LeetCode #1514 — MEDIUM

Path with Maximum Probability

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Constraints:

2 <= n <= 10^4
0 <= start, end < n
start != end
0 <= a, b < n
a != b
0 <= succProb.length == edges.length <= 2*10^4
0 <= succProb[i] <= 1
There is at most one edge between every two nodes.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

3
[[0,1],[1,2],[0,2]]
[0.5,0.5,0.2]
0
2

Example 2

3
[[0,1],[1,2],[0,2]]
[0.5,0.5,0.3]
0
2

Example 3

3
[[0,1]]
[0.5]
0
2

Core Insight

What unlocks the optimal approach

Multiplying probabilities will result in precision errors.
Take log probabilities to sum up numbers instead of multiplying them.
Use Dijkstra's algorithm to find the minimum path between the two nodes after negating all costs.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1514: Path with Maximum Probability
class Solution {
    public double maxProbability(
        int n, int[][] edges, double[] succProb, int start_node, int end_node) {
        List<Pair<Integer, Double>>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int i = 0; i < edges.length; ++i) {
            var e = edges[i];
            int a = e[0], b = e[1];
            double p = succProb[i];
            g[a].add(new Pair<>(b, p));
            g[b].add(new Pair<>(a, p));
        }
        double[] dist = new double[n];
        dist[start_node] = 1;
        PriorityQueue<Pair<Integer, Double>> pq
            = new PriorityQueue<>(Comparator.comparingDouble(p -> - p.getValue()));
        pq.offer(new Pair<>(start_node, 1.0));
        while (!pq.isEmpty()) {
            var p = pq.poll();
            int a = p.getKey();
            double w = p.getValue();
            if (dist[a] > w) {
                continue;
            }
            for (var e : g[a]) {
                int b = e.getKey();
                double pab = e.getValue();
                double wab = w * pab;
                if (wab > dist[b]) {
                    dist[b] = wab;
                    pq.offer(new Pair<>(b, wab));
                }
            }
        }
        return dist[end_node];
    }
}

// Accepted solution for LeetCode #1514: Path with Maximum Probability
func maxProbability(n int, edges [][]int, succProb []float64, start_node int, end_node int) float64 {
	g := make([][]pair, n)
	for i, e := range edges {
		a, b := e[0], e[1]
		p := succProb[i]
		g[a] = append(g[a], pair{p, b})
		g[b] = append(g[b], pair{p, a})
	}
	pq := hp{{1, start_node}}
	dist := make([]float64, n)
	dist[start_node] = 1
	for len(pq) > 0 {
		p := heap.Pop(&pq).(pair)
		w, a := p.p, p.a
		if dist[a] > w {
			continue
		}
		for _, e := range g[a] {
			b, p := e.a, e.p
			if nw := w * p; nw > dist[b] {
				dist[b] = nw
				heap.Push(&pq, pair{nw, b})
			}
		}
	}
	return dist[end_node]
}

type pair struct {
	p float64
	a int
}
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].p > h[j].p }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(x any)        { *h = append(*h, x.(pair)) }
func (h *hp) Pop() (x any)      { a := *h; x = a[len(a)-1]; *h = a[:len(a)-1]; return }

# Accepted solution for LeetCode #1514: Path with Maximum Probability
class Solution:
    def maxProbability(
        self,
        n: int,
        edges: List[List[int]],
        succProb: List[float],
        start_node: int,
        end_node: int,
    ) -> float:
        g: List[List[Tuple[int, float]]] = [[] for _ in range(n)]
        for (a, b), p in zip(edges, succProb):
            g[a].append((b, p))
            g[b].append((a, p))
        pq = [(-1, start_node)]
        dist = [0] * n
        dist[start_node] = 1
        while pq:
            w, a = heappop(pq)
            w = -w
            if dist[a] > w:
                continue
            for b, p in g[a]:
                if (t := w * p) > dist[b]:
                    dist[b] = t
                    heappush(pq, (-t, b))
        return dist[end_node]

// Accepted solution for LeetCode #1514: Path with Maximum Probability
struct Solution;
use std::cmp::Ordering;
use std::collections::BinaryHeap;
use std::collections::HashMap;

#[derive(PartialEq)]
struct State {
    id: usize,
    prob: f64,
}

impl State {
    fn new(id: usize, prob: f64) -> Self {
        State { id, prob }
    }
}

impl Eq for State {}

impl PartialOrd for State {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        self.prob.partial_cmp(&other.prob)
    }
}

impl Ord for State {
    fn cmp(&self, other: &Self) -> Ordering {
        self.partial_cmp(other).unwrap()
    }
}

impl Solution {
    fn max_probability(
        n: i32,
        edges: Vec<Vec<i32>>,
        succ_prob: Vec<f64>,
        start: i32,
        end: i32,
    ) -> f64 {
        let n = n as usize;
        let start = start as usize;
        let end = end as usize;
        let mut adj: Vec<HashMap<usize, f64>> = vec![HashMap::new(); n];
        for (i, edge) in edges.into_iter().enumerate() {
            let u = edge[0] as usize;
            let v = edge[1] as usize;
            let p = succ_prob[i];
            adj[u].insert(v, p);
            adj[v].insert(u, p);
        }
        let mut probs: Vec<f64> = vec![0.0; n];
        let mut queue: BinaryHeap<State> = BinaryHeap::new();
        queue.push(State::new(start, 1.0));
        probs[start] = 1.0;
        while let Some(parent) = queue.pop() {
            if parent.id == end {
                return parent.prob;
            }
            for (&child_id, &prob) in &adj[parent.id] {
                let new_prob = parent.prob * prob;
                if new_prob > probs[child_id] {
                    probs[child_id] = new_prob;
                    queue.push(State::new(child_id, new_prob));
                }
            }
        }
        0.0
    }
}

#[test]
fn test() {
    use assert_approx_eq::assert_approx_eq;
    let n = 3;
    let edges = vec_vec_i32![[0, 1], [1, 2], [0, 2]];
    let succ_prob = vec![0.5, 0.5, 0.2];
    let start = 0;
    let end = 2;
    let res = 0.25;
    assert_approx_eq!(
        Solution::max_probability(n, edges, succ_prob, start, end),
        res
    );

    let n = 3;
    let edges = vec_vec_i32![[0, 1], [1, 2], [0, 2]];
    let succ_prob = vec![0.5, 0.5, 0.3];
    let start = 0;
    let end = 2;
    let res = 0.3;
    assert_approx_eq!(
        Solution::max_probability(n, edges, succ_prob, start, end),
        res
    );

    let n = 3;
    let edges = vec_vec_i32![[0, 1]];
    let succ_prob = vec![0.5];
    let start = 0;
    let end = 2;
    let res = 0.00000;
    assert_approx_eq!(
        Solution::max_probability(n, edges, succ_prob, start, end),
        res
    );
}

// Accepted solution for LeetCode #1514: Path with Maximum Probability
function maxProbability(
    n: number,
    edges: number[][],
    succProb: number[],
    start_node: number,
    end_node: number,
): number {
    const pq = new PriorityQueue<number[]>((a, b) => b[0] - a[0]);
    const g: [number, number][][] = Array.from({ length: n }, () => []);
    for (let i = 0; i < edges.length; ++i) {
        const [a, b] = edges[i];
        g[a].push([b, succProb[i]]);
        g[b].push([a, succProb[i]]);
    }
    const dist = Array.from({ length: n }, () => 0);
    dist[start_node] = 1;
    pq.enqueue([1, start_node]);
    while (!pq.isEmpty()) {
        const [w, a] = pq.dequeue();
        if (dist[a] > w) {
            continue;
        }
        for (const [b, p] of g[a]) {
            const nw = w * p;
            if (nw > dist[b]) {
                dist[b] = nw;
                pq.enqueue([nw, b]);
            }
        }
    }
    return dist[end_node];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × log m)

Space

O(m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.