# Find alphabetical order such that words can be considered sorted

Given an array of words, find any alphabetical order in the English alphabet such that the given words can be considered sorted (increasing), if there exists such an order, otherwise output impossible.

Examples:

Input : words[] = {"zy", "ab"} Output : zabcdefghijklmnopqrstuvwxy Basically we need to make sure that 'z' comes before 'a'. Input : words[] = {"geeks", "gamers", "coders", "everyoneelse"} Output : zyxwvutsrqponmlkjihgceafdb Input : words[] = {"marvel", "superman", "spiderman", "batman" Output : zyxwvuptrqonmsbdlkjihgfeca

**Naive approach:** The brute-force approach would be to check all the possible orders, and check if any of them satisfy the given order of words. Considering there are **26** alphabets in the English language, there are **26!** number of permutations that can be valid orders. Considering we check every pair for verifying an order, the complexity of this approach goes to **O(26!*N^2)**, which is well beyond practically preferred time complexity.

**Using topological sort:** This solution requires knowledge of Graphs and its representation as adjacency lists, DFS and Topological sorting.

In our required order, it is required to print letters such that each letter must be followed by the letters that are placed in lower priority than them. It seems somewhat similar to what topological sort is defined as – In topological sorting, we need to print a vertex before its adjacent vertices. Let’s define each letter in the alphabet as nodes in a standard directed graph. **A** is said to be connected to **B** (A—>B) if **A** precedes **B** in the order. The algorithm can be formulated as follows:

- If
**n**is**1**, then any order is valid. - Take the first two words. Identify the first different letter (at the same index of the words) in the words. The letter in the first word will precede the letter in the second word.
- If there exists no such letter, then the first string must be smaller in length than the second string.
- Assign the second word to the first word and input the third word into the second word. Repeat
**2**,**3**and**4****(n-1)**times. - Run a DFS traversal in topological order.
- Check if all the nodes are visited. In topological order, if there are cycles in the graph, the nodes in the cycles remain not visited, since it is not possible to visit these nodes after visiting every node adjacent to it. In such a case, order does not exist. In this case, it means that the order in our list contradicts itself.

`/* CPP program to find an order of alphabets ` `so that given set of words are considered ` `sorted */` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `#define MAX_CHAR 26 ` ` ` `void` `findOrder(vector<string> v) ` `{ ` ` ` `int` `n = v.size(); ` ` ` ` ` `/* If n is 1, then any order works */` ` ` `if` `(n == 1) { ` ` ` `cout << ` `"abcdefghijklmnopqrstuvwxyz"` `; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `/* Adjacency list of 26 characters*/` ` ` `vector<` `int` `> adj[MAX_CHAR]; ` ` ` ` ` `/* Array tracking the number of edges that are ` ` ` `inward to each node*/` ` ` `vector<` `int` `> in(MAX_CHAR, 0); ` ` ` ` ` `// Traverse through all words in given array ` ` ` `string prev = v[0]; ` ` ` ` ` `/* (n-1) loops because we already acquired the ` ` ` `first word in the list*/` ` ` `for` `(` `int` `i = 1; i < n; ++i) { ` ` ` `string s = v[i]; ` ` ` ` ` `/* Find first such letter in the present string that is different ` ` ` `from the letter in the previous string at the same index*/` ` ` `int` `j; ` ` ` `for` `(j = 0; j < min(prev.length(), s.length()); ++j) ` ` ` `if` `(s[j] != prev[j]) ` ` ` `break` `; ` ` ` ` ` `if` `(j < min(prev.length(), s.length())) { ` ` ` ` ` `/* The letter in the previous string precedes the one ` ` ` `in the present string, hence add the letter in the present ` ` ` `string as the child of the letter in the previous string*/` ` ` `adj[prev[j] - ` `'a'` `].push_back(s[j] - ` `'a'` `); ` ` ` ` ` `/* The number of inward pointing edges to the node representing ` ` ` `the letter in the present string increases by one*/` ` ` `in[s[j] - ` `'a'` `]++; ` ` ` ` ` `/* Assign present string to previous string for the next ` ` ` `iteration. */` ` ` `prev = s; ` ` ` `continue` `; ` ` ` `} ` ` ` ` ` `/* If there exists no such letter then the string length of ` ` ` `the previous string must be less than or equal to the ` ` ` `present string, otherwise no such order exists*/` ` ` `if` `(prev.length() > s.length()) { ` ` ` `cout << ` `"Impossible"` `; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `/* Assign present string to previous string for the next ` ` ` `iteration */` ` ` `prev = s; ` ` ` `} ` ` ` ` ` `/* Topological ordering requires the source nodes ` ` ` `that have no parent nodes*/` ` ` `stack<` `int` `> stk; ` ` ` `for` `(` `int` `i = 0; i < MAX_CHAR; ++i) ` ` ` `if` `(in[i] == 0) ` ` ` `stk.push(i); ` ` ` ` ` `/* Vector storing required order (anyone that satisfies) */` ` ` `vector<` `char` `> out; ` ` ` ` ` `/* Array to keep track of visited nodes */` ` ` `bool` `vis[26]; ` ` ` `memset` `(vis, ` `false` `, ` `sizeof` `(vis)); ` ` ` ` ` `/* Standard DFS */` ` ` `while` `(!stk.empty()) { ` ` ` ` ` `/* Acquire present character */` ` ` `char` `x = stk.top(); ` ` ` `stk.pop(); ` ` ` ` ` `/* Mark as visited */` ` ` `vis[x] = ` `true` `; ` ` ` ` ` `/* Insert character to output vector */` ` ` `out.push_back(x + ` `'a'` `); ` ` ` ` ` `for` `(` `int` `i = 0; i < adj[x].size(); ++i) { ` ` ` `if` `(vis[adj[x][i]]) ` ` ` `continue` `; ` ` ` ` ` `/* Since we have already included the present ` ` ` `character in the order, the number edges inward ` ` ` `to this child node can be reduced*/` ` ` `in[adj[x][i]]--; ` ` ` ` ` `/* If the number of inward edges have been removed, ` ` ` `we can include this node as a source node*/` ` ` `if` `(in[adj[x][i]] == 0) ` ` ` `stk.push(adj[x][i]); ` ` ` `} ` ` ` `} ` ` ` ` ` `/* Check if all nodes(alphabets) have been visited. ` ` ` `Order impossible if any one is unvisited*/` ` ` `for` `(` `int` `i = 0; i < MAX_CHAR; ++i) ` ` ` `if` `(!vis[i]) { ` ` ` `cout << ` `"Impossible"` `; ` ` ` `return` `; ` ` ` `} ` ` ` ` ` `for` `(` `int` `i = 0; i < out.size(); ++i) ` ` ` `cout << out[i]; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `vector<string> v{ ` `"efgh"` `, ` `"abcd"` `}; ` ` ` `findOrder(v); ` ` ` `return` `0; ` `} ` |

Output :

zyxwvutsrqponmlkjihgfeadcb

The complexity of this approach is **O(N*|S|) + O(V+E)**, where **|V|**=26 (number of nodes is the same as number of alphabets) and **|E|**<**N** (since at most 1 edge is created for each word as input). Hence overall complexity is **O(N*|S|+N)**. **|S|** represents the length of each word.