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I’ve tried to approach this challange with binary search alghoritm, but then I realised when the egg breaks on 50th floor we have 49 more possible floors, so I needed some other way around.
So after some reading on the net, because I wasn’t quite sure how to do it, I’ve managed to write this simple solution for this challange: explanation
function eggDrop(n, d = 1){
// Write your code here
if (n <= 0) {
return d-1;
} else if (n => 1) {
return eggDrop(n-d, ++d);
}
}
console.log(eggDrop(2));
// Leave this here so we can test your code
module.exports = eggDrop;
function eggDrop(n){
// Write your code here
return bld(n);
}
// Binary Search
function bs(n) {
// First egg drop is from the middle
// And then depending on the result
// It's brute force for the rest
return Math.ceil(n / 2);
}
// Divide a Little, Conquer a Bit
function dlcb(n) {
// The egg is dropped in decimal floors of n
// And if the egg breaks in any of those 10
// The worst case scenario would be between
// that last egg drop and the previous decimals
// until the previous egg drop
return n > 9 ? 10 + (Math.floor(n * .09)) : n;
}
// Balancing Linear Drops
function bld(n) {
// Similar to the previous one, divide and conquer
// Except it starts from it's square root and
// goes up by that square root - 1 on each drop
// If the egg breaks then you basically sum twice
// that square root that was consistent all over
// because of the iteration of square root - 1 on
// each drop. That's how you get 14 drops
// out of 100 floors as worst case scenario
return Math.round(Math.sqrt(n*2));
}
console.log(eggDrop(100));
// Leave this here so we can test your code
module.exports = eggDrop;
) below.
) to up-vote the contribution!