DEV Community is a community of 783,060 amazing developers

We're a place where coders share, stay up-to-date and grow their careers.

Aroup Goldar Dhruba

Posted on • Updated on

LeetCode: Jewels and Stones

Problem Statement

You're given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.

The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".

Example

Example 1:

Input: J = "aA", S = "aAAbbbb"
Output: 3

Example 2:

Input: J = "z", S = "ZZ"
Output: 0

Note:

• S and J will consist of letters and have length at most 50.
• The characters in J are distinct.

Solution Thought Process

This is a well-known hash problem. First, we take the jewel string and record the frequencies in the hash set. Then we go through the S string to find out if this is a jewel by checking the set one by one, increasing the count of the result by one.

We can find the items in the unordered_set in O(1) time.

Solution

class Solution {
public:
int numJewelsInStones(string J, string S) {
unordered_set<char>jewels;
int result = 0;
for(int i=0;i<J.size();i++)
{
jewels.insert(J[i]);
}
for(int i=0;i<S.size();i++)
{
if(jewels.find(S[i])!=jewels.end())
{
result++;
}
}
return result;
}
};

Complexity

Time Complexity: O(m+n) where m = length of J, n = length of S

Space Complexity: O(m) where m = length of J