JavaScript Challenge - Egg Dropper

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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;

For 2 eggs.

function eggDrop(n){
  return Math.round(Math.sqrt(n / 2) * 2);
1 Like

Works for any positive integer, negative integers, 1, and 0, and even responds with -1 if ‘n’ is not provided:

const eggDrop = (n = -1, drops = 0, floor = 0) => {
  if (floor >= n) return drops
  floor += drops
  return eggDrop(n, drops, floor)
function eggDrop(n){
  return Math.floor(Math.sqrt(n/2) * 2)
console.log(eggDrop(n)) // logs 14

The floored square root of 50 multiplied by 2 gives us the the number of floors to omit for the first drop of the first egg.

1. egg

  1. drop: 14th floor → doesn’t break +13 drops left between 14th floor and 27th floor if egg breaks at 27th floor
  2. drop: 27th floor → doesn’t break +12
  3. drop 39th floor → doesn’t break + 11
  4. drop: 50th floor → doesn’t break +10
  5. drop: 60th floor → doesn’t break +9
  6. drop 69th floor → doesn’t break +8
  7. drop 77th floor → doesn’t break +7
  8. drop 84th floor → doesn’t break +6
  9. drop 90th floor → doesn’t break +5
  10. drop 95th floor → doesn’t break +4
  11. drop 99th floor → breaks

2. egg
12. drop 96th floor → doesn’t break
13. drop 97th floor → doesn’t break
14. drop 98th floor → doesn’t break ( f===99 )
14. drop 98th floor → breaks ( f===98 )


Solution with a plus.

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;

My code is :slight_smile:
function eggDrop(n) {
// Base case: if there is only 1 floor, no need to drop the egg
if (n === 1) {
return 1;

// Base case: if there are only 2 floors, only 1 drop is needed
if (n === 2) {
return 2;

// Create a 2D array to store the results of subproblems
const dp = Array.from({ length: n + 1 }, () => new Array(3).fill(0));

// Fill in the base cases
for (let i = 0; i <= n; i++) {
dp[i][1] = i;

// Fill in the remaining array using dynamic programming
for (let i = 1; i <= n; i++) {
for (let j = 2; j <= 2; j++) {
dp[i][j] = Infinity;

  // Drop the egg from each floor and choose the minimum number of drops
  for (let k = 1; k <= i; k++) {
    const maxDrops = 1 + Math.max(dp[k - 1][j - 1], dp[i - k][j]);
    dp[i][j] = Math.min(dp[i][j], maxDrops);


// The minimum number of drops required is stored in dp[n][2]
return dp[n][2];


// Leave this here so we can test your code
module.exports = eggDrop;