Problem
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
/**
* @param {number[]} usageLimits
* @return {number}
*/
var maxIncreasingGroups = function(usageLimits) {
var count = 0;
var num = 0;
var minCount = 1;
usageLimits.sort((a, b) => a - b);
for (var i = 0; i < usageLimits.length; i++) {
count += usageLimits[i];
if (count >= minCount) {
num++;
minCount += num + 1;
}
}
return num;
};
Explain:
Math.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).