Given n eggs and k floors, find the minimum number of trials needed in worst case to find the floor below which all floors are safe. A floor is safe if dropping an egg from it does not break the egg. Please see n eggs and k floors. for complete statements

**Example**

Input : n = 2, k = 10 Output : 4 We first try from 4-th floor. Two cases arise, (1) If egg breaks, we have one egg left so we need three more trials. (2) If egg does not break, we try next from 7-th floor. Again two cases arise. We can notice that if we choose 4th floor as first floor, 7-th as next floor and 9 as next of next floor, we never exceed more than 4 trials. Input : n = 2. k = 100 Output : 14

We have discussed the problem for 2 eggs and k floors. We have also discussed a dynamic programming solution to find the solution. The dynamic programming solution is based on below recursive nature of the problem. Let us look at the discussed recursive formula from a different perspective.**How many floors we can cover with x trials?**

When we drop an egg, two cases arise.

- If egg breaks, then we are left with x-1 trials and n-1 eggs.
- If egg does not break, then we are left with x-1 trials and n eggs

Let maxFloors(x, n) be the maximum number of floors that we can cover with x trials and n eggs. From above two cases, we can write. maxFloors(x, n) = maxFloors(x-1, n-1) + maxFloors(x-1, n) + 1 For all x >= 1 and n >= 1 Base cases : We can't cover any floor with 0 trials or 0 eggs maxFloors(0, n) = 0 maxFloors(x, 0) = 0 Since we need to cover k floors, maxFloors(x, n) >= k ----------(1) The above recurrence simplifies to following, Refer this for proof. maxFloors(x, n) = ∑^{x}C_{i}1 <= i <= n ----------(2) Here C represents Binomial Coefficient. From above two equations, we can say. ∑^{x}C_{j}>= k 1 <= i <= n Basically we need to find minimum value of x that satisfies above inequality. We can find such x using Binary Search.

## C++

`// C++ program to find minimum ` `// number of trials in worst case. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `// Find sum of binomial coefficients xCi ` `// (where i varies from 1 to n). ` `int` `binomialCoeff(` `int` `x, ` `int` `n) ` `{ ` ` ` `int` `sum = 0, term = 1; ` ` ` `for` `(` `int` `i = 1; i <= n; ++i) { ` ` ` `term *= x - i + 1; ` ` ` `term /= i; ` ` ` `sum += term; ` ` ` `} ` ` ` `return` `sum; ` `} ` `// Do binary search to find minimum ` `// number of trials in worst case. ` `int` `minTrials(` `int` `n, ` `int` `k) ` `{ ` ` ` `// Initialize low and high as 1st ` ` ` `// and last floors ` ` ` `int` `low = 1, high = k; ` ` ` `// Do binary search, for every mid, ` ` ` `// find sum of binomial coefficients and ` ` ` `// check if the sum is greater than k or not. ` ` ` `while` `(low < high) { ` ` ` `int` `mid = (low + high) / 2; ` ` ` `if` `(binomialCoeff(mid, n) < k) ` ` ` `low = mid + 1; ` ` ` `else` ` ` `high = mid; ` ` ` `} ` ` ` `return` `low; ` `} ` `/* Driver code*/` `int` `main() ` `{ ` ` ` `cout << minTrials(2, 10); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find minimum ` `// number of trials in worst case.` `class` `Geeks {` `// Find sum of binomial coefficients xCi` `// (where i varies from 1 to n). If the sum` `// becomes more than K` `static` `int` `binomialCoeff(` `int` `x, ` `int` `n, ` `int` `k)` `{` ` ` `int` `sum = ` `0` `, term = ` `1` `;` ` ` `for` `(` `int` `i = ` `1` `; i <= n && sum < k; ++i) {` ` ` `term *= x - i + ` `1` `;` ` ` `term /= i;` ` ` `sum += term;` ` ` `}` ` ` `return` `sum;` `}` `// Do binary search to find minimum ` `// number of trials in worst case.` `static` `int` `minTrials(` `int` `n, ` `int` `k)` `{` ` ` `// Initialize low and high as 1st ` ` ` `//and last floors` ` ` `int` `low = ` `1` `, high = k;` ` ` `// Do binary search, for every mid, ` ` ` `// find sum of binomial coefficients and ` ` ` `// check if the sum is greater than k or not.` ` ` `while` `(low < high) {` ` ` `int` `mid = (low + high) / ` `2` `;` ` ` `if` `(binomialCoeff(mid, n, k) < k)` ` ` `low = mid + ` `1` `;` ` ` `else` ` ` `high = mid;` ` ` `}` ` ` `return` `low;` `}` `/* Driver code*/` `public` `static` `void` `main(String args[])` `{` ` ` `System.out.println(minTrials(` `2` `, ` `10` `));` `}` `}` `// This code is contributed by ankita_saini` |

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## Python3

`# Python3 program to find minimum ` `# number of trials in worst case.` `# Find sum of binomial coefficients ` `# xCi (where i varies from 1 to n). ` `# If the sum becomes more than K` `def` `binomialCoeff(x, n, k):` ` ` `sum` `=` `0` `;` ` ` `term ` `=` `1` `;` ` ` `i ` `=` `1` `;` ` ` `while` `(i <` `=` `n ` `and` `sum` `< k): ` ` ` `term ` `*` `=` `x ` `-` `i ` `+` `1` `;` ` ` `term ` `/` `=` `i;` ` ` `sum` `+` `=` `term;` ` ` `i ` `+` `=` `1` `;` ` ` `return` `sum` `;` `# Do binary search to find minimum ` `# number of trials in worst case.` `def` `minTrials(n, k):` ` ` `# Initialize low and high as ` ` ` `# 1st and last floors` ` ` `low ` `=` `1` `;` ` ` `high ` `=` `k;` ` ` `# Do binary search, for every ` ` ` `# mid, find sum of binomial ` ` ` `# coefficients and check if ` ` ` `# the sum is greater than k or not.` ` ` `while` `(low < high):` ` ` `mid ` `=` `int` `((low ` `+` `high) ` `/` `2` `);` ` ` `if` `(binomialCoeff(mid, n, k) < k):` ` ` `low ` `=` `mid ` `+` `1` `;` ` ` `else` `:` ` ` `high ` `=` `mid;` ` ` `return` `int` `(low);` `# Driver Code` `print` `(minTrials(` `2` `, ` `10` `));` `# This code is contributed ` `# by mits` |

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## C#

`// C# program to find minimum ` `// number of trials in worst case.` `using` `System;` `class` `GFG ` `{` `// Find sum of binomial coefficients` `// xCi (where i varies from 1 to n). ` `// If the sum becomes more than K` `static` `int` `binomialCoeff(` `int` `x, ` ` ` `int` `n, ` `int` `k)` `{` ` ` `int` `sum = 0, term = 1;` ` ` `for` `(` `int` `i = 1; ` ` ` `i <= n && sum < k; ++i) ` ` ` `{` ` ` `term *= x - i + 1;` ` ` `term /= i;` ` ` `sum += term;` ` ` `}` ` ` `return` `sum;` `}` `// Do binary search to find minimum ` `// number of trials in worst case.` `static` `int` `minTrials(` `int` `n, ` `int` `k)` `{` ` ` `// Initialize low and high` ` ` `// as 1st and last floors` ` ` `int` `low = 1, high = k;` ` ` `// Do binary search, for every ` ` ` `// mid, find sum of binomial ` ` ` `// coefficients and check if the` ` ` `// sum is greater than k or not.` ` ` `while` `(low < high) ` ` ` `{` ` ` `int` `mid = (low + high) / 2;` ` ` `if` `(binomialCoeff(mid, n, k) < k)` ` ` `low = mid + 1;` ` ` `else` ` ` `high = mid;` ` ` `}` ` ` `return` `low;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `Console.WriteLine(minTrials(2, 10));` `}` `}` `// This code is contributed` `// by Akanksha Rai(Abby_akku)` |

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## PHP

`<?php` `// PHP program to find minimum ` `// number of trials in worst case.` `// Find sum of binomial coefficients ` `// xCi (where i varies from 1 to n). ` `// If the sum becomes more than K` `function` `binomialCoeff(` `$x` `, ` `$n` `, ` `$k` `)` `{` ` ` `$sum` `= 0; ` `$term` `= 1;` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `&& ` ` ` `$sum` `< ` `$k` `; ++` `$i` `) ` ` ` `{` ` ` `$term` `*= ` `$x` `- ` `$i` `+ 1;` ` ` `$term` `/= ` `$i` `;` ` ` `$sum` `+= ` `$term` `;` ` ` `}` ` ` `return` `$sum` `;` `}` `// Do binary search to find minimum ` `// number of trials in worst case.` `function` `minTrials(` `$n` `, ` `$k` `)` `{` ` ` `// Initialize low and high as ` ` ` `// 1st and last floors` ` ` `$low` `= 1; ` `$high` `= ` `$k` `;` ` ` `// Do binary search, for every ` ` ` `// mid, find sum of binomial ` ` ` `// coefficients and check if ` ` ` `// the sum is greater than k or not.` ` ` `while` `(` `$low` `< ` `$high` `) ` ` ` `{` ` ` `$mid` `= (` `$low` `+ ` `$high` `) / 2;` ` ` `if` `(binomialCoeff(` `$mid` `, ` `$n` `, ` `$k` `) < ` `$k` `)` ` ` `$low` `= ` `$mid` `+ 1;` ` ` `else` ` ` `$high` `= ` `$mid` `;` ` ` `}` ` ` `return` `(int)` `$low` `;` `}` `// Driver Code` `echo` `minTrials(2, 10);` `// This code is contributed ` `// by Akanksha Rai(Abby_akku)` `?>` |

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**Output**

4

**Time Complexity : **O(n Log k)

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