Codewars - Unique In Order <6kyu>

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Author: Doe Hoon LEE

Codewars Problem [JavaScript]

Solved on Apr. 3rd, 2020

Implement the function unique_in_order which takes as argument a sequence

and returns a list of items without any elements with the same value next to each other

and preserving the original order of elements. For example:

uniqueInOrder(‘AAAABBBCCDAABBB’) == [‘A’, ‘B’, ‘C’, ‘D’, ‘A’, ‘B’]

uniqueInOrder(‘ABBCcAD’) == [‘A’, ‘B’, ‘C’, ‘c’, ‘A’, ‘D’]

uniqueInOrder([1,2,2,3,3]) == [1,2,3]

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Test_01

var uniqueInOrder = function(iterable) {
    var arr = [];
    for (var i=0;i<iterable.length;i++) {

    // check if there is back to back letter
    if (iterable[i] !== iterable[i+1]) {

        // if not, push the current value to arr
      arr.push(iterable[i]);
    }
  }
  return arr;
}

Test_02

var uniqueInOrder = function(iterable){
  return (typeof iterable == "object") ? iterable.filter((v,i,a) => a[i] !== a[i+1]) : iterable.split("").filter((v,i,a) => a[i] !== a[i+1]);
}

1. check given input type -> object/array or string

(typeof iterable == "object")

2. filter and return a new array with values that are not given back to back

filter((v,i,a) => a[i] !== a[a+1])

• Please NOTE that filter() method returns a new array with elements that PASS the test

• Here are some similar methods

methods	Description	Returns	Argument
every()	EVERY item has to pass the test	TRUE/FALSE
some()	SOME item has to pass the test	TRUE/FALSE	function
includes()	checks if the array has certain string	TRUE/FALSE	string
filter()	check for elements that pass the test	new array
map()	similar to filter(), but modifies items	new array

• Please NOTE that every(), some(), includes() methods return true or false !
• filter(), map() return new array