Leaky Cauldron
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Cutting Ribbons

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题目(LeetCode #1891)

You are given an integer array ribbons, where ribbons[i] represents the length of the ith ribbon, and an integer k. You may cut any of the ribbons into any number of segments of positive integer lengths, or perform no cuts at all.

For example, if you have a ribbon of length 4, you can:
Keep the ribbon of length 4,
Cut it into one ribbon of length 3 and one ribbon of length 1,
Cut it into two ribbons of length 2,
Cut it into one ribbon of length 2 and two ribbons of length 1, or
Cut it into four ribbons of length 1.
Your goal is to obtain k ribbons of all the same positive integer length. You are allowed to throw away any excess ribbon as a result of cutting.

Return the maximum possible positive integer length that you can obtain k ribbons of, or 0 if you cannot obtain k ribbons of the same length.

Example 1:

Input: ribbons = [9,7,5], k = 3
Output: 5
Explanation:

  • Cut the first ribbon to two ribbons, one of length 5 and one of length 4.
  • Cut the second ribbon to two ribbons, one of length 5 and one of length 2.
  • Keep the third ribbon as it is.

Now you have 3 ribbons of length 5.

Example 2:

Input: ribbons = [7,5,9], k = 4
Output: 4
Explanation:

  • Cut the first ribbon to two ribbons, one of length 4 and one of length 3.
  • Cut the second ribbon to two ribbons, one of length 4 and one of length 1.
  • Cut the third ribbon to three ribbons, two of length 4 and one of length 1.

Now you have 4 ribbons of length 4.

Example 3:

Input: ribbons = [5,7,9], k = 22
Output: 0
Explanation: You cannot obtain k ribbons of the same positive integer length.

Constraints:

  • 1 <= ribbons.length <= 105
  • 1 <= ribbons[i] <= 105
  • 1 <= k <= 109

题解

采用二分法求解，绳子的最小长度为 0，最大长度为绳长最大值。根据当前绳长 m 所能得到的绳子段数总和与 k 进行比较来判断边界是否满足条件。

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class Solution {
    public int maxLength(int[] ribbons, int k) {
        if (ribbons == null || ribbons.length == 0) return 0;

        int l = 0, r = 0;
        for (int ribbon : ribbons) r = Math.max(r, ribbon);

        while (l < r) {
            int m = l + (r - l + 1) / 2;
            if (check(ribbons, k, m)) l = m;
            else r = m - 1; 
        }

        return l;
    }

    private boolean check(int[] ribbons, int k, int m) {
        int sum = 0;
        for (int ribbon : ribbons) sum += ribbon / m;
        return sum >= k;
    }
}