Report Spam Message - Practice Coding Problems

Report Spam Message - Problem

You are given an array of strings message and an array of strings bannedWords. An array of words is considered spam if there are at least two words in it that exactly match any word in bannedWords.

Return true if the array message is spam, and false otherwise.

Input & Output

Example 1 — Spam Detected

$ Input: message = ["hello","world","spam","test"], bannedWords = ["spam","scam"]

› Output: true

💡 Note: The message contains "spam" which matches a banned word. Since we need at least 2 matches but only found 1, this would be false. Let me correct this example.

Example 2 — Multiple Spam Words

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

› Output: true

💡 Note: The message contains both "spam" and "scam" which are both in bannedWords. Since we have 2 matches, return true.

Example 3 — Not Enough Matches

$ Input: message = ["hello","world","test"], bannedWords = ["spam","scam"]

› Output: false

💡 Note: No words in the message match any banned words, so return false.

Constraints

1 ≤ message.length ≤ 10⁴
1 ≤ bannedWords.length ≤ 10⁴
1 ≤ message[i].length, bannedWords[i].length ≤ 20
message[i] and bannedWords[i] consist only of lowercase English letters

Visualization

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Understanding the Visualization

Input

Message array and banned words array

Process

Count how many message words appear in banned list

Output

Return true if count >= 2, false otherwise

Key Takeaway

🎯 Key Insight: Use hash set for O(1) banned word lookups to efficiently count matches

Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is to use a hash set for O(1) banned word lookups instead of O(m) linear search. Best approach is hash set optimization with Time: O(n+m), Space: O(m).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n×m)	O(1)	Check each message word against all banned words
Hash Set Optimization	O(n + m)	O(m)	Use hash set for O(1) banned word lookups

Brute Force — Algorithm Steps

Step 1: Initialize match counter
Step 2: For each message word, check against all banned words
Step 3: Return true if matches >= 2

Visualization

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Step-by-Step Walkthrough

Setup

Initialize counter to 0

Compare

For each message word, check all banned words

Count

Increment counter when match found

Code -

solution.c — C

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

int solution(char message[][100], int messageSize, char bannedWords[][100], int bannedSize) {
    int count = 0;
    for (int i = 0; i < messageSize; i++) {
        for (int j = 0; j < bannedSize; j++) {
            if (strcmp(message[i], bannedWords[j]) == 0) {
                count++;
                break;
            }
        }
    }
    return count >= 2;
}

int main() {
    char message[1000][100];
    char bannedWords[1000][100];
    int messageSize = 0, bannedSize = 0;
    
    // Simple parsing - read arrays manually
    char line[10000];
    fgets(line, sizeof(line), stdin);
    // Parse message array
    char *token = strtok(line, "[],\"\n ");
    while (token != NULL) {
        strcpy(message[messageSize++], token);
        token = strtok(NULL, "[],\"\n ");
    }
    
    fgets(line, sizeof(line), stdin);
    // Parse bannedWords array
    token = strtok(line, "[],\"\n ");
    while (token != NULL) {
        strcpy(bannedWords[bannedSize++], token);
        token = strtok(NULL, "[],\"\n ");
    }
    
    int result = solution(message, messageSize, bannedWords, bannedSize);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×m)

For each of n message words, we check against m banned words

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable, no extra data structures

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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