Question

2790. Maximum Number of Groups With Increasing Length

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Description

You are given a 0-indexed array usageLimits of length n.

Your task is to create groups using numbers from 0 to n - 1, ensuring that each number, i, is used no more than usageLimits[i] times in total across all groups. You must also satisfy the following conditions:

• Each group must consist of distinct numbers, meaning that no duplicate numbers are allowed within a single group.
• Each group (except the first one) must have a length strictly greater than the previous group.

Return _an integer denoting the maximum number of groups you can create while satisfying these conditions._

 

Example 1:

Input: usageLimits = [1,2,5]
Output: 3
Explanation: In this example, we can use 0 at most once, 1 at most twice, and 2 at most five times.
One way of creating the maximum number of groups while satisfying the conditions is: 
Group 1 contains the number [2].
Group 2 contains the numbers [1,2].
Group 3 contains the numbers [0,1,2]. 
It can be shown that the maximum number of groups is 3. 
So, the output is 3. 

Example 2:

Input: usageLimits = [2,1,2]
Output: 2
Explanation: In this example, we can use 0 at most twice, 1 at most once, and 2 at most twice.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [0].
Group 2 contains the numbers [1,2].
It can be shown that the maximum number of groups is 2.
So, the output is 2. 

Example 3:

Input: usageLimits = [1,1]
Output: 1
Explanation: In this example, we can use both 0 and 1 at most once.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [0].
It can be shown that the maximum number of groups is 1.
So, the output is 1. 

 

Constraints:

• 1 <= usageLimits.length <= 105
• 1 <= usageLimits[i] <= 109

Solutions

Solution 1

class Solution:
  def maxIncreasingGroups(self, usageLimits: List[int]) -> int:
    usageLimits.sort()
    k, n = 0, len(usageLimits)
    for i in range(n):
      if usageLimits[i] > k:
        k += 1
        usageLimits[i] -= k
      if i + 1 < n:
        usageLimits[i + 1] += usageLimits[i]
    return k

class Solution {
  public int maxIncreasingGroups(List<Integer> usageLimits) {
    Collections.sort(usageLimits);
    int k = 0;
    long s = 0;
    for (int x : usageLimits) {
      s += x;
      if (s > k) {
        ++k;
        s -= k;
      }
    }
    return k;
  }
}

class Solution {
public:
  int maxIncreasingGroups(vector<int>& usageLimits) {
    sort(usageLimits.begin(), usageLimits.end());
    int k = 0;
    long long s = 0;
    for (int x : usageLimits) {
      s += x;
      if (s > k) {
        ++k;
        s -= k;
      }
    }
    return k;
  }
};

func maxIncreasingGroups(usageLimits []int) int {
  sort.Ints(usageLimits)
  s, k := 0, 0
  for _, x := range usageLimits {
    s += x
    if s > k {
      k++
      s -= k
    }
  }
  return k
}

function maxIncreasingGroups(usageLimits: number[]): number {
  usageLimits.sort((a, b) => a - b);
  let k = 0;
  let s = 0;
  for (const x of usageLimits) {
    s += x;
    if (s > k) {
      ++k;
      s -= k;
    }
  }
  return k;
}

Solution 2

class Solution:
  def maxIncreasingGroups(self, usageLimits: List[int]) -> int:
    usageLimits.sort()
    k = s = 0
    for x in usageLimits:
      s += x
      if s > k:
        k += 1
        s -= k
    return k