# Intuition

To solve this problem, we can iterate over each stone and check if it matches any of the jewels. If it does, we increment a counter.

# Approach

We can use two nested loops to compare each stone with each jewel. Alternatively, we can use a hash set to store the jewels and look up each stone in constant time.

# Complexity

Time complexity:

The time complexity of the nested loops approach is $$O(nm)$$

where n is the length of stones and m is the length of jewels. The time complexity of the hash set approach is $$O(n+m)$$

where n is the length of stones and m is the length of jewels.Space complexity:

The space complexity of the nested loops approach is $$O(1)$$

as we do not use any extra space. The space complexity of the hash set approach is $$O(m)$$

where m is the length of jewels.

# Code

```
class Solution {
public int numJewelsInStones(String jewels, String stones) {
int count = 0;
for (char stone : stones.toCharArray()) {
for (char jewel : jewels.toCharArray()) {
if (stone == jewel) {
count++;
continue;
}
}
}
return count;
}
}
```

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