**Dictionary Nomad - Amazon Top Interview Questions**

### Problem Statement :

You are given a list of strings dictionary and two strings start and end. You want to reach from start to end by modifying one character at a time and making sure each resulting word is also in the dictionary. Words are case-sensitive. Return the minimum number of steps it would take to reach end. Return -1 if it's not possible. Constraints 0 ≤ n * m ≤ 300,000 where n is the length of dictionary and m is the length of the longest string Example 1 Input dictionary = ["day", "say", "soy"] start = "soy" end = "day" Output 3 Explanation We can take this path: ["soy", "say", "day"]. Example 2 Input dictionary = ["day", "soy"] start = "soy" end = "day" Output -1 Explanation There's no way to change 1 character to reach "day".

### Solution :

` ````
Solution in C++ :
int solve(vector<string>& dictionary, string start, string end) {
unordered_set<string> seen{start}, dict(dictionary.begin(), dictionary.end());
queue<string> q{{start}};
int steps = 0;
while (!q.empty()) {
int sz = q.size();
steps++;
while (sz-- > 0) {
auto& p = q.front();
if (p == end) return steps;
q.pop();
for (int i = 0; i < p.size(); i++) {
char old = p[i];
for (char c = 'a'; c <= 'z'; c++) {
p[i] = c;
if (dict.count(p) && !seen.count(p)) {
seen.insert(p);
q.push(p);
}
}
p[i] = old;
}
}
}
return -1;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(String[] dictionary, String start, String end) {
Set<String> dict = new HashSet<>();
for (String d : dictionary) dict.add(d);
Set<String> visited = new HashSet<>();
Deque<String> q = new ArrayDeque<>();
q.add(start);
int steps = 0;
while (!q.isEmpty()) {
steps++;
int sz = q.size();
for (int i = 0; i < sz; i++) {
String cur = q.remove();
if (visited.contains(cur))
continue;
visited.add(cur);
if (cur.equals(end))
return steps;
// try to generate all possible strings for next level
for (int j = 0; j < cur.length(); j++) {
// for each possition, we try [a-z] chars
for (char c = 'a'; c <= 'z'; c++) {
String newstr =
cur.substring(0, j) + c + cur.substring(j + 1, cur.length());
if (dict.contains(newstr))
q.offer(newstr);
}
}
}
}
return -1;
}
}
/*
build a bi directional graph, node u and v will have an edge if u and v have only
one character difference
add start and end too in graph
find shorted path to reach from start to end using bfs then
challenge: how to build graph efficiently?
*/
```

` ````
Solution in Python :
class Solution:
def solve(self, dictionary, start, end):
words, visited = set(dictionary), set()
q = deque()
q.append(start)
level = 1
while len(q) > 0:
size = len(q)
for i in range(size):
cur = q.popleft()
if cur == end:
return level
for idx in range(len(cur)):
for c in range(97, 97 + 26):
new_word = cur[:idx] + chr(c) + cur[idx + 1 :]
if new_word not in visited and new_word in words:
q.append(new_word)
visited.add(new_word)
level += 1
return -1
```

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