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This checks whether one string has all the chars in another string to make up that exact string. Counts matter so for, for example, str1 = 'abcdgbff' str2 ='aaabb' should return false because str1 only has one 'a' and 'b'.

How can I optimize my code further?

function scramble(str1, str2) {
  for (let i of str1) {
    if (str2.indexOf(i) != -1) {
      str2 = str2.slice(0, (str2.indexOf(i))) + str2.slice(str2.indexOf(i) + 1);
    }
  }
  return str2.length === 0;
}

// Example:
console.log(scramble('abcdgbff', 'aaabb')); // false

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If your input strings are of length n, your current implementation has a cubic runtime complexity O(n³). This could become an issue for longer strings.

I recommend to split counting character frequencies and comparing those counts into two separate steps with nearly linear complexity O(n):

// Count characters in 'str':
function frequencies(str) {
  let result = {};
  for (let chr of str) {
    result[chr] = (result[chr] || 0) + 1;
  }
  return result;
}

// Check if 'str1' contains all characters of 'str2':
function scramble(str1, str2) {
  let frequencies1 = frequencies(str1);
  let frequencies2 = frequencies(str2);
  return Object.entries(frequencies2).every(([chr2, frequency2]) =>
    frequencies1[chr2] >= frequency2
  );
}

// Examples:
console.log(scramble("abc", "cba"));
console.log(scramble("aa", "aaaa"));
console.log(scramble("123", "234"));

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I can't program in JavaScript but i will tell you, your dealing with strings, so if you look at the type of input and know that the third letter into a word is usualy a vowl, but the last is usually a consent, then you can try to make your code error/False statement as fast as possible, if you choose to measure length then chose Consent positions to compare first.

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