Leetcode #1733: Minimum Number of People to Teach
In this guide, we solve Leetcode #1733 Minimum Number of People to Teach in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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 ith user knows, and friendships[i] = [ui, vi] denotes a friendship between the users ui and vi.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Array, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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)
Complexity
The time complexity is . The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.