383. Ransom Note
easyGiven two strings ransomNote and magazine, decide whether ransomNote can be constructed from the letters of magazine (each letter used at most once). Drop-in letter-count problem — one of the gentlest hash-counting warm-ups in the catalog.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
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.
Constraints
1 <= ransomNote.length, magazine.length <= 10^5ransomNote and magazine consist of lowercase English letters.
Examples
Example 1
ransomNote = 'a', magazine = 'b'falseExample 2
ransomNote = 'aa', magazine = 'ab'falseExample 3
ransomNote = 'aa', magazine = 'aab'trueSolve it now
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Hints
Progressive — try the first before opening the next.
Hint 1
If ransomNote.length > magazine.length, immediate false.
Hint 2
Count letters in magazine; for each letter in ransomNote, decrement. If any count goes negative, return false.
Hint 3
For lowercase-only, a length-26 array is faster than a hash map with the same logic.
Solution approach
Reveal approach
If len(ransomNote) > len(magazine), return false. Build a length-26 count array from magazine: count[c - 'a'] += 1 for each char. Walk ransomNote: for each char, decrement count[c - 'a']; if it goes below 0, return false. If you finish the walk without going negative, return true. Single linear scan in each direction. O(n + m) time, O(1) space (26-entry array is constant). The hash-map version is the same algorithm with a dict; slightly slower constants. The 'len check first' is the cheap O(1) early-exit.
Complexity
- Time
- O(n + m)
- Space
- O(1)
Related patterns
- hash-table
- counting
- string
Related problems
Asked at
Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).
- Amazon
- Apple
- Microsoft
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