Report Spam Message - Practice Coding Problems

Report Spam Message - Problem

Spam Detection Challenge

You're building a content moderation system for a messaging platform. Given an array of strings message representing words in a message and an array of strings bannedWords containing prohibited terms, determine if the message should be flagged as spam.

A message is considered spam if it contains at least two words that exactly match any word in the bannedWords list. The same banned word can appear multiple times and each occurrence counts toward the spam threshold.

Goal: Return true if the message is spam, false otherwise.

Example:
• Message: ["hello", "world", "spam", "spam"]
• Banned: ["spam", "bad"]
• Result: true (contains 2 instances of "spam")

Input & Output

example_1.py — Basic Spam Detection

$ Input: message = ["hello", "world", "leetcode"], bannedWords = ["world", "hello"]

› Output: true

💡 Note: The message contains "hello" and "world", both of which are in bannedWords. Since we found 2 banned words, the message is spam.

example_2.py — Not Enough Banned Words

$ Input: message = ["hello", "programming", "fun"], bannedWords = ["world", "programming"]

› Output: false

💡 Note: Only "programming" from the message matches bannedWords. Since we need at least 2 matches and only found 1, the message is not spam.

example_3.py — Duplicate Banned Words Count

$ Input: message = ["spam", "spam", "spam"], bannedWords = ["spam"]

› Output: true

💡 Note: The word "spam" appears 3 times in the message and matches the banned word. Since 3 ≥ 2, the message is spam. Each occurrence counts separately.

Constraints

1 ≤ message.length ≤ 10⁴
1 ≤ bannedWords.length ≤ 10⁴
1 ≤ message[i].length, bannedWords[i].length ≤ 2000
message[i] and bannedWords[i] consist only of lowercase English letters
No duplicate words within bannedWords array

Visualization

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Understanding the Visualization

Setup Scanner

Create hash set of prohibited items (banned words) for instant recognition

Scan Items

Check each message word against the prohibited items list

Count Violations

Increment counter each time a prohibited item is found

Trigger Alert

Flag as spam immediately when 2 prohibited items detected

Key Takeaway

🎯 Key Insight: Hash set provides O(1) lookups, and early termination saves unnecessary processing once spam is detected!

Asked in

G Google 42 a Amazon 38 f Meta 35 ⊞ Microsoft 31

The optimal solution uses a hash set to store banned words for O(1) lookup time. We iterate through the message once, checking each word against the banned set and counting matches. The algorithm returns true immediately when 2 or more banned words are found, achieving O(n + m) time complexity with early termination optimization.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Hash Table (Optimal)	O(n + m)	O(m)	Build hash set and count matches simultaneously in one iteration
Two-Pass Hash Table	O(n + m)	O(m)	Build hash set first, then count matches in message
Brute Force (Nested Loops)	O(n × m)	O(1)	Check each message word against every banned word

One-Pass Hash Table (Optimal) — Algorithm Steps

Create hash set from bannedWords for O(1) lookups
Initialize counter to 0
For each word in message, check if it's in the banned set
If banned word found, increment counter
Return true immediately when counter reaches 2
Return false if we finish without finding 2 matches

Visualization

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Step-by-Step Walkthrough

Create Hash Set

Build hash set from banned words for O(1) lookups

Scan & Count

Check each message word and increment counter for matches

Early Exit

Return true immediately when 2 banned words found

Complete Scan

Return false if fewer than 2 banned words found

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

#define MAX_WORDS 1000
#define MAX_LEN 100

bool reportSpam(char message[][MAX_LEN], int msgLen, char bannedWords[][MAX_LEN], int bannedLen) {
    int count = 0;
    
    for (int i = 0; i < msgLen; i++) {
        for (int j = 0; j < bannedLen; j++) {
            if (strcmp(message[i], bannedWords[j]) == 0) {
                count++;
                if (count >= 2) {
                    return true;
                }
                break;
            }
        }
    }
    
    return false;
}

int main() {
    char message[MAX_WORDS][MAX_LEN], bannedWords[MAX_WORDS][MAX_LEN];
    int msgLen, bannedLen;
    // Parse input
    printf("%s\n", reportSpam(message, msgLen, bannedWords, bannedLen) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

O(m) to create hash set, O(n) to check message words, but can terminate early

✓ Linear Growth

Space Complexity

O(m)

Hash set stores unique banned words

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Hash Table (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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