771. Jewels and Stones

DifficultyEasy

Description

You're 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".

Example 1:

Input: jewels = "aA", stones = "aAAbbbb"
Output: 3

Example 2:

Input: jewels = "z", stones = "ZZ"
Output: 0

Constraints:

1 <= jewels.length, stones.length <= 50
jewels and stones consist of only English letters.
All the characters of jewels are unique.

Solutions

Solution 1: Hash Table or Array

Thinking

Count how many stones are jewels. Length \(50\): put jewels in a set and test each stone. Case matters.

We can first use a hash table or array \(s\) to record all types of jewels. Then traverse all the stones, and if the current stone is a jewel, increment the answer by one.

The time complexity is \(O(m+n)\), and the space complexity is \(O(|\Sigma|)\), where \(m\) and \(n\) are the lengths of the strings \(\textit{jewels}\) and \(\textit{stones}\) respectively, and \(\Sigma\) is the character set, which in this problem is the set of all uppercase and lowercase English letters.

Python3JavaC++GoTypeScriptRustJavaScriptC

class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        s = set(jewels)
        return sum(c in s for c in stones)

class Solution {
    public int numJewelsInStones(String jewels, String stones) {
        int[] s = new int[128];
        for (char c : jewels.toCharArray()) {
            s[c] = 1;
        }
        int ans = 0;
        for (char c : stones.toCharArray()) {
            ans += s[c];
        }
        return ans;
    }
}

class Solution {
public:
    int numJewelsInStones(string jewels, string stones) {
        int s[128] = {0};
        for (char c : jewels) {
            s[c] = 1;
        }
        int ans = 0;
        for (char c : stones) {
            ans += s[c];
        }
        return ans;
    }
};

func numJewelsInStones(jewels string, stones string) (ans int) {
    s := [128]int{}
    for _, c := range jewels {
        s[c] = 1
    }
    for _, c := range stones {
        ans += s[c]
    }
    return
}

function numJewelsInStones(jewels: string, stones: string): number {
    const s = new Set([...jewels]);
    let ans = 0;
    for (const c of stones) {
        s.has(c) && ans++;
    }
    return ans;
}

use std::collections::HashSet;
impl Solution {
    pub fn num_jewels_in_stones(jewels: String, stones: String) -> i32 {
        let mut s = jewels.as_bytes().iter().collect::<HashSet<&u8>>();
        let mut ans = 0;
        for c in stones.as_bytes() {
            if s.contains(c) {
                ans += 1;
            }
        }
        ans
    }
}

/**
 * @param {string} jewels
 * @param {string} stones
 * @return {number}
 */
var numJewelsInStones = function (jewels, stones) {
    const s = new Set(jewels.split(''));
    return stones.split('').reduce((prev, val) => prev + s.has(val), 0);
};

int numJewelsInStones(char* jewels, char* stones) {
    int set[128] = {0};
    for (int i = 0; jewels[i]; i++) {
        set[jewels[i]] = 1;
    }
    int ans = 0;
    for (int i = 0; stones[i]; i++) {
        set[stones[i]] && ans++;
    }
    return ans;
}

771. Jewels and Stones

Description

Solutions

Solution 1: Hash Table or Array

Comments