lintcode:Copy Books

Given an array A of integer with size of n( means n books and number of pages of each book) and k people to copy the book. You must distribute the continuous id books to one people to copy. (You can give book A[1],A[2] to one people, but you cannot give book A[1], A[3] to one people, because book A[1] and A[3] is not continuous.) Each person have can copy one page per minute. Return the number of smallest minutes need to copy all the books.

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Example

Given array A = [3,2,4], k = 2.

Return 5( First person spends 5 minutes to copy book 1 and book 2 and second person spends 4 minutes to copy book 3. )

Challenge

Could you do this in O(n*k) time ?

class Solution {public:    /**     * @param pages: a vector of integers     * @param k: an integer     * @return: an integer     */int copyBooks(vector<int> &pages, int k) {    // write your code here    if (pages.size() == 0 || k == 0)        return 0;            if (k > pages.size())        k = pages.size();        vector<vector<int>> dp(k, vector<int>(pages.size()));    vector<int> auxPages(pages.size());    auxPages[0] = pages[0]; //页面的累加和    dp[0][0] = pages[0];    for (int i=1; i<pages.size(); i++)    {        auxPages[i] = auxPages[i-1]+pages[i];        dp[0][i] = auxPages[i];    }        //dp[i][j] i个人抄写书，抄到以j下标结尾的书时的最少耗时        for (int i=1; i<k; i++)    {        for (int j=i; j<pages.size(); j++)        {            int targetValue = INT_MAX;            for (int m=i; m<=j; m++)            {                //i-1人抄到下标为m-1的书，剩下的一人抄从m到j的书                targetValue = min(max(dp[i-1][m-1], auxPages[j]-auxPages[m-1]), targetValue);            }            dp[i][j] = targetValue;        }    }        return dp[k-1][pages.size()-1];}};