Copy Books

ID: 437; medium; 书籍复印

LintCode 炼码

Solution 1 (Java)

public class Solution {
    /**
     * @param pages: an array of integers
     * @param k: An integer
     * @return: an integer
     */
    public int copyBooks(int[] pages, int k) {
        if (pages == null || pages.length == 0 || k <= 0)
            return 0;
        
        int left = 0, right = Integer.MAX_VALUE;

        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (isTimeFeasible(pages, k, mid)) {
                right = mid;
            } else {
                left = mid;
            }
        }

        return isTimeFeasible(pages, k, left) ? left : right;
    }

    private boolean isTimeFeasible(int[] pages, int k, int time) {
        int numOfCopiers = 0, remain = 0;
        for (int page : pages) {
            if (page > time) return false;
            if (page > remain) {
                numOfCopiers++;
                remain = time;
            }
            remain -= page;
        }
        return numOfCopiers <= k;
    }
}

Notes

• The check of a feasible time here is crucial. Given a time, how do we know the minimum number of copiers needed to complete the job?

  • If there is one book that takes longer than time to be copied, there is no way this time is feasible.

  • Else in the general case, count the minimum number of copier needed. If this minimum number is larger than k, then the time is not feasible.

Solution 2 (Java)

public class Solution {
    /**
     * @param pages: an array of integers
     * @param k: An integer
     * @return: an integer
     */
    public int copyBooks(int[] pages, int k) {
        if (pages == null || pages.length == 0 || k <= 0)
            return 0;
        
        int left = pages[0], right = 0, mid;
        for (int page : pages) {
            left = Math.max(left, page);
            right += page;
        }

        while (left + 1 < right) {
            mid = left + (right - left) / 2;
            if (isTimeFeasible(pages, k, mid)) {
                right = mid;
            } else {
                left = mid;
            }
        }

        return isTimeFeasible(pages, k, left) ? left : right;
    }

    private boolean isTimeFeasible(int[] pages, int k, int time) {
        int numOfCopiers = 1, pageSum = 0;
        for (int page : pages) {
            if (page > time) return false;
            if (pageSum + page <= time) {
                pageSum += page;
            } else {
                numOfCopiers++;
                pageSum = page;
            }
        }
        
        return numOfCopiers <= k;
    }
}

Notes

• We can further reduce the initial range by setting left to be the maximum page number and right to be the sum of the pages array. The reason is that the lower bound of the time is achieved by having k = pages.length and each person copies one book; the minimum time now would be the time to copy the book with largest page count. Then, the upper bound of the time is achieved by having only 1 copier to copy all the books by himself; so the time would be the sum of all the pages counts.

• Also, this solution uses a slightly different method to determine if time is feasible.