LeetCode #3310 — MEDIUM

Remove Methods From Project

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

You are maintaining a project that has n methods numbered from 0 to n - 1.

You are given two integers n and k, and a 2D integer array invocations, where invocations[i] = [a_i, b_i] indicates that method a_i invokes method b_i.

There is a known bug in method k. Method k, along with any method invoked by it, either directly or indirectly, are considered suspicious and we aim to remove them.

A group of methods can only be removed if no method outside the group invokes any methods within it.

Return an array containing all the remaining methods after removing all the suspicious methods. You may return the answer in any order. If it is not possible to remove all the suspicious methods, none should be removed.

Example 1:

Input: n = 4, k = 1, invocations = [[1,2],[0,1],[3,2]]

Output: [0,1,2,3]

Explanation:

Method 2 and method 1 are suspicious, but they are directly invoked by methods 3 and 0, which are not suspicious. We return all elements without removing anything.

Example 2:

Input: n = 5, k = 0, invocations = [[1,2],[0,2],[0,1],[3,4]]

Output: [3,4]

Explanation:

Methods 0, 1, and 2 are suspicious and they are not directly invoked by any other method. We can remove them.

Example 3:

Input: n = 3, k = 2, invocations = [[1,2],[0,1],[2,0]]

Output: []

Explanation:

All methods are suspicious. We can remove them.

Constraints:

1 <= n <= 10⁵
0 <= k <= n - 1
0 <= invocations.length <= 2 * 10⁵
invocations[i] == [a_i, b_i]
0 <= a_i, b_i <= n - 1
a_i != b_i
invocations[i] != invocations[j]

Step 01

Brute Force Baseline

Problem summary: You are maintaining a project that has n methods numbered from 0 to n - 1. You are given two integers n and k, and a 2D integer array invocations, where invocations[i] = [ai, bi] indicates that method ai invokes method bi. There is a known bug in method k. Method k, along with any method invoked by it, either directly or indirectly, are considered suspicious and we aim to remove them. A group of methods can only be removed if no method outside the group invokes any methods within it. Return an array containing all the remaining methods after removing all the suspicious methods. You may return the answer in any order. If it is not possible to remove all the suspicious methods, none should be removed.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

4
1
[[1,2],[0,1],[3,2]]

Example 2

5
0
[[1,2],[0,2],[0,1],[3,4]]

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Use DFS from node <code>k</code>.
Mark all the nodes visited from node <code>k</code>, and then check if they can be visited from the other nodes.

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 #3310: Remove Methods From Project
class Solution {
    private boolean[] suspicious;
    private boolean[] vis;
    private List<Integer>[] f;
    private List<Integer>[] g;

    public List<Integer> remainingMethods(int n, int k, int[][] invocations) {
        suspicious = new boolean[n];
        vis = new boolean[n];
        f = new List[n];
        g = new List[n];
        Arrays.setAll(f, i -> new ArrayList<>());
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : invocations) {
            int a = e[0], b = e[1];
            f[a].add(b);
            f[b].add(a);
            g[a].add(b);
        }
        dfs(k);
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i] && !vis[i]) {
                dfs2(i);
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (!suspicious[i]) {
                ans.add(i);
            }
        }
        return ans;
    }

    private void dfs(int i) {
        suspicious[i] = true;
        for (int j : g[i]) {
            if (!suspicious[j]) {
                dfs(j);
            }
        }
    }

    private void dfs2(int i) {
        vis[i] = true;
        for (int j : f[i]) {
            if (!vis[j]) {
                suspicious[j] = false;
                dfs2(j);
            }
        }
    }
}

// Accepted solution for LeetCode #3310: Remove Methods From Project
func remainingMethods(n int, k int, invocations [][]int) []int {
	suspicious := make([]bool, n)
	vis := make([]bool, n)
	f := make([][]int, n)
	g := make([][]int, n)

	for _, e := range invocations {
		a, b := e[0], e[1]
		f[a] = append(f[a], b)
		f[b] = append(f[b], a)
		g[a] = append(g[a], b)
	}

	var dfs func(int)
	dfs = func(i int) {
		suspicious[i] = true
		for _, j := range g[i] {
			if !suspicious[j] {
				dfs(j)
			}
		}
	}

	dfs(k)

	var dfs2 func(int)
	dfs2 = func(i int) {
		vis[i] = true
		for _, j := range f[i] {
			if !vis[j] {
				suspicious[j] = false
				dfs2(j)
			}
		}
	}

	for i := 0; i < n; i++ {
		if !suspicious[i] && !vis[i] {
			dfs2(i)
		}
	}

	var ans []int
	for i := 0; i < n; i++ {
		if !suspicious[i] {
			ans = append(ans, i)
		}
	}

	return ans
}

# Accepted solution for LeetCode #3310: Remove Methods From Project
class Solution:
    def remainingMethods(
        self, n: int, k: int, invocations: List[List[int]]
    ) -> List[int]:
        def dfs(i: int):
            suspicious[i] = True
            for j in g[i]:
                if not suspicious[j]:
                    dfs(j)

        def dfs2(i: int):
            vis[i] = True
            for j in f[i]:
                if not vis[j]:
                    suspicious[j] = False
                    dfs2(j)

        f = [[] for _ in range(n)]
        g = [[] for _ in range(n)]
        for a, b in invocations:
            f[a].append(b)
            f[b].append(a)
            g[a].append(b)
        suspicious = [False] * n
        dfs(k)

        vis = [False] * n
        ans = []
        for i in range(n):
            if not suspicious[i] and not vis[i]:
                dfs2(i)
        return [i for i in range(n) if not suspicious[i]]

// Accepted solution for LeetCode #3310: Remove Methods From Project
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3310: Remove Methods From Project
// class Solution {
//     private boolean[] suspicious;
//     private boolean[] vis;
//     private List<Integer>[] f;
//     private List<Integer>[] g;
// 
//     public List<Integer> remainingMethods(int n, int k, int[][] invocations) {
//         suspicious = new boolean[n];
//         vis = new boolean[n];
//         f = new List[n];
//         g = new List[n];
//         Arrays.setAll(f, i -> new ArrayList<>());
//         Arrays.setAll(g, i -> new ArrayList<>());
//         for (var e : invocations) {
//             int a = e[0], b = e[1];
//             f[a].add(b);
//             f[b].add(a);
//             g[a].add(b);
//         }
//         dfs(k);
//         for (int i = 0; i < n; ++i) {
//             if (!suspicious[i] && !vis[i]) {
//                 dfs2(i);
//             }
//         }
//         List<Integer> ans = new ArrayList<>();
//         for (int i = 0; i < n; ++i) {
//             if (!suspicious[i]) {
//                 ans.add(i);
//             }
//         }
//         return ans;
//     }
// 
//     private void dfs(int i) {
//         suspicious[i] = true;
//         for (int j : g[i]) {
//             if (!suspicious[j]) {
//                 dfs(j);
//             }
//         }
//     }
// 
//     private void dfs2(int i) {
//         vis[i] = true;
//         for (int j : f[i]) {
//             if (!vis[j]) {
//                 suspicious[j] = false;
//                 dfs2(j);
//             }
//         }
//     }
// }

// Accepted solution for LeetCode #3310: Remove Methods From Project
function remainingMethods(n: number, k: number, invocations: number[][]): number[] {
    const suspicious: boolean[] = Array(n).fill(false);
    const vis: boolean[] = Array(n).fill(false);
    const f: number[][] = Array.from({ length: n }, () => []);
    const g: number[][] = Array.from({ length: n }, () => []);

    for (const [a, b] of invocations) {
        f[a].push(b);
        f[b].push(a);
        g[a].push(b);
    }

    const dfs = (i: number) => {
        suspicious[i] = true;
        for (const j of g[i]) {
            if (!suspicious[j]) {
                dfs(j);
            }
        }
    };

    dfs(k);

    const dfs2 = (i: number) => {
        vis[i] = true;
        for (const j of f[i]) {
            if (!vis[j]) {
                suspicious[j] = false;
                dfs2(j);
            }
        }
    };

    for (let i = 0; i < n; i++) {
        if (!suspicious[i] && !vis[i]) {
            dfs2(i);
        }
    }

    return Array.from({ length: n }, (_, i) => i).filter(i => !suspicious[i]);
}

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 + m)

Space

O(n + 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.