875. Koko Eating Bananas

Problem Statement

Koko loves to eat bananas. There are n piles of bananas, the ith pile has piles[i] bananas. The guards have gone and will come back in h hours.

Koko can decide her bananas-per-hour eating speed of k. Each hour, she chooses some pile of bananas and eats k bananas from that pile. If the pile has less than k bananas, she eats all of them instead and will not eat any more bananas during this hour.

Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.

Return the minimum integer k such that she can eat all the bananas within h hours.

Example 1:

Input: piles = [3,6,7,11], h = 8
Output: 4

Example 2:

Input: piles = [30,11,23,4,20], h = 5
Output: 30

Example 3:

Input: piles = [30,11,23,4,20], h = 6
Output: 23

Constraints:

1 <= piles.length <= 104
piles.length <= h <= 109
1 <= piles[i] <= 109

Intuition

Approach:
He can eat a maximum of max(array) bananas
range is 1->max(array)

Do BS on this answer space to find the min

Links

https://leetcode.com/problems/koko-eating-bananas/description/

Video Links

Approach 1:

C++

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {        
        int size=piles.size(), ans, max=-1;
        
        for(int i=0;i<size;i++){
            if(max<piles[i])
                max=piles[i];
        }

        if(size==h) return max;
        if(size==1){
            return ceil(piles[0]/double(h));
        }

        int low=1, high=max;

        while(low<=high){
            int mid=low+(high-low)/2;
            double count=0;

            for(int i=0; i<size; i++){
                // Replacement for ceil function
                count += (piles[i] + mid -1)/mid;
            }

            if(count>h){
                low=mid+1;
            }
            else { //count<=h
                high=mid-1;
                ans = mid;
            }
        }

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++