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My favorite solution although it might take awhile to compile with different constraints.
I wonder if there's a way to "clean up" my solution, which uses the lengths of both arrays as constraints.
function getTotalX(a, b) {
let maxIntA = Math.max(...a);
let minIntB = Math.min(...b);
let intsBetweenArrays = [];
let newArrA = [];
let finalArr = [];
// let isDivisible = Boolean;
if (maxIntA > minIntB) {
return 0;
} else {
for (let i = maxIntA; i <= minIntB; i++) {
intsBetweenArrays.push(i);
}
for (let i = 0; i < intsBetweenArrays.length; i++) {
if (a.every(num => intsBetweenArrays[i] % num === 0)) {
newArrA.push(intsBetweenArrays[i]);
}
}
for (let i = 0; i < newArrA.length; i++) {
if (b.every(num => num % newArrA[i] === 0)) {
finalArr.push(newArrA[i]);
}
}
}
return finalArr.length;
Between Two Sets
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My javascript solution:
Thats one of the most elegant solutions I have ever seen but can you explain to a novice what (a.every(int => (x % int == 0))) means?
It means this: For every value (
int) in the array (a), isxevenly divisible by that value?For Javascript, [].every(function()) checks if all elements in an array pass a test which is the parameterized function as condition.
This is brilliant! Thanks!
This might be a silly question but why
x <= 100?this is based on the constraints of the problem. we are told that each value in both arrays are between 1 < x <= 100
ah yes, thanks. I am not use to seeing the constraints accounted for in the code like that.
You're right, why x <= 100? Excution time may be reduced using max = Math.max(...b) instead of 100 ...
for x > max => b.every(int => (int % x == 0)) return false
My favorite solution although it might take awhile to compile with different constraints.
I wonder if there's a way to "clean up" my solution, which uses the lengths of both arrays as constraints.
function getTotalX(a, b) {
}
The most elegant solution i've seen. Bravo