Minimize Maximum -- Cutting Ribbons

Published: Jul 24, 2022

Medium Binary Search Array


This is the fourth minimize-maximum algorithm problem. This problem has a variation compared to the basic, splitting array problem. If we look around other minimize maximum problems, it is very close to the divide chocolate problem. Even though this has another variation, we can solve using the same manner. Again, the problem is the type of minimize-maximum.

Problem Description

You are given an integer array ribbons, where ribbons[i] represents the length of the i-th 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.

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 of cutting ribbons of length 4

  • use as is
  • cut into 3 and 1
  • cut into 2 and 2
  • cut into 2, 1 and 1
  • cut into 1, 1, 1 and 1


example 1:
Input: ribbons = [9,7,5], k = 3
Answer: 5
Explanation: ribbons subarrays after cutting each element: [[5,4],[5,2],[5]].
5 is the minimized largest length when the array is splitted.
example 2:
Input: ribbons = [7,5,9], k = 4
Answer: 4
Explanation: ribbons subarrays after cutting each element: [[4,3],[4,1],[4,4,1]].
4 is the minimized largest length when the array is splitted.
example 3:
Input: ribbons = [5,7,9], k = 22
Answer: 0
Explanation: sum of all ribbons are 21. Even the shortest length of 1 won't make 22 ribbons.


Each element of the given ribbons array should be divided into smaller lengths based on a certain criteria. The criteria is a total number of divided ribbons. It can be found by counting the number of ribbons in subarrays.

So, we will count how many ribbons are cut out based on some value (middle value). The smallest value (left) can be 0, and the largest (right) would be the sum of all ribbons in the given array. Here, we should be careful to choose the starting right value. Since the sum might goes to very large, sometime, it triggers a runtime error. Because of this reason, the solution chose 100010 as a big enough number. The middle value is set to (left + right + 1) / 2 since the length of each ribbon should exceed the exact middle value.

Next step is to adjust the left or right values. If the number of ribbons is less than the specified value, the middle value should be smaller to create more ribbons. If the number of ribbons is greater than the specified value, the middle value should be greater to create less ribbons. Here’s a bit of variation in updating left value. When it is updated, the left will have the middle value instead of middle + 1.

When the binary search is over, we can find the answer.


class CuttingRibbons
  bool check(vector<int> &ribbons, int k, int mid)
    int count = 0;
    for (int i = 0; i < ribbons.size(); ++i)
      count += ribbons[i] / mid;
    return count < k;
  int maxLength(vector<int> &ribbons, int k)
    int n = ribbons.size();
    int left = 0, right = 100010;  // this is big enough to get a correct answer
    while (left < right)
      int mid = (left + right + 1) / 2;
      int count = 0;
      if (check(ribbons, k, mid))
        right = mid - 1;
        left = mid;
    return left;


  • Time: O(n log(s)) – s: difference between 1 and max value in the given array.
  • Space: O(1)