**Fix Flight Itinerary - Google Top Interview Questions**

### Problem Statement :

You are given a list of uppercase alphabet strings itinerary and a two-dimensional list of uppercase alphabet strings edges. itinerary contains the list of airports you visited in order but the airport names may be misspelled. Each element in edges contains [source, dest] meaning there is a flight that goes from source to dest. Every airport has the same length of 3. Return the minimum number of characters you can change in itinerary such that the itinerary becomes valid. You can assume that there is a solution. Constraints n * m ≤ 200,000 where n is the length of itinerary, m is the length of edges Example 1 Input itinerary = ["YYZ", "SFO", "JFK"] edges = [ ["YYZ", "SEA"], ["SEA", "JAM"], ["SEA", "JFL"] ] Output 3 Explanation We change "SFO" to "SEA" for 2 character changes and "JFK" to 'JFL" for 1 character change. In total, we made 3 character changes. Example 2 Input itinerary = ["YYZ", "SFO", "JFK"] edges = [ ["YYZ", "SFO"], ["SFO", "JFK"] ] Output 0 Explanation We can go directly from "YYZ" to "SFO" to "JFK". So no characters need to be changed.

### Solution :

` ````
Solution in C++ :
int solve(vector<string>& arr, vector<vector<string>>& edge) {
int n = arr.size();
int m = edge.size();
set<string> st;
for (auto it : edge) {
st.insert(it[0]);
st.insert(it[1]);
}
map<string, int> mp;
vector<string> str;
int cnt = 0;
for (string s : st) {
str.push_back(s);
mp[s] = cnt;
cnt += 1;
}
int sz = str.size();
vector<vector<int>> adj(sz);
for (auto it : edge) {
int u = mp[it[0]], v = mp[it[1]];
adj[u].push_back(v);
}
queue<array<int, 2>> q;
for (int i = 0; i < sz; i++) {
cnt = 0;
for (int j = 0; j < 3; j++) {
if (str[i][j] != arr[0][j]) cnt += 1;
}
q.push({cnt, i});
}
cnt = 0;
int ans = INT_MAX;
vector<int> dis(sz, INT_MAX);
while (cnt < n) {
cnt += 1;
set<int> id;
while (q.size()) {
array<int, 2> tmp = q.front();
q.pop();
if (cnt == n) {
ans = min(ans, tmp[0]);
continue;
}
for (int a : adj[tmp[1]]) {
int chk = 0;
for (int i = 0; i < 3; i++) {
if (arr[cnt][i] != str[a][i]) {
chk += 1;
}
}
dis[a] = min(dis[a], tmp[0] + chk);
id.insert(a);
}
}
for (int x : id) {
q.push({dis[x], x});
dis[x] = INT_MAX;
}
}
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
private Map<String, Integer> portMap = new LinkedHashMap();
private Map<String, List<String>> graph = new HashMap();
private int index = 0;
private int infi = Integer.MAX_VALUE / 2;
private int ans = Integer.MAX_VALUE;
private Set<String> ports = new HashSet();
public int solve(String[] itinerary, String[][] edges) {
for (String[] edge : edges) {
String u = edge[0];
String v = edge[1];
// incoming edges
graph.computeIfAbsent(v, k -> new ArrayList()).add(u);
if (!portMap.containsKey(u)) {
portMap.put(u, index++);
}
if (!portMap.containsKey(v)) {
portMap.put(v, index++);
}
ports.add(u);
ports.add(v);
}
List<String> portList = new ArrayList(ports);
int[][] dp = new int[portList.size()][itinerary.length];
for (int i = 0; i < portList.size(); i++) {
String port = portList.get(i);
dp[portMap.get(port)][0] = getDiff(port, itinerary[0]);
}
for (int idx = 1; idx < itinerary.length; idx++) {
for (int j = 0; j < portList.size(); j++) {
String port = portList.get(j);
int diff = getDiff(port, itinerary[idx]);
if (graph.get(port) == null) {
dp[portMap.get(port)][idx] = infi;
} else {
int min = Integer.MAX_VALUE;
for (String prevPort : graph.get(port)) {
min = Math.min(min, dp[portMap.get(prevPort)][idx - 1] + diff);
}
dp[portMap.get(port)][idx] = min;
}
}
}
for (int i = 0; i < portList.size(); i++) {
ans = Math.min(ans, dp[portMap.get(portList.get(i))][itinerary.length - 1]);
}
return ans;
}
private int getDiff(String str1, String str2) {
int count = 0;
for (int i = 0; i < str1.length(); i++) {
if (str1.charAt(i) != str2.charAt(i))
++count;
}
return count;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, itinerary, edges):
n = len(itinerary)
def diff(a, b):
"""The cost to change the airport name `a` to `b` is the number of different characters."""
return sum(x != y for x, y in zip(a, b))
# We will use dynamic programming.
# dp[port] is the minimum cost to reach airport `port`.
# We can freely change the first airport in the itinerary,
# incurring only the cost of the name change itself.
dp = dict()
airports = set(e[0] for e in edges)
for p in airports:
dp[p] = diff(itinerary[0], p)
for i in range(1, n):
nextdp = dict()
for p1, p2 in edges:
# If it's possible to reach p1 at the previous index...
if p1 in dp:
# then it's now possible to reach p2 at the current index.
# We incur an additional cost to change `itinerary[i]` to `p2`.
cost = dp[p1] + diff(itinerary[i], p2)
if p2 not in nextdp or cost < nextdp[p2]:
nextdp[p2] = cost
dp = nextdp
# Choose the lowest cost among all possible final airports.
return min(dp.values())
```

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