained to form a circle | Set 2

Given an array of strings, find if the given strings can be chained to form a circle. A string X can be put before another string Y in a circle if the last character of X is the same as the first character of Y.

**Examples: **

Input: arr[] = {"geek", "king"} Output: Yes, the given strings can be chained. Note that the last character of first string is same as first character of second string and vice versa is also true. Input: arr[] = {"for", "geek", "rig", "kaf"} Output: Yes, the given strings can be chained. The strings can be chained as "for", "rig", "geek" and "kaf" Input: arr[] = {"aab", "bac", "aaa", "cda"} Output: Yes, the given strings can be chained. The strings can be chained as "aaa", "aab", "bac" and "cda" Input: arr[] = {"aaa", "bbb", "baa", "aab"}; Output: Yes, the given strings can be chained. The strings can be chained as "aaa", "aab", "bbb" and "baa" Input: arr[] = {"aaa"}; Output: Yes Input: arr[] = {"aaa", "bbb"}; Output: No Input : arr[] = ["abc", "efg", "cde", "ghi", "ija"] Output : Yes These strings can be reordered as, “abc”, “cde”, “efg”, “ghi”, “ija” Input : arr[] = [“ijk”, “kji”, “abc”, “cba”] Output : No

## We strongly recommend that you click here and practice it, before moving on to the solution.

We have discussed one approach to this problem in the below post.

Find if an array of strings can be chained to form a circle | Set 1

In this post, another approach is discussed. We solve this problem by treating this as a graph problem, where vertices will be the first and last character of strings, and we will draw an edge between two vertices if they are the first and last character of the same string, so a number of edges in the graph will be same as the number of strings in the array.

Graph representation of some string arrays are given in the below diagram,

Now it can be clearly seen after graph representation that if a loop among graph vertices is possible then we can reorder the strings otherwise not. As in the above diagram’s example, a loop can be found in the first and third array of string but not in the second array of string. Now to check whether **this graph can have a loop which goes through all the vertices**, we’ll check two conditions,

- Indegree and Outdegree of each vertex should be the same.
- The graph should be strongly connected.

The first condition can be checked easily by keeping two arrays, in and out for each character. For checking whether a graph is having a loop which goes through all vertices is the same as checking complete directed graph is strongly connected or not because if it has a loop which goes through all vertices then we can reach to any vertex from any other vertex that is, the graph will be strongly connected and the same argument can be given for reverse statement also.

Now for checking the second condition we will just run a DFS from any character and visit all reachable vertices from this, now if the graph has a loop then after this one DFS all vertices should be visited, if all vertices are visited then we will return true otherwise false so **visiting all vertices in a single DFS flags a possible ordering among strings**.

## C++

`// C++ code to check if cyclic order is possible among strings` `// under given constrainsts` `#include <bits/stdc++.h>` `using` `namespace` `std;` `#define M 26` `// Utility method for a depth first search among vertices` `void` `dfs(vector<` `int` `> g[], ` `int` `u, vector<` `bool` `> &visit)` `{` ` ` `visit[u] = ` `true` `;` ` ` `for` `(` `int` `i = 0; i < g[u].size(); ++i)` ` ` `if` `(!visit[g[u][i]])` ` ` `dfs(g, g[u][i], visit);` `}` `// Returns true if all vertices are strongly connected` `// i.e. can be made as loop` `bool` `isConnected(vector<` `int` `> g[], vector<` `bool` `> &mark, ` `int` `s)` `{` ` ` `// Initialize all vertices as not visited` ` ` `vector<` `bool` `> visit(M, ` `false` `);` ` ` `// perform a dfs from s` ` ` `dfs(g, s, visit);` ` ` `// now loop through all characters` ` ` `for` `(` `int` `i = 0; i < M; i++)` ` ` `{` ` ` `/* I character is marked (i.e. it was first or last` ` ` `character of some string) then it should be` ` ` `visited in last dfs (as for looping, graph` ` ` `should be strongly connected) */` ` ` `if` `(mark[i] && !visit[i])` ` ` `return` `false` `;` ` ` `}` ` ` `// If we reach that means graph is connected` ` ` `return` `true` `;` `}` `// return true if an order among strings is possible` `bool` `possibleOrderAmongString(string arr[], ` `int` `N)` `{` ` ` `// Create an empty graph` ` ` `vector<` `int` `> g[M];` ` ` `// Initialize all vertices as not marked` ` ` `vector<` `bool` `> mark(M, ` `false` `);` ` ` `// Initialize indegree and outdegree of every` ` ` `// vertex as 0.` ` ` `vector<` `int` `> in(M, 0), out(M, 0);` ` ` `// Process all strings one by one` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` `// Find first and last characters` ` ` `int` `f = arr[i].front() - ` `'a'` `;` ` ` `int` `l = arr[i].back() - ` `'a'` `;` ` ` `// Mark the characters` ` ` `mark[f] = mark[l] = ` `true` `;` ` ` `// increase indegree and outdegree count` ` ` `in[l]++;` ` ` `out[f]++;` ` ` `// Add an edge in graph` ` ` `g[f].push_back(l);` ` ` `}` ` ` `// If for any character indegree is not equal to` ` ` `// outdegree then ordering is not possible` ` ` `for` `(` `int` `i = 0; i < M; i++)` ` ` `if` `(in[i] != out[i])` ` ` `return` `false` `;` ` ` `return` `isConnected(g, mark, arr[0].front() - ` `'a'` `);` `}` `// Driver code to test above methods` `int` `main()` `{` ` ` `// string arr[] = {"abc", "efg", "cde", "ghi", "ija"};` ` ` `string arr[] = {` `"ab"` `, ` `"bc"` `, ` `"cd"` `, ` `"de"` `, ` `"ed"` `, ` `"da"` `};` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `if` `(possibleOrderAmongString(arr, N) == ` `false` `)` ` ` `cout << ` `"Ordering not possible\n"` `;` ` ` `else` ` ` `cout << ` `"Ordering is possible\n"` `;` ` ` `return` `0;` `}` |

## Java

`// Java code to check if cyclic order is` `// possible among strings under given constrainsts` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG{` ` ` `// Return true if an order among strings is possible ` `public` `static` `boolean` `possibleOrderAmongString(` ` ` `String s[], ` `int` `n)` `{` ` ` `int` `m = ` `26` `;` ` ` `boolean` `mark[] = ` `new` `boolean` `[m];` ` ` `int` `in[] = ` `new` `int` `[` `26` `];` ` ` `int` `out[] = ` `new` `int` `[` `26` `];` ` ` ` ` `ArrayList<` ` ` `ArrayList<Integer>> adj = ` `new` `ArrayList<` ` ` `ArrayList<Integer>>();` ` ` `for` `(` `int` `i = ` `0` `; i < m; i++)` ` ` `adj.add(` `new` `ArrayList<>());` ` ` ` ` `// Process all strings one by one` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` ` ` `// Find first and last characters` ` ` `int` `f = (` `int` `)(s[i].charAt(` `0` `) - ` `'a'` `);` ` ` `int` `l = (` `int` `)(s[i].charAt(` ` ` `s[i].length() - ` `1` `) - ` `'a'` `);` ` ` ` ` `// Mark the characters` ` ` `mark[f] = mark[l] = ` `true` `;` ` ` ` ` `// Increase indegree and outdegree count` ` ` `in[l]++;` ` ` `out[f]++;` ` ` ` ` `// Add an edge in graph` ` ` `adj.get(f).add(l);` ` ` `}` ` ` ` ` `// If for any character indegree is not equal to` ` ` `// outdegree then ordering is not possible` ` ` `for` `(` `int` `i = ` `0` `; i < m; i++)` ` ` `{` ` ` `if` `(in[i] != out[i])` ` ` `return` `false` `;` ` ` `}` ` ` `return` `isConnected(adj, mark,` ` ` `s[` `0` `].charAt(` `0` `) - ` `'a'` `);` `}` `// Returns true if all vertices are strongly` `// connected i.e. can be made as loop` `public` `static` `boolean` `isConnected(` ` ` `ArrayList<ArrayList<Integer>> adj,` ` ` `boolean` `mark[], ` `int` `src)` `{` ` ` `boolean` `visited[] = ` `new` `boolean` `[` `26` `];` ` ` ` ` `// Perform a dfs from src` ` ` `dfs(adj, visited, src);` ` ` `for` `(` `int` `i = ` `0` `; i < ` `26` `; i++)` ` ` `{` ` ` ` ` `/* I character is marked (i.e. it was first or` ` ` `last character of some string) then it should` ` ` `be visited in last dfs (as for looping, graph` ` ` `should be strongly connected) */` ` ` `if` `(mark[i] && !visited[i])` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `// If we reach that means graph is connected` ` ` `return` `true` `;` `}` `// Utility method for a depth first` `// search among vertices` `public` `static` `void` `dfs(ArrayList<ArrayList<Integer>> adj,` ` ` `boolean` `visited[], ` `int` `src)` `{` ` ` `visited[src] = ` `true` `;` ` ` `for` `(` `int` `i = ` `0` `; i < adj.get(src).size(); i++)` ` ` `if` `(!visited[adj.get(src).get(i)])` ` ` `dfs(adj, visited, adj.get(src).get(i));` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `String s[] = { ` `"ab"` `, ` `"bc"` `, ` `"cd"` `, ` `"de"` `, ` `"ed"` `, ` `"da"` `};` ` ` `int` `n = s.length;` ` ` ` ` `if` `(possibleOrderAmongString(s, n))` ` ` `System.out.println(` `"Ordering is possible"` `);` ` ` `else` ` ` `System.out.println(` `"Ordering is not possible"` `);` `}` `}` `// This code is contributed by parascoding` |

## Python3

`# Python3 code to check if` `# cyclic order is possible` `# among strings under given` `# constrainsts` `M ` `=` `26` `# Utility method for a depth` `# first search among vertices` `def` `dfs(g, u, visit):` ` ` `visit[u] ` `=` `True` ` ` `for` `i ` `in` `range` `(` `len` `(g[u])):` ` ` `if` `(` `not` `visit[g[u][i]]):` ` ` `dfs(g, g[u][i], visit)` `# Returns true if all vertices` `# are strongly connected i.e.` `# can be made as loop` `def` `isConnected(g, mark, s):` ` ` `# Initialize all vertices` ` ` `# as not visited` ` ` `visit ` `=` `[` `False` `for` `i ` `in` `range` `(M)]` ` ` `# Perform a dfs from s` ` ` `dfs(g, s, visit)` ` ` `# Now loop through` ` ` `# all characters` ` ` `for` `i ` `in` `range` `(M):` ` ` `# I character is marked` ` ` `# (i.e. it was first or last` ` ` `# character of some string)` ` ` `# then it should be visited` ` ` `# in last dfs (as for looping,` ` ` `# graph should be strongly` ` ` `# connected) */` ` ` `if` `(mark[i] ` `and` `(` `not` `visit[i])):` ` ` `return` `False` ` ` ` ` `# If we reach that means` ` ` `# graph is connected` ` ` `return` `True` `# return true if an order among` `# strings is possible` `def` `possibleOrderAmongString(arr, N):` ` ` `# Create an empty graph` ` ` `g ` `=` `{}` ` ` `# Initialize all vertices` ` ` `# as not marked` ` ` `mark ` `=` `[` `False` `for` `i ` `in` `range` `(M)]` ` ` `# Initialize indegree and` ` ` `# outdegree of every` ` ` `# vertex as 0.` ` ` `In ` `=` `[` `0` `for` `i ` `in` `range` `(M)]` ` ` `out ` `=` `[` `0` `for` `i ` `in` `range` `(M)]` ` ` `# Process all strings` ` ` `# one by one` ` ` `for` `i ` `in` `range` `(N):` ` ` `# Find first and last` ` ` `# characters` ` ` `f ` `=` `(` `ord` `(arr[i][` `0` `]) ` `-` ` ` `ord` `(` `'a'` `))` ` ` `l ` `=` `(` `ord` `(arr[i][` `-` `1` `]) ` `-` ` ` `ord` `(` `'a'` `))` ` ` `# Mark the characters` ` ` `mark[f] ` `=` `True` ` ` `mark[l] ` `=` `True` ` ` `# Increase indegree` ` ` `# and outdegree count` ` ` `In[l] ` `+` `=` `1` ` ` `out[f] ` `+` `=` `1` ` ` `if` `f ` `not` `in` `g:` ` ` `g[f] ` `=` `[]` ` ` `# Add an edge in graph` ` ` `g[f].append(l)` ` ` `# If for any character` ` ` `# indegree is not equal to` ` ` `# outdegree then ordering` ` ` `# is not possible` ` ` `for` `i ` `in` `range` `(M):` ` ` `if` `(In[i] !` `=` `out[i]):` ` ` `return` `False` ` ` ` ` `return` `isConnected(g, mark,` ` ` `ord` `(arr[` `0` `][` `0` `]) ` `-` ` ` `ord` `(` `'a'` `))` `# Driver code` `arr ` `=` `[` `"ab"` `, ` `"bc"` `,` ` ` `"cd"` `, ` `"de"` `,` ` ` `"ed"` `, ` `"da"` `]` `N ` `=` `len` `(arr)` `if` `(possibleOrderAmongString(arr, N) ` `=` `=` ` ` `False` `):` ` ` `print` `(` `"Ordering not possible"` `)` `else` `:` ` ` `print` `(` `"Ordering is possible"` `)` `# This code is contributed by avanitrachhadiya2155` |

**Output:**

Ordering is possible

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.