37. Path with Maximum Probability
You are given an undirected weighted graph of n
nodes (0-indexed), represented by an edge list where edges[i] = [a, b]
is an undirected edge connecting the nodes a
and b
with a probability of success of traversing that edge succProb[i]
.
Given two nodes start
and end
, find the path with the maximum probability of success to go from start
to end
and return its success probability.
If there is no path from start
to end
, return 0. Your answer will be accepted if it differs from the correct answer by at most 1e-5.
Example 1:
Input: n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2
Output: 0.25000
Explanation: There are two paths from start to end, one having a probability of success = 0.2 and the other has 0.5 * 0.5 = 0.25.
Example 2:
Input: n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start = 0, end = 2
Output: 0.30000
Example 3:
Input: n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2
Output: 0.00000
Explanation: There is no path between 0 and 2.
Solution: (Dijkstra With MaxHeap)
class Solution
{
public:
void findPath(vector<pair<int, double>> v[], int node, vector<bool> &vs, vector<double> &wt)
{
priority_queue<pair<double, int>> pq;
pq.push({1, node});
while (!pq.empty())
{
int n = pq.top().second;
double w = pq.top().first;
pq.pop();
if (vs[n])
{
continue;
}
vs[n] = 1;
for (int i = 0; i < v[n].size(); i++)
{
int in = v[n][i].first;
double iwt = v[n][i].second;
if (w * iwt > wt[in])
{
wt[in] = w * iwt;
pq.push({wt[in],in});
}
}
}
return;
}
double maxProbability(int n, vector<vector<int>> &edges, vector<double> &succProb, int start, int end)
{
vector<pair<int, double>> v[n];
vector<bool> vs(n, false);
vector<double> wt(n, INT_MIN);
for (int i = 0; i < edges.size(); i++)
{
int x = edges[i][0];
int y = edges[i][1];
v[x].push_back({y, succProb[i]});
v[y].push_back({x, succProb[i]});
}
findPath(v, start, vs, wt);
if(wt[end] == INT_MIN){
return 0;
}
return wt[end];
}
};
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