# Combine each element of an array with the ones after it

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”]

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']) );```