771. Jewels and Stones
easyCount how many characters in one string also appear in another. A classic hash-set warm-up that maps cleanly to membership-test fundamentals.
By Sam K., Founder, InterviewChamp.AI · Last verified
Problem
You are given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels. Letters are case sensitive, so 'a' is considered a different type of stone from 'A'.
Constraints
1 <= jewels.length, stones.length <= 50jewels and stones consist of only English letters.All the characters of jewels are unique.
Examples
Example 1
jewels = "aA", stones = "aAAbbbb"3Explanation: Three stones ('a', 'A', 'A') match characters in jewels.
Example 2
jewels = "z", stones = "ZZ"0Explanation: Case-sensitive: 'Z' is not in jewels.
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Hints
Progressive — try the first before opening the next.
Hint 1
A naive nested loop is O(n*m) — fine for the given constraints but worth recognising the inefficiency.
Hint 2
Put the jewels characters into a set, then iterate stones once and increment a counter on membership.
Hint 3
Sets give O(1) average lookup, so the total cost drops to O(n + m).
Solution approach
Reveal approach
Build a set from jewels, then count how many characters in stones are in that set. O(n + m) time where n = len(jewels) and m = len(stones). O(n) extra space for the set. The set is tiny in practice (at most 52 letters), so this is essentially constant memory.
Complexity
- Time
- O(n + m)
- Space
- O(n)
Related patterns
- hash-set
- counting
Related problems
Asked at
Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).
- Amazon
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