**Unique Paths to Go Home - Google Top Interview Questions**

### Problem Statement :

You are given a two-dimensional list of integers edges where each element contains [u, v, distance] representing a weighted undirected graph. You are currently at node 0 and your home is the largest node. You can go from u to v if it's immediately connected and the shortest distance from u to home is larger than the shortest distance from v to home. Return the number of unique paths possible to go from node 0 to home. Mod the result by 10 ** 9 + 7. Constraints 1 ≤ n ≤ 100,000 where n is the length of edges 0 ≤ distance Example 1 Input edges = [ [0, 1, 1], [1, 2, 1], [2, 3, 1], [1, 3, 2] ] Output 2 Explanation There are two unique paths to go home: We can go 0 to 1 then 1 to 2 then 2 to 3 We can go 0 to 1 then 1 to 3 Example 2 Input edges = [ [0, 1, 1], [0, 2, 1], [1, 2, 2] ] Output 1 Explanation There is one unique path to go home: go 0 to 2 We can't go through 0 -> 1 -> 2 because the shortest distance from 0 to 2 is smaller than 1 to 2

### Solution :

` ````
Solution in C++ :
long long dp[100005];
long long dist[100005];
int n;
vector<pair<int, int>> edges[100005];
int solve(vector<vector<int>>& graph) {
// begin graph initialization
n = 0;
for (auto& edge : graph) {
n = max(n, edge[0]);
n = max(n, edge[1]);
}
n++;
for (int i = 0; i < n; i++) edges[i].clear();
for (auto& edge : graph) {
edges[edge[0]].emplace_back(edge[1], edge[2]);
edges[edge[1]].emplace_back(edge[0], edge[2]);
}
// end graph initialization
// begin dijkstra
for (int i = 0; i < n; i++) dist[i] = 1e18;
dist[n - 1] = 0;
priority_queue<pair<long long, long long>> q;
q.emplace(0, n - 1);
vector<int> orderv;
while (q.size()) {
auto [w, v] = q.top();
q.pop();
w *= -1;
if (dist[v] != w) continue;
orderv.push_back(v);
for (auto edge : edges[v]) {
if (dist[edge.first] > dist[v] + edge.second) {
dist[edge.first] = dist[v] + edge.second;
q.emplace(-dist[edge.first], edge.first);
}
}
}
// end dijkstra
// begin path counting
for (int i = 0; i < n; i++) dp[i] = 0;
dp[n - 1] = 1;
const int MOD = 1000000007;
for (int currv : orderv) {
for (auto& edge : edges[currv]) {
if (dist[edge.first] < dist[currv]) {
dp[currv] += dp[edge.first];
dp[currv] %= MOD;
}
}
}
// end path counting
return dp[0];
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] edges) {
int N = 0;
for (int[] edge : edges) N = Math.max(N, Math.max(edge[0], edge[1]));
N++;
ArrayList<State>[] graph = new ArrayList[N];
for (int i = 0; i < N; i++) graph[i] = new ArrayList<State>();
for (int[] e : edges) {
long w = (long) e[2];
graph[e[0]].add(new State(e[1], w));
graph[e[1]].add(new State(e[0], w));
}
// Dijkstra's Algorithm
long[] distances = new long[N];
Arrays.fill(distances, Long.MAX_VALUE);
distances[N - 1] = 0L;
PriorityQueue<State> pq = new PriorityQueue<State>();
State s = new State(N - 1, 0L);
pq.add(s);
while (!pq.isEmpty()) {
s = pq.poll();
for (State e : graph[s.node]) {
if (distances[e.node] > distances[s.node] + e.dist) {
distances[e.node] = distances[s.node] + e.dist;
pq.add(new State(e.node, distances[e.node]));
}
}
}
State[] vals = new State[N];
for (int i = 0; i < N; i++) vals[i] = new State(i, distances[i]);
Arrays.sort(vals);
long[] ans = new long[N];
ans[N - 1] = 1L;
for (int i = 1; i < N; i++) {
int u = vals[i].node;
for (State e : graph[u]) {
int v = e.node;
if (distances[u] > distances[v])
ans[u] = (ans[u] + ans[v]) % 1000000007;
}
}
return (int) ans[0];
}
static class State implements Comparable<State> {
int node;
long dist;
public State(int n, long d) {
node = n;
dist = d;
}
public int compareTo(State s) {
return Long.compare(dist, s.dist);
}
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, edges):
MOD = 10 ** 9 + 7
graph = defaultdict(dict)
for u, v, w in edges:
graph[u][v] = graph[v][u] = w
# Find distance from home to nodes with Dijkstra
home = max(graph)
dist = defaultdict(lambda: float("inf"))
dist[home] = 0
pq = [[0, home]]
while pq:
d, node = heappop(pq)
if dist[node] < d:
continue
for nei, w in graph[node].items():
if d + w < dist[nei]:
heappush(pq, [d + w, nei])
dist[nei] = d + w
# DP on topological order
nodes = sorted(graph, key=dist.__getitem__)
dp = defaultdict(int)
dp[home] = 1
for u in nodes:
for v in graph[u]:
if dist[v] > dist[u]:
dp[v] += dp[u]
dp[v] %= MOD
return dp[0]
```

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