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day-32.cpp
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day-32.cpp
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/*
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 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.
*/
// O(N) & O(N) solution using hash map
class Solution {
public:
int numJewelsInStones(string J, string S) {
int result = 0;
unordered_map<char, int> hashMap;
for (char ch : S) {
hashMap[ch] += 1;
}
for (char ch : J) {
result += hashMap[ch];
}
return result;
}
};