LeetCode #1733 — MEDIUM

Minimum Number of People to Teach

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

The Problem

Problem Statement

On a social network consisting of m users and some friendships between users, two users can communicate with each other if they know a common language.

You are given an integer n, an array languages, and an array friendships where:

There are n languages numbered 1 through n,
languages[i] is the set of languages the i^th user knows, and
friendships[i] = [u_i, v_i] denotes a friendship between the users u_i and v_i.

You can choose one language and teach it to some users so that all friends can communicate with each other. Return the minimum number of users you need to teach.

Note that friendships are not transitive, meaning if x is a friend of y and y is a friend of z, this doesn't guarantee that x is a friend of z.

Example 1:

Input: n = 2, languages = [[1],[2],[1,2]], friendships = [[1,2],[1,3],[2,3]]
Output: 1
Explanation: You can either teach user 1 the second language or user 2 the first language.

Example 2:

Input: n = 3, languages = [[2],[1,3],[1,2],[3]], friendships = [[1,4],[1,2],[3,4],[2,3]]
Output: 2
Explanation: Teach the third language to users 1 and 3, yielding two users to teach.

Constraints:

2 <= n <= 500
languages.length == m
1 <= m <= 500
1 <= languages[i].length <= n
1 <= languages[i][j] <= n
1 <= u_i < v_i <= languages.length
1 <= friendships.length <= 500
All tuples (u_i,v_i) are unique
languages[i] contains only unique values

Step 01

Brute Force Baseline

Problem summary: On a social network consisting of m users and some friendships between users, two users can communicate with each other if they know a common language. You are given an integer n, an array languages, and an array friendships where: There are n languages numbered 1 through n, languages[i] is the set of languages the ith user knows, and friendships[i] = [ui, vi] denotes a friendship between the users ui and vi. You can choose one language and teach it to some users so that all friends can communicate with each other. Return the minimum number of users you need to teach. Note that friendships are not transitive, meaning if x is a friend of y and y is a friend of z, this doesn't guarantee that x is a friend of z.

Baseline thinking

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

Pattern signal: Array · Hash Map · Greedy

Example 1

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

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

You can just use brute force and find out for each language the number of users you need to teach
Note that a user can appear in multiple friendships but you need to teach that user only once

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 #1733: Minimum Number of People to Teach
class Solution {
    public int minimumTeachings(int n, int[][] languages, int[][] friendships) {
        Set<Integer> s = new HashSet<>();
        for (var e : friendships) {
            int u = e[0], v = e[1];
            if (!check(u, v, languages)) {
                s.add(u);
                s.add(v);
            }
        }
        if (s.isEmpty()) {
            return 0;
        }
        int[] cnt = new int[n + 1];
        for (int u : s) {
            for (int l : languages[u - 1]) {
                ++cnt[l];
            }
        }
        int mx = 0;
        for (int v : cnt) {
            mx = Math.max(mx, v);
        }
        return s.size() - mx;
    }

    private boolean check(int u, int v, int[][] languages) {
        for (int x : languages[u - 1]) {
            for (int y : languages[v - 1]) {
                if (x == y) {
                    return true;
                }
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #1733: Minimum Number of People to Teach
func minimumTeachings(n int, languages [][]int, friendships [][]int) int {
	check := func(u, v int) bool {
		for _, x := range languages[u-1] {
			for _, y := range languages[v-1] {
				if x == y {
					return true
				}
			}
		}
		return false
	}
	s := map[int]bool{}
	for _, e := range friendships {
		u, v := e[0], e[1]
		if !check(u, v) {
			s[u], s[v] = true, true
		}
	}
	if len(s) == 0 {
		return 0
	}
	cnt := make([]int, n+1)
	for u := range s {
		for _, l := range languages[u-1] {
			cnt[l]++
		}
	}
	return len(s) - slices.Max(cnt)
}

# Accepted solution for LeetCode #1733: Minimum Number of People to Teach
class Solution:
    def minimumTeachings(
        self, n: int, languages: List[List[int]], friendships: List[List[int]]
    ) -> int:
        def check(u: int, v: int) -> bool:
            for x in languages[u - 1]:
                for y in languages[v - 1]:
                    if x == y:
                        return True
            return False

        s = set()
        for u, v in friendships:
            if not check(u, v):
                s.add(u)
                s.add(v)
        cnt = Counter()
        for u in s:
            for l in languages[u - 1]:
                cnt[l] += 1
        return len(s) - max(cnt.values(), default=0)

// Accepted solution for LeetCode #1733: Minimum Number of People to Teach
use std::collections::{HashMap, HashSet};

impl Solution {
    pub fn minimum_teachings(n: i32, languages: Vec<Vec<i32>>, friendships: Vec<Vec<i32>>) -> i32 {
        fn check(u: usize, v: usize, languages: &Vec<Vec<i32>>) -> bool {
            for &x in &languages[u - 1] {
                for &y in &languages[v - 1] {
                    if x == y {
                        return true;
                    }
                }
            }
            false
        }

        let mut s: HashSet<usize> = HashSet::new();
        for edge in friendships.iter() {
            let u = edge[0] as usize;
            let v = edge[1] as usize;
            if !check(u, v, &languages) {
                s.insert(u);
                s.insert(v);
            }
        }

        let mut cnt: HashMap<i32, i32> = HashMap::new();
        for &u in s.iter() {
            for &l in &languages[u - 1] {
                *cnt.entry(l).or_insert(0) += 1;
            }
        }

        let mx = cnt.values().cloned().max().unwrap_or(0);
        (s.len() as i32) - mx
    }
}

// Accepted solution for LeetCode #1733: Minimum Number of People to Teach
function minimumTeachings(n: number, languages: number[][], friendships: number[][]): number {
    function check(u: number, v: number): boolean {
        for (const x of languages[u - 1]) {
            for (const y of languages[v - 1]) {
                if (x === y) {
                    return true;
                }
            }
        }
        return false;
    }

    const s = new Set<number>();
    for (const [u, v] of friendships) {
        if (!check(u, v)) {
            s.add(u);
            s.add(v);
        }
    }

    const cnt = new Map<number, number>();
    for (const u of s) {
        for (const l of languages[u - 1]) {
            cnt.set(l, (cnt.get(l) || 0) + 1);
        }
    }

    return s.size - Math.max(0, ...cnt.values());
}

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 log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.