Javascript - Generating all combinations of elements in a single array (in pairs)

A simple way would be to do a double for loop over the array where you skip the first i elements in the second loop.

let array = ["apple", "banana", "lemon", "mango"];
let results = [];

// Since you only want pairs, there's no reason
// to iterate over the last element directly
for (let i = 0; i < array.length - 1; i++) {
  // This is where you'll capture that last value
  for (let j = i + 1; j < array.length; j++) {
    results.push(`${array[i]} ${array[j]}`);
  }
}

console.log(results);

Rewritten with ES5:

var array = ["apple", "banana", "lemon", "mango"];
var results = [];

// Since you only want pairs, there's no reason
// to iterate over the last element directly
for (var i = 0; i < array.length - 1; i++) {
  // This is where you'll capture that last value
  for (var j = i + 1; j < array.length; j++) {
    results.push(array[i] + ' ' + array[j]);
  }
}

console.log(results);

Although solutions have been found, I post here an algorithm for general case to find all combinations size n of m (m>n) elements. In your case, we have n=2 and m=4.

const result = [];
result.length = 2; //n=2

function combine(input, len, start) {
  if(len === 0) {
    console.log( result.join(" ") ); //process here the result
    return;
  }
  for (let i = start; i <= input.length - len; i++) {
    result[result.length - len] = input[i];
    combine(input, len-1, i+1 );
  }
}

const array = ["apple", "banana", "lemon", "mango"];    
combine( array, result.length, 0);

Here are some functional programming solutions:

Using EcmaScript2019's flatMap:

var array = ["apple", "banana", "lemon", "mango"];

var result = array.flatMap(
    (v, i) => array.slice(i+1).map( w => v + ' ' + w )
);

console.log(result);

Before the introduction of flatMap (my answer in 2017), you would go for reduce or [].concat(...) in order to flatten the array:

var array = ["apple", "banana", "lemon", "mango"];

var result = array.reduce( (acc, v, i) =>
    acc.concat(array.slice(i+1).map( w => v + ' ' + w )),
[]);

console.log(result);

Or:

var array = ["apple", "banana", "lemon", "mango"];

var result = [].concat(...array.map( 
    (v, i) => array.slice(i+1).map( w => v + ' ' + w ))
);

console.log(result);

In my case, I wanted to get the combinations as follows, based on the size range of the array:

function getCombinations(valuesArray: String[])
{

var combi = [];
var temp = [];
var slent = Math.pow(2, valuesArray.length);

for (var i = 0; i < slent; i++)
{
    temp = [];
    for (var j = 0; j < valuesArray.length; j++)
    {
        if ((i & Math.pow(2, j)))
        {
            temp.push(valuesArray[j]);
        }
    }
    if (temp.length > 0)
    {
        combi.push(temp);
    }
}

combi.sort((a, b) => a.length - b.length);
console.log(combi.join("\n"));
return combi;
}

Example:

// variable "results" stores an array with arrays string type
let results = getCombinations(['apple', 'banana', 'lemon', ',mango']);

Output in console:

enter image description here

The function is based on the logic of the following documentation, more information in the following reference: https://www.w3resource.com/javascript-exercises/javascript-function-exercise-3.php

if ((i & Math.pow(2, j)))

Each bit of the first value is compared with the second, it is taken as valid if it matches, otherwise it returns zero and the condition is not met.

enter image description here