LT 183 wood cut

/*
Given n pieces of wood with length L[i] (integer array). Cut them into small pieces to guarantee you could have equal or more than k pieces with the same length. What is the longest length you can get from the n pieces of wood? Given L & k, return the maximum length of the small pieces.

Example

Example 1

Input:

L = [232, 124, 456]

k = 7

Output: 114

Explanation: We can cut it into 7 pieces if any piece is 114cm long, however we can't cut it into 7 pieces if any piece is 115cm long.

Example 2

Input:

L = [1, 2, 3]

k = 7

Output: 0

Explanation: It is obvious we can't make it.*/

public class Solution {
    /** 
     *@param L: Given n pieces of wood with length L[i]
     *@param k: An integer
     *return: The maximum length of the small pieces.
     */
    public int woodCut(int[] L, int k) {
        // write your code here
        
        if (L == null || L.length == 0 || k <= 0) {
            return 0;
        }
        
        int start = 1;
        int end = 1;
        for (int wood : L) {
            end = Math.max(wood, end);
        }
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (getPieces(L, mid) >= k) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        
        if (getPieces(L, end) >= k) {
            return end;
        }
        if (getPieces(L, start) >= k) {
            return start;
        }
        return 0;
    }
    
    public int getPieces(int[] L, int length) {
        int count = 0;
        for (int wood : L) {
            count += wood / length;
        }
        return count;
    }
}

Analysis:
Binary search for the result. The min length is 1, and the max length is the Max(L[i]). So we can try every possible length and calculate how many pieces can we cut? If the number is greater than k, means the the number we tried might be too small. Otherwise, the number is too large. So we can do it via binary search. The total time complexity would be O(nlogAns)

http://buttercola.blogspot.com/2019/03/lintcode-183-wood-cut.html