javascript 771.珠宝和石头

Posted 2021-05-03

// basic solution
// time complexity: sort O(nlogn), map + filter O(n^2)
// total time complexity: O(n^2)


/**
 * @param {string} J
 * @param {string} S
 * @return {number}
 */
var numJewelsInStones = function(J, S) {
  const sortedS = _.sortBy(S);

  return _.map(J, j => {
    return _.filter(sortedS, s => s === j).length;
  })
  .reduce((len1, len2) => len1 + len2)
};

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