Report Spam Message

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, but since we need at least 2 matches and only found 1, this should return false. However, this example appears to be incomplete or incorrect.

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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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

✓ Brute Force

⏱️ Time: O(n×m) Space: O(1)

For each word in the message, iterate through all banned words to check for matches. Count the total matches and return true if count >= 2.

Hash Set Optimization

⏱️ Time: O(n + m) Space: O(m)

Convert banned words array to a hash set for constant time lookups. Then iterate through message words, counting matches using the hash set.

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 parseArray(char line[], char arr[][100]) {
    int count = 0;
    int inQuote = 0;
    int start = -1;
    int len = strlen(line);
    
    for (int i = 0; i < len; i++) {
        if (line[i] == '"') {
            if (!inQuote) {
                inQuote = 1;
                start = i + 1;
            } else {
                inQuote = 0;
                int wordLen = i - start;
                strncpy(arr[count], &line[start], wordLen);
                arr[count][wordLen] = '\0';
                count++;
            }
        }
    }
    return count;
}

int main() {
    char message[1000][100];
    char bannedWords[1000][100];
    char line[10000];
    
    fgets(line, sizeof(line), stdin);
    int messageSize = parseArray(line, message);
    
    fgets(line, sizeof(line), stdin);
    int bannedSize = parseArray(line, bannedWords);
    
    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

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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