2790. Maximum Number of Groups With Increasing Length

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.

class Solution {
    bool helper(vector<int>& A, int k) {
        int need = 0, x = k;
        for(int i = A.size() - 1; i >= 0; i--,x = max(x - 1, 0)) {
            if(A[i] == x) continue;
            if(A[i] > x) {
                int now = A[i] - x;
                need = max(0, need - now);
            } else {
                need += x - A[i];
            }

        }
        return need == 0;
    }
public:
    int maxIncreasingGroups(vector<int>& A) {
        sort(begin(A), end(A));
        int l = 0, r = A.size(), res = l;
        while(l <= r) {
            int m = l + (r - l) / 2;
            bool ok = helper(A, m);
            if(ok) {
                res = max(res, m);
                l = m + 1;
            } else r= m - 1;
        }
        return res;
    }
};