Okay, so the purpose of this blog is explain my thought process in building my solution for this kata. https://www.codewars.com/kata/54bf1c2cd5b56cc47f0007a1/javascript
Basically the idea is to count the number of distinct duplicates. So if you’ve got a string aabBb that would be 2 duplicates. Note that the solution is case-insensitive.
Okay, so here’s the function below. I start out with three inital arrays. resArr has a value because of the way I am comparing and counting later along. First, I am iterating over the lower case text and pushing the text element (here t) to array. There are duplicates so that each element of array is now an array containing one letter or multiple of the same letters. The next step is to get only arrays with a length of more than one because these are where the duplicates occur.
Next I sort resArr and compare to see if the first letter of each array within resArr is equal or not. If they are not equal I add it to the result array. Because of the comparing from the last item to the non-existent last item plus 1, I need to start with resArr out with some value, in this case 1. Finally I return the array based on length of result.
function duplicateCount(text){
// here are three arrays to start
let array = []
//let resArr = [1]
let result = []
// next I convert the text to lower case since case does not matter for counting
text = text.toLowerCase()
for (let i=0; i<text.length; i++) {
array.push(text.split('').filter(t => t===text[i]))
}
array = array.filter(arr=>arr.length>1)
array.forEach( arr => {
if (!resArr.includes(arr)) {
resArr.push(arr)
}
})
resArr = resArr.sort()
for (let i=0; i<resArr.length-1; i++) {
if (resArr[i][0]!==resArr[i+1][0]) {
result.push(resArr[i])
}
}
if (result.length===0){
return 0
} else {
return result.length
}
}