LintCode 183. Wood Cut

183.Wood Cut

Given n pieces of wood with lengthL[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

ForL=[232, 124, 456],k=7, return114.

Challenge

O(n log Len), where Len is the longest length of the wood.

Notice

You couldn't cut wood into float length.

If you couldn't get >=_k _pieces, return0.

Thoughts:

1. Binary Search on the length by testing the resulted cutting pieces

Code: T: Nlog(max(L[i])), S: O(1)

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

        int l = 1, r = 0;
        for (int len : L){
            r = Math.max(r, len);

        }
        if(k == 0) return r;

        while (l <= r){
            int mid = l + ((r - l) >> 1);
            if (count(L, mid) >= k) l = mid + 1;
            else r = mid - 1;
        }


        // if (count(L,l) >= k) return l;
        if (count(L,r) >= k) return r;

        return 0;

    }

    private int count(int[] L, int query){
        int sum = 0;
        if (query == 0) return 0; // in case of dividing with 0
        for (int len : L){
            sum+= len/query;
        }
        return sum;
    }
}

出来之后看的标准的做法是用二分法加greedy去找, N * log max 

当场想的办法是用dp, K^2 * N, 具体的思路就是： 

dp[i][j] = 把前0 - i 根木头切 j段的最大长度 j < k 

dp[i][j] = max(min(dp[i - 1][a], wood[i] / (j - a)) for a < j)