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 <= 1051 <= usageLimits[i] <= 109
Solution
C++
class Solution {
public:
int maxIncreasingGroups(std::vector<int>& usageLimits) {
int N = usageLimits.size();
sort(usageLimits.rbegin(), usageLimits.rend());
int left = 0, right = usageLimits.size();
auto good = [&](int k) -> bool {
vector<int> A = usageLimits;
int credit = 0;
for (int i = 0; i < N; i++) {
if (A[i] >= k) {
A[i] -= k;
if (credit > 0) credit -= A[i];
} else {
credit += k - A[i];
}
if (k > 0) {
k -= 1;
}
}
return credit <= 0;
};
while (left < right) {
int mid = left + (right - left + 1) / 2;
if (good(mid)) {
left = mid;
} else {
right = mid - 1;
}
}
return left;
}
};
Python3
class Solution:
def maxIncreasingGroups(self, usageLimits: List[int]) -> int:
N = len(usageLimits)
usageLimits.sort(reverse = 1)
left, right = 0, N
def good(k):
A = usageLimits[:]
credit = 0
for x in A:
if x >= k:
x -= k
if credit > 0:
credit -= x
else:
credit += k - x
if k > 0:
k -= 1
return credit <= 0
while left < right:
mid = left + (right - left + 1) // 2
if good(mid):
left = mid
else:
right = mid - 1
return left