Eat Bananas in K Hours - Facebook Top Interview Questions


Problem Statement :


You are given a list of integers piles and an integer k.

piles[i] represents the number of bananas on pile i. On each hour, you can choose any pile and eat r number of bananas in that pile. 

If you pick a pile with fewer than r bananas, it still takes an hour to eat the pile.

Return the minimum r required such that you can eat all the bananas in less than or equal to k hours.

Constraints

n ≤ 100,000 where n is the length of piles

n ≤ k

Example 1

Input

piles = [6, 4, 3]

k = 5

Output

3

Explanation

At r = 3 bananas per hour, we can eat the first pile in 2 hours, eat the second in 2 hours, and eat the last 
pile in 1 hour.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& piles, int k) {
    int s = 1, e = *max_element(piles.begin(), piles.end());
    auto check = [&piles, &k](int r) {
        int count = 0;
        for (int p : piles) {
            count += (p + r - 1) / r;
        }
        return count <= k;
    };

    while (s <= e) {
        int m = s + (e - s) / 2;
        if (check(m)) {
            e = m - 1;
        } else {
            s = m + 1;
        }
    }
    return s;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] A, int k) {
        int lo = 1;
        int hi = Integer.MAX_VALUE;
        int r = -1;

        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2; // speed

            if (timeToEat(A, mid) <= k) { // valid, try to eat slower, go left.
                hi = mid - 1;
                r = mid;
            } else { // couldn't eat, need to eat faster, go right.
                lo = mid + 1;
            }
        }

        return r;
    }

    public int timeToEat(int[] A, int speed) {
        int r = 0;

        for (int x : A) {
            r += x / speed;
            if (x % speed != 0)
                r++;
        }

        return r;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, piles, k):
        lo = 1
        hi = max(piles)

        while lo <= hi:
            r = (lo + hi) // 2

            if self.bananas_hour(piles, r) <= k:
                hi = r - 1
            else:
                lo = r + 1
        return lo

    def bananas_hour(self, piles, max_bananas_hour):
        count_hours = 0

        for pile in piles:
            count_hours += pile // max_bananas_hour

            if pile % max_bananas_hour != 0:
                count_hours += 1

        return count_hours


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