How do you approach Ransom Note?

You need to know whether the magazine has enough of each letter.

What is the time complexity of Ransom Note?

Time complexity: O(m + n) - one pass over each string. Space complexity: O(1) - at most 26 keys.

What are common mistakes on Ransom Note?

Storing the current value before checking for its complement, which lets an element pair with itself. Reaching for a sort to deduplicate when a set does it in one pass - that swap quietly costs you the O(n). Using a plain object as the map when keys may be numbers or __proto__; a real hash-map type keeps key identity intact.

gitGood.dev

← Back to problems

Ransom Note

EasyData Structures

Description

Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine, and false otherwise.

Each letter in magazine can only be used once in ransomNote.

Example 1

Input: ransomNote = "a", magazine = "b"

Output: false

Example 2

Input: ransomNote = "aa", magazine = "ab"

Output: false

Example 3

Input: ransomNote = "aa", magazine = "aab"

Output: true

Constraints

- 1 <= ransomNote.length, magazine.length <= 10^5 - ransomNote and magazine consist of lowercase English letters.

Ransom Note: approach, complexity and pitfalls

An easy arrays & hashing problem, graded against 7 test cases (4 of them hidden).

The pattern: Arrays & Hashing

Hash maps and sets to trade memory for time - the most common first instinct in interviews.

Reach for it when you see: "Have I seen this before?", counting occurrences, or an O(n²) scan you want to make O(n).

More Arrays & Hashing problems →

If you are stuck

Show the 3-step hint ladder

You need to know whether the magazine has enough of each letter.
Count the occurrences of each letter in the magazine first.
Walk through the ransom note, decrementing counts - if a count goes below zero, return false.

The optimal approach

Show the approach: Frequency Count

Count each letter in `magazine` using a hash map (or a 26-slot array, since input is lowercase letters). Then iterate over `ransomNote`, decrementing the count for each letter. If any letter's count drops below zero, the magazine doesn't have enough of that letter.

Time: O(m + n) - one pass over each string
Space: O(1) - at most 26 keys

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

Common Arrays & Hashing mistakes

These apply to the pattern as a whole, not just this problem.

Storing the current value before checking for its complement, which lets an element pair with itself.
Reaching for a sort to deduplicate when a set does it in one pass - that swap quietly costs you the O(n).
Using a plain object as the map when keys may be numbers or `__proto__`; a real hash-map type keeps key identity intact.

← Back to problems

Ransom Note

EasyData Structures

Description

Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine, and false otherwise.

Each letter in magazine can only be used once in ransomNote.

Example 1

Input: ransomNote = "a", magazine = "b"

Output: false

Example 2

Input: ransomNote = "aa", magazine = "ab"

Output: false

Example 3

Input: ransomNote = "aa", magazine = "aab"

Output: true

Constraints

- 1 <= ransomNote.length, magazine.length <= 10^5 - ransomNote and magazine consist of lowercase English letters.

Ransom Note: approach, complexity and pitfalls

An easy arrays & hashing problem, graded against 7 test cases (4 of them hidden).

The pattern: Arrays & Hashing

Hash maps and sets to trade memory for time - the most common first instinct in interviews.

Reach for it when you see: "Have I seen this before?", counting occurrences, or an O(n²) scan you want to make O(n).

More Arrays & Hashing problems →

If you are stuck

Show the 3-step hint ladder

You need to know whether the magazine has enough of each letter.
Count the occurrences of each letter in the magazine first.
Walk through the ransom note, decrementing counts - if a count goes below zero, return false.

The optimal approach

Show the approach: Frequency Count

Time: O(m + n) - one pass over each string
Space: O(1) - at most 26 keys

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

Common Arrays & Hashing mistakes

These apply to the pattern as a whole, not just this problem.

Storing the current value before checking for its complement, which lets an element pair with itself.
Reaching for a sort to deduplicate when a set does it in one pass - that swap quietly costs you the O(n).
Using a plain object as the map when keys may be numbers or `__proto__`; a real hash-map type keeps key identity intact.