LeetCode #797 — MEDIUM

All Paths From Source to Target

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

The Problem

Problem Statement

Given a directed acyclic graph (DAG) of n nodes labeled from 0 to n - 1, find all possible paths from node 0 to node n - 1 and return them in any order.

The graph is given as follows: graph[i] is a list of all nodes you can visit from node i (i.e., there is a directed edge from node i to node graph[i][j]).

Example 1:

Input: graph = [[1,2],[3],[3],[]]
Output: [[0,1,3],[0,2,3]]
Explanation: There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Example 2:

Input: graph = [[4,3,1],[3,2,4],[3],[4],[]]
Output: [[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]

Constraints:

n == graph.length
2 <= n <= 15
0 <= graph[i][j] < n
graph[i][j] != i (i.e., there will be no self-loops).
All the elements of graph[i] are unique.
The input graph is guaranteed to be a DAG.

Step 01

Brute Force Baseline

Problem summary: Given a directed acyclic graph (DAG) of n nodes labeled from 0 to n - 1, find all possible paths from node 0 to node n - 1 and return them in any order. The graph is given as follows: graph[i] is a list of all nodes you can visit from node i (i.e., there is a directed edge from node i to node graph[i][j]).

Baseline thinking

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

Pattern signal: Backtracking

Example 1

[[1,2],[3],[3],[]]

Example 2

[[4,3,1],[3,2,4],[3],[4],[]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #797: All Paths From Source to Target
class Solution {
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        int n = graph.length;
        Queue<List<Integer>> queue = new ArrayDeque<>();
        queue.offer(Arrays.asList(0));
        List<List<Integer>> ans = new ArrayList<>();
        while (!queue.isEmpty()) {
            List<Integer> path = queue.poll();
            int u = path.get(path.size() - 1);
            if (u == n - 1) {
                ans.add(path);
                continue;
            }
            for (int v : graph[u]) {
                List<Integer> next = new ArrayList<>(path);
                next.add(v);
                queue.offer(next);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #797: All Paths From Source to Target
func allPathsSourceTarget(graph [][]int) [][]int {
	var path []int
	path = append(path, 0)
	var ans [][]int

	var dfs func(i int)
	dfs = func(i int) {
		if i == len(graph)-1 {
			ans = append(ans, append([]int(nil), path...))
			return
		}
		for _, j := range graph[i] {
			path = append(path, j)
			dfs(j)
			path = path[:len(path)-1]
		}
	}

	dfs(0)
	return ans
}

# Accepted solution for LeetCode #797: All Paths From Source to Target
class Solution:
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        n = len(graph)
        q = deque([[0]])
        ans = []
        while q:
            path = q.popleft()
            u = path[-1]
            if u == n - 1:
                ans.append(path)
                continue
            for v in graph[u]:
                q.append(path + [v])
        return ans

// Accepted solution for LeetCode #797: All Paths From Source to Target
impl Solution {
    fn dfs(i: usize, path: &mut Vec<i32>, res: &mut Vec<Vec<i32>>, graph: &Vec<Vec<i32>>) {
        path.push(i as i32);
        if i == graph.len() - 1 {
            res.push(path.clone());
        }
        for j in graph[i].iter() {
            Self::dfs(*j as usize, path, res, graph);
        }
        path.pop();
    }

    pub fn all_paths_source_target(graph: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        let mut res = Vec::new();
        Self::dfs(0, &mut vec![], &mut res, &graph);
        res
    }
}

// Accepted solution for LeetCode #797: All Paths From Source to Target
function allPathsSourceTarget(graph: number[][]): number[][] {
    const ans: number[][] = [];

    const dfs = (path: number[]) => {
        const curr = path.at(-1)!;
        if (curr === graph.length - 1) {
            ans.push([...path]);
            return;
        }

        for (const v of graph[curr]) {
            path.push(v);
            dfs(path);
            path.pop();
        }
    };

    dfs([0]);

    return ans;
}

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(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.