JavaScript How do you merge certain items in a JavaScript array?

Assuming the elements will always be single letters, you can merge the elements, then match on either bs or non-bs:

ab.join('').match(/(b+|.)/g)

const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];

let output = ab.join('').match(/(b+|.)/g);

console.log(output);

Answer:1

Using Array#reduce you could do something like this.

I'm assuming the first two characters in your solution were a typo.

const data = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];

const res = data
.reduce((a,c)=>{
  const lastIndex = a.length - 1;
  if(a[lastIndex] && a[lastIndex].includes('b') && c === 'b') a[lastIndex] += c;
  else a.push(c);
  return a;
}, []);

console.log(res);

Answer:2

I don't know how to give an explanation for this but using reduce you can do it like this:

const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];

function merge(arr) {
  return arr.reduce((acc,cur, index) => {
    if(arr[index - 1] === 'b' && cur !== 'a') {
      acc[acc.length - 1] = acc[acc.length - 1] + cur;
      return acc;
    }
    acc.push(cur);
    return acc;
  },[]);
}

console.log(merge(ab))

Answer:3

Here's what you're after:

var ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];

var index = 0;
var groupedArray = [];
var firstOccurence = true;
var groupedString = "";
var grouping = false;
for (var a = 0; a < ab.length; a++) {

  if (ab[a] == "a") {
    if (grouping) {
      grouping = false;
      firstOccurence = true;
      groupedArray.push(groupedString);
    }
    groupedArray.push(ab[a]);
  } else {
    if (firstOccurence) {
      groupedString = "";
      firstOccurence = false;
    }
    groupedString += ab[a];
    grouping = true;
  }
}
console.log(groupedArray);

Answer:4

If you just want to merge b then you could use reduce like this:

If the current item and the previous item are b, then append it to the last accumulator item. Else, push it to the accumulator

const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a']

const output = ab.reduce((acc, c, i, arr) => {
  arr[i-1] === "b" && c === "b" 
    ? acc[acc.length - 1] += c
    : acc.push(c)
    
  return acc;
},[])

console.log(output)

Answer:5

You can just map through the ab array and if the current element is a, push it to a new array but if the current element is b, check if the previous element is b or not, and if it is, merge the current element to the previous element else just push the b to the new array.

const ab = ['a', 'a', 'b', 'b', 'b', 'a', 'b', 'b', 'a'];

let arr = [];

ab.map((e,i) => {
  if(i > 0) { // check if element is the first one or not
    if(e == "b") { 
      if(ab[i - 1].indexOf('b') > -1) { // check if prev element is "b" or not
      	arr[arr.length-1] += e; // merge if prev element is "b"
      } else {
      	arr.push(e);
      }
    } else {
      arr.push(e);
    }
  } else {
  	arr.push(e);
  }
});

console.log(arr);

Answer:6

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