How To Write A Function That Returns True If A Portion Of Str1 Can Be Rearranged To Str2?

I am having trouble with below question. I basically have to write a code/function that returns true if a portion of str1 can be rearraged to str2. Write function scramble(str1,str

Solution 1:

You could use a hash table with the count of the letters and check with count and decrement the count.

This proposal does not mutate the arrays.

functionscramble(str1, str2) {
    var count = Object.create(null);

    Array.prototype.forEach.call(str1, function(a) {
        count[a] = (count[a] || 0) + 1;
    });

    returnArray.prototype.every.call(str2, function(a) {
        return count[a]--;
    });
}

console.log(scramble('rkqodlw', 'world'));              // trueconsole.log(scramble('cedewaraaossoqqyt', 'codewars')); // trueconsole.log(scramble('katas', 'steak'));                // falseconsole.log(scramble('', 'o'));                // false

Solution 2:

Here is the function with some tests:

functionscramble(str1, str2) {
  var l = str2.length;
  for (var i = 0; i < l; i++) {
    if (str1.indexOf(str2[i]) > -1) {
      str1 = str1.replace(str2[i], '');
    } else {
      returnfalse;
    }
  }
  returntrue;
}

functiontest(str1, str2) {
  console.log('testing "'+str1+'" w/ "'+str2+'": '+(scramble(str1, str2) ? 'true' : 'false'));
}

test('rkqodlw', 'world');
test('cedewaraaossoqqyt', 'codewars');
test('katas', 'steak');

The tests are returning:

testing "rkqodlw" w/ "world": true
testing "cedewaraaossoqqyt" w/ "codewars": true
testing "katas" w/ "steak": false

The function checks if every char of str2 is in str1 and removes it from str1 so that a char from str1 doesn't count twice.

Solution 3:

Split the strings into arrays, and check if every character in the second array is inside the first array.

You probably want to splice of characters as you go, to account for mulitiples of the same character

functionscramble(str1, str2) {
    var [arr1, arr2] = [str1.split(''), str2.split('')];
    return arr2.every(x=>arr1.indexOf(x)===-1?false:arr1.splice(arr1.indexOf(x),1));
}

console.log( scramble('rkqwodlw', 'world') );     // trueconsole.log( scramble('mgoaon', 'moon') );        // trueconsole.log( scramble('oijhnnassduda', 'moon') ); // false, only one "o"console.log( scramble('test', 'unicorn') );       // false

Solution 4:

function scramble(str1, str2) {
var [a,b,c] = [str1.split(''),str2.split(''),[]];
for (let i = 0; i < b.length; i++) {
    if (a.indexOf(b[i]) !== -1) {c.push(b[i]), a.splice(a.indexOf(b[i]), 1);}
}
return b.join('') === c.join('');}