Ransom Note

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Ever seen a movie where characters cut out random letters from magazines to spell out a secret note? In Ransom Note, you're given two strings, ransomNote and magazine. Your task is to return true if you can construct the note using the characters available in the magazine. Each letter in the magazine can only be used once, just like a real clipping!

The "secret sauce" here is a Frequency Map. Think of it as your character inventory. First, count how many of each letter you have in the magazine. As you walk through your note, you "spend" those counts. If you try to use a letter you don’t have, or one you’ve already run out of, the note becomes impossible to finish! This approach gives you a linear time complexity of O(n + m), which is much faster than searching for each individual letter. It’s an elegant, fast, and professional way to handle inventory-style lookup problems!

For text and token tasks, be precise with index movement and substring boundaries. Most hidden failures come from partial-match handling and boundary cuts, so keep comparisons explicit and avoid assumptions about implicit separators or formatting not guaranteed by the input contract.

Examples

Example 1

Input

ransomNote = "a", magazine = "b"

Output

false

Explanation

The needed character does not exist in magazine.

Example 2

Input

ransomNote = "aa", magazine = "ab"

Output

false

Explanation

There is only one a available.

Example 3

Input

ransomNote = "aa", magazine = "aab"

Output

true

Explanation

The magazine provides enough characters.

Algorithm Flow

Recommendation Algorithm Flow for Ransom Note - Budibadu

Best Answers

java

import java.util.*;
class Solution {
    public boolean ransom_note(String ransomNote, String magazine) {
        Map<Character, Integer> count = new HashMap<>();
        for (char ch : magazine.toCharArray()) count.put(ch, count.getOrDefault(ch, 0) + 1);
        for (char ch : ransomNote.toCharArray()) {
            if (!count.containsKey(ch) || count.get(ch) == 0) return false;
            count.put(ch, count.get(ch) - 1);
        }
        return true;
    }
}

javascript

function ransom_note(ransomNote, magazine) {
    const count = {};
    for (const ch of magazine) count[ch] = (count[ch] || 0) + 1;
    for (const ch of ransomNote) {
        if (!count[ch]) return false;
        count[ch]--;
    }
    return true;
}

php

<?php
function ransom_note($ransomNote, $magazine) {
    $count = [];
    foreach (str_split($magazine) as $ch) {
        $count[$ch] = ($count[$ch] ?? 0) + 1;
    }
    foreach (str_split($ransomNote) as $ch) {
        if (!isset($count[$ch]) || $count[$ch] === 0) return false;
        $count[$ch]--;
    }
    return true;
}
?>

python

def ransom_note(ransomNote: str, magazine: str) -> bool:
    count = {}
    for ch in magazine:
        count[ch] = count.get(ch, 0) + 1
    for ch in ransomNote:
        if count.get(ch, 0) == 0:
            return False
        count[ch] -= 1
    return True

rust

use std::collections::HashMap;
fn ransom_note(ransom_note: String, magazine: String) -> bool {
    let mut count: HashMap<char, i32> = HashMap::new();
    for ch in magazine.chars() {
        *count.entry(ch).or_insert(0) += 1;
    }
    for ch in ransom_note.chars() {
        match count.get_mut(&ch) {
            Some(value) if *value > 0 => *value -= 1,
            _ => return false,
        }
    }
    true
}

typescript

function ransom_note(ransomNote: string, magazine: string): boolean {
    var count: { [key: string]: number } = {};
    var i: number;
    for (i = 0; i < magazine.length; i++) {
        var m = magazine.charAt(i);
        count[m] = (count[m] || 0) + 1;
    }
    for (i = 0; i < ransomNote.length; i++) {
        var r = ransomNote.charAt(i);
        if (!count[r]) return false;
        count[r]--;
    }
    return true;
}

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Ransom Note

Examples

Algorithm Flow

Best Answers

Solutions Coming Soon

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Contains Duplicate

Valid Anagram

Two Sum Indices

Word Frequency Counter

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