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771. Jewels and Stones

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

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• 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;
      }
  };
  

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

• 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;
  }
  
  

• /**
   * @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);
  };
  
  

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

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