I want to collect all subarrays for further computation efficiently in javascript. I'm not sure this is possible, but it seems for a subarray sum kadane's formula is o(n) which is more efficient than other methods. But I'm not sure I how I can store the array at each step.
Similar to this quora question, for me the pseudo code was not enough. Thanks for the further breakdown.
another meta link
an example in action of this for [3, 3, 9, 9, 5]
[3], [9], [5], [9, 5], [9, 3], [9, 9], [3, 3],
[3, 9, 9], [3, 3, 9], [9, 9, 5], [3, 3, 9, 9],
[3, 9, 9, 5], [3, 3, 9, 9, 5]
This is fairly simple to do: https://jsfiddle.net/j1LuvxLq/
All you do is iterate possible lenghts and starting points and just print out the subsets. Complexity is O(n²) where n is the length of the original array. No way to improve it thought because that's the order of how many subsets there are.
var set = [3, 3, 9, 9, 5].join('')
var set_length = set.length
var subsets = []
for (var length_of_subset = 1; length_of_subset <= set_length; length_of_subset++) {
for (var start_of_subset = 0; start_of_subset <= set_length - length_of_subset; start_of_subset++) {
var current_subset = set.substring(start_of_subset, start_of_subset + length_of_subset)
if(subsets.indexOf(current_subset) == -1) {
subsets.push(current_subset.split(''))
}
}
}
// print the subsets out
for (s in subsets) {
$('body').append(subsets[s].join(', ') + '<br>')
}
Alternative, possibly nicer solution would be to use dynamic programming. Start with 3 and either remove last element or add next element. Check it out here: https://jsfiddle.net/p82fcs4m/
var set = [3, 3, 9, 9, 5].join('')
var subsets = []
take(set[0], set.substring(1))
function take(chosen, left) {
if(subsets.indexOf(chosen) != -1) {
return
}
subsets.push(chosen)
if (chosen.length > 1) {
take(chosen.substring(1), left)
}
if (left.length > 0) {
take(chosen.concat(left[0]), left.substring(1))
}
}
$('body').append(subsets.join('<br>'))
I had done a work previously to calculate all combinations of amino acids total molecular weight. If you neglect the empty one you should have 2^n - 1 sub arrays. So there is no O(n) here. I've got two methods as binary and recursive.
function getSubArrays(arr){
var len = arr.length,
subs = Array(Math.pow(2,len)).fill();
return subs.map((_,i) => { var j = -1,
k = i,
res = [];
while (++j < len ) {
k & 1 && res.push(arr[j]);
k = k >> 1;
}
return res;
}).slice(1);
}
console.log(JSON.stringify(getSubArrays([1,2,3,4,5])));
function getSubArrays(arr){
if (arr.length === 1) return [arr];
else {
subarr = getSubArrays(arr.slice(1));
return subarr.concat(subarr.map(e => e.concat(arr[0])), [[arr[0]]]);
}
}
console.log(JSON.stringify(getSubArrays([1,2,3,4,5])));
I couldn't manage to get subarrays of an array with more than 23 items though.
Here are the performances. To be on the safe side i try with 22 items, first with recursive and then with binary route.
function getSubArrays(arr){
if (arr.length === 1) return [arr];
else {
subarr = getSubArrays(arr.slice(1));
return subarr.concat(subarr.map(e => e.concat(arr[0])), [[arr[0]]]);
}
}
var aminos = Array(22).fill().map((_,i) => i+1),
subarrs = [],
ps = 0,
pe = 0;
ps = performance.now();
subarrs = getSubArrays(aminos);
pe = performance.now();
console.log("recursive route took: " + (pe-ps) + "msec to produce " + subarrs.length + " sub arrays");
function getSubArrays(arr){
var len = arr.length,
subs = Array(Math.pow(2,len)).fill();
return subs.map((_,i) => { var j = -1,
k = i,
res = [];
while (++j < len ) {
k & 1 && res.push(arr[j]);
k = k >> 1;
}
return res;
}).slice(1);
}
var aminos = Array(22).fill().map((_,i) => i+1),
subarrs = [],
ps = 0,
pe = 0;
ps = performance.now();
subarrs = getSubArrays(aminos);
pe = performance.now();
console.log("binary route took: " + (pe-ps) + "msec to produce " + subarrs.length + " sub arrays");
