Combine each element of an array with the ones after it

Question

I am trying to combine items in an array, with every item below it. It should make a set of the current character and each character below it, and iteratively walk down the array. For example, if I have an array like this:

var myArray = ['A','B','C','D']

I would like an output like this:

AB AC AD BC BD CD

The code I have is getting me close, but I am having a hard time figuring out the rest. Here is what I have so far:

var myArray = ['A', 'B', 'C', 'D']
var sql_parts = []
var string = "";
for (var i = 0; i < myArray.length; i++) {
  recurse_function(string, i)
}
console.log(sql_parts)

function recurse_function(string_val, count) {
  if ((myArray.length - count) == 0) {
    return string_val;
  } else {
    string_val += myArray[count]
    sql_parts.push(string_val)
    recurse_function(string_val, count + 1)
  }
}

But this produces:

["A", "AB", "ABC", "ABCD", "B", "BC", "BCD", "C", "CD", "D"]

Where do you specify that the resulting strings should only be 2 characters long? — Barmar, Aug 29 '21 at 16:52
@Barmar Thanks for the clarifying question. I will edit the original question to reflect that requirement — jason, Aug 29 '21 at 16:54

score 1 · Accepted Answer · answered Aug 29 '21 at 16:55

Here is one solution:

Define the recursive function to take the array and an empty list initially to store the combinations
The base condition is when the array is empty or has one element
Otherwise, Remove the first element "start"
Iterate over the array to store its combinations with its following elements
Recur again with array and combinations updated

function recurse_function(array, combinations = []) {
  if(array.length <= 1) return combinations;
  const start = array.shift();
  for(let i = 0; i < array.length; i++) combinations.push(`${start}${array[i]}`);
  return recurse_function(array, combinations);
}

console.log( recurse_function(['A','B','C','D']) );

score 1 · Answer 2 · answered Aug 29 '21 at 16:56

1

var myArray = ['A','B','C','D']
var sql_parts = []
for(var i =0; i< myArray.length; i++){
  var a = myArray[i]
  for(var j = i+1; j<myArray.length; j++){
    var b=myArray[j]
    var c= a+b
    sql_parts.push(c)
  }
}

answered Aug 29 '21 at 16:56

Subhashis Pandey

1,473
1
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16

score 1 · Answer 3 · answered Aug 29 '21 at 17:22

I think this is something that you're looking for, it's a function that is very cheap on resources. Its not recursive (why you now would need something like that for this simple scenario)

var myArray = ['A','B','C','D']
let [a, b, iter] = [0, 1, 2]
let result = []

for (;a < myArray.length; a++) {
  for (;b < myArray.length; b++) {
      result.push(myArray[a]+myArray[b])
  } 
  b = iter++
}

console.log(result)

customcommander · Answer 4 · 2021-08-30T06:17:37.090

1

I don't think you need to recurse in this particular case. You can simply combine the current element with the following ones with flatMap:

['A','B','C','D'].flatMap((x, i, xs) => xs.slice(i+1).map(y => x+y));
//=> ["AB", "AC", "AD", "BC", "BD", "CD"]

edited Aug 30 '21 at 06:17

answered Aug 29 '21 at 21:42

customcommander

17,580
5
58
84

score 0 · Answer 5 · answered Aug 29 '21 at 17:07

0

var myArray = ['A', 'B', 'C', 'D'];
var result = [];
myArray.forEach((item, index) => {
  myArray.forEach((item2, index2) => (index < index2 ? result.push(item + item2) : ''));
});
console.log(result);

answered Aug 29 '21 at 17:07

Pramod Kumar

96
1
2
7

Combine each element of an array with the ones after it

5 Answers5