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jewels-and-stones.dart (688B)

      1 //For each stone check if the given index is the same character as
      2 //a jewel. If it is add one to the number of jewels returned.
      3 //This algorithm is a O(n + m) time complexity where n is the length of the stones and m is the length of the jewels.
      4 //Runtime: 249ms Beats: 83.33%
      5 //Memory: 141MB Beats: 40.48%
      6 class Solution {
      7   int numJewelsInStones(String jewels, String stones) {
      8       Set <String> vals = {};
      9       for(int i = 0 ; i < jewels.length ; ++i){
     10           vals.add(jewels[i]);
     11       }
     12       int return_val = 0;
     13       for(int i = 0 ; i < stones.length ; ++i){
     14         if(jewels.contains(stones[i])){
     15           return_val += 1;
     16         }
     17       }
     18       return return_val;
     19   }
     20 }