I have two arrays:
arr1 = [0,1,1,0]
arr2 = [a,b,c,d]
I would like to find the values and corresponding indices of arr2[ i ] where i is such that arr1[ i ] != 0 without looping through each position in arr2. What efficient techniques could do something like the following:
arr2.forEach( ( 'element with index such that arr1[index] != 0') => {} );
** EDIT ** The initial posting of this question wasn't clear about needing to record the indices of the elements that met the condition.
Another solution to this is using reduce() over the arra1:
const arr1 = [0,1,1,0,1];
const arr2 = ["a","b","c","d","e"];
let res = arr1.reduce(
(acc, v, idx) => v ? acc.concat({val: arr2[idx], idx}) : acc,
[]
);
console.log(res);EDIT: getting the indices too is easy - updated the answer
You say you don't want to loop through arr2 but what you could do is loop through arr1 and collect the respective value of arr2 everytime you get a hit like so:
let result = []
arr1.forEach((value, idx) => {
if (value !== 0) result.push({index: idx, value: arr2[idx]})
}
Though this seems like kind of a hack around your requirements. Other than this I don't think you have another choice but to loop through either array (optimization: only loop through the smallest one).
If you want the complete result you have to check every element of arr2 for the condition because if you don't there could still be some elements missing from your result.
Simply use filter:
arr1 = [0,1,1,0]
arr2 = ['a','b','c','d']
console.log( arr2.filter((item, index) => arr1[index] !== 0) )
You can reduce array 1 to the indexes you want and when you need the values you can maps them to their value in array 2
let arr1 = [0, 1, 1, 0];
let arr2 = ['a', 'b', 'c', 'd'];
let indexes = arr1.reduce((results, val, index) => {
if (val) {
results.push(index);
}
return results;
}, []);
console.log(indexes);
let vals = indexes.map(idx => arr2[idx]);
console.log(vals);Try $.each():
$(function() {
var arr1 = [0, 1, 1, 0];
var arr2 = ['a', 'b', 'c', 'd'];
$.each(arr1, function(k, v) {
if (!v) {
console.log(arr2[k]);
}
});
});<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>Obviously, arrays must be of equal size or arr1 can be smaller than arr2.
Hope that helps.
for-loop overarr1...for(...) if (arr1[i] === 0) doSomething(arr2[i])O(n log n)followed by a binary search ofO(log n), whereas a loop isO(n). I'm not saying your suggestion is wrong, just that it would require very specific circumstances for that to be the most efficient.