How to get all combinations of array elements after removing each element

Question

I want to get all combinations of all array elements after removing each element

For example:

input: [6,5,3,4]

output should be: [[5,3,4],[6,3,4],[6,5,4],[6,5,3]]

Without using any libraries

I worked in this code but did not output the right results

 var arrs = [[]]; function doIt(arr) { for (var i = 0; i < arr.length; i++) { arrs[i] = arr.filter(function (item) { return item != i }); } console.log(JSON.stringify(arrs)); } doIt([4, 3, 1]);

Answer 1

You could use a nested approach by iterating the given array and map arrays without the element of the outer index.

 var array = [6, 5, 3, 4], result = array.map((_, i) => array.filter((_, j) => i !== j)); console.log(result);

In your code, you need to check against the index of the filter callback instead of the value.

 var arrs = [[]]; function doIt(arr) { for (var i = 0; i < arr.length; i++) { arrs[i] = arr.filter(function (item, j) { // ^ use index instead of value return j != i; // ^ here }); } console.log(arrs); } doIt([4, 3, 1]);

Answer 2

Simple forEach + filter :

Just compare the indexes to filter the next element.

 var result = []; var array = [6,5,3,4]; array.forEach((_, i) => result.push(array.filter((_, j) => i !== j))); console.log(result);

Answer 3

No need to complex it, it's realy a simple algorithm.

 var input = [6,5,3,4]; var output = []; for (i = 0; i < input.length; ++i) { temp = []; for (j = 0 ; j < input.length; ++j) { if (i != j) temp.push(input[j]); } output.push(temp); } console.log(output);